1. Short Exact Sequences 1.1. Definition

Home , Exact sequence, Module homomorphism

1. Short exact sequences ′ ′ f1 f2 ′ f1 ′ f2 ′ 1.1. Deﬁnition. Let 0 A1 A2 A3 0 and 0 A1 A2 A3 0 be two short exact sequence of R→-modules.−→ We−→ call these→ short exact→ sequences−→ −→E and E→′. Suppose h : A A′ are R-module homomorphisms such that the following diagram commutes: j j → j

f1 f2 0 / A1 / A2 / A3 / 0 (1)

h1 h2 h3 f ′ f ′ ′ 1 ′ 2 ′ 0 / A1 / A2 / A3 / 0

′ Then one says that h = (h1, h2, h3) is a homomorphism from E to E . We say that h is an ′ isomorphism from E to E if each hj is an isomorphism. Let A and C be two R-modules. Let i : A A C be the inclusion and p : A C C be the projection. Then one has an exact sequence→ ⊕ ⊕ → p 0 A i A C C 0. (2) → −→ ⊕ −→ → We say that a short exact sequence is split if it is isomorphic to an exact sequence of the form (2).

1.2. Lemma. Let (1) be a homomorphism of short exact sequences. If h1 and h3 are isomorphism, then so is h2.

j q 1.3. Theorem (Splitting criteria). Let 0 A B C 0 be a short exact sequence of R-modules. Then the following are equivalent:→ −→ −→ → (a) The exact sequence is split. (b) There exists k : B A such that k j = id . → ◦ A (c) There exists r : C B such that q r = idC . → ′ ◦ ′ (d) There exists a submodule C B such that q C′ : C C is an isomorphism. Further, if C′ is any submodule⊆ as in (d), then B| = j(A→) C′. In particular, the short exact sequence above is split, then B A C. ⊕ ≃ ⊕ Proof. (b) = (d) : Define C′ = ker(k) and verify that q ′ : C′ C is an isomorphism. ⇒ |C → (d) = (c) : Define r : C B to be the inverse of the isomorphism q ′ . ⇒ → |C (c) = (a) : Define a homomorphism h2 : A C B by h(a, c) = j(a)+ r(c). One verifies that⇒ the diagram below commutes. ⊕ → i p 0 / A / A C / C / 0 ⊕ id h2 id j q 0 / A / B / C / 0 Lemma 1.2 implies that h2 is an isomorphism. Thus (c) implies (a). Finally (a) = (b) is left as an exercise. ⇒ Let C′ be any submodule as in (d). Let b B. Then q(b rqb)=0, so(b rqb) i(A). So writing b = (b rqb)+ rqb, we see that B∈ = i(A)+ C′.− If b i(A) C′,− then b∈= i(a) for some a A.− so q(b) = qi(a) = 0. Since b C′ and the restriction∈ ∩ of q to C′ is an isomorphism,∈ it follows that b =0. So B = i(A) ∈C′. 1 ⊕ 2. Hom’s and Tensors References: Dummit and Foote (Ch. 10.4 - 10.5), Atiyah Macdonald (Ch. 2). Let R be a commutative ring with 1. 2.1. Exercise: Let M ′,M,M ′′ be R-modules. Given an R-module N and an R-module homomorphism f : M ′ M, one obtains an R-module homomorphism f ∗ : Hom(M, N) Hom(M ′, N) given by f→∗(α)= α f. → Let ◦ f g M ′ M M ′′ 0 (3) −→ −→ → be a sequence of R-modules and homomorphisms. Then the following are equivalent: (a) The sequence (3) is exact. g∗ f ∗ (b) The sequence 0 Hom(M ′′, N) Hom(M, N) Hom(M ′, N) is exact for every R-module N. → −→ −→ 2.2. Definition. Let M,N,P be R-modules. Let α : M N P be a map. If m M, × → ∈ then let αm,• : N P denote that map αm,•(n)= α(m, n). If n N, then let α•,n : M P denote that map →α (m) = α(m, n). Say that α : M N P∈is a R-bilinear map if →α •,n × → m,• and α•,n are R-linear maps for all m M and n N. Let Bilinear(M ,N,P ) denote the set of all R-bilinear maps from M N∈ P . ∈ × × → 2.3. Exercise: Verify that there are isomorphisms i : Bilinear(M ,N,P ) Hom (M, Hom (N,P )) (4) M × ≃ R R given by α αm,•. These isomorphisms are natural in the sense that for all R-module homomorphism7→ φ : M M ′, the following diagram commutes: → iM Bilinear(M N,P ) / HomR(M, HomR(N,P )) O × O ∗ ∗ (φ×idN ) φ

′ ′ Bilinear(M N,P ) / HomR(M , HomR(N,P )) × iM′ Here φ∗ and (φ id )∗ are the maps induced by the functoriality of Hom (see 2.1). × N 2.4. Definition. Let M and N be R-modules. Then there exists a pair (T,g) where T is an R-module and g : M N T is an R-bilinear map satisfying the following universal property: Given any R-module× →P and an R-bilinear map α : M N P , there exists a unique R-module homomorphism α¯ : T P such that α =α ¯ g. × → One verifies that such a pair (T,g) is necessarily→ unique upto unique◦ isomorphism. We write T =(M R N) and call it the tensor product of M and N over R. We write M N instead of M ⊗N, when there is no chance of confusion. ⊗ ⊗R 2.5. Construction of the tensor product: Let F be the free module on the set M N with a basis indexed by all ordered pairs (m, n), with m M and n N. Let S be× the submodule of F generated by all expressions of the following∈ form: ∈ (m + m , n) (m , n) (m , n), r(m, n) (mr, n), 1 2 − 1 − 2 − (m, n + n ) (m, n ) (m, n ), r(m, n) (m, rn). 1 2 − 1 − 2 − 2 Define M N = F/S. The image of (m, n) F in M N is denoted by m n. So ⊗R ∈ ⊗R ⊗ by construction, M R N consists of finite expressions of the form j rj(mj nj) and the following properties⊗ hold: P ⊗ (m + m ) n = m n + m n, m (n + n )= m n + m n , 1 2 ⊗ 1 ⊗ 2 ⊗ ⊗ 1 2 ⊗ 1 ⊗ 2 r(m n)=(mr) n = m (rn). ⊗ ⊗ ⊗ 2.6. Lemma. Let φ : M M ′ and ψ : N N ′ be R-module homomorphisms. Then we get a map φ ψ : M N →M ′ N ′. Let (M→ N,g) and (M ′ N ′,g′) be the tensor products. Then there× exists× a unique→ R×-module homomorphism⊗ φ ψ⊗: M M ′ N N ′ such that g′ (φ ψ)=(φ ψ) g. ⊗ ⊗ → ⊗ ◦Let ×be the category⊗ ◦ of R-modules. The construction (M, N) (M N) and (φ, ψ) (φ ψC) defines a functor from the category to . 7→ ⊗ 7→ ⊗ C × C C 2.7. Lemma. Let M,N,P be R-modules: Show that there are the following natural isomorphisms: (a) R R M M. (b) (M⊗ N)≃ P M (N P ). (c) M ⊗(N ⊗P ) ≃ (M⊗ N)⊗ (M P ). ⊗ ⊕ ≃ ⊗ ⊕ ⊗

2.8. Remark. The above lemma says that the category of R-modules with the operations and behave like a ring where R plays the role of anC unit. ⊕ ⊗ 2.9. Lemma. There are natural isomorphisms M N N M induced by (m n) (n m). ⊗ ≃ ⊗ × 7→ ⊗ 2.10. Exercise: Let (M N,g) be the tensor product of the R-modules M and N. Show that the universal property⊗ of (M N,g) implies that we have isomorphisms ⊗ g∗ : Hom (M N,P ) Bilinear(M N,P ) R ⊗ ≃ × given by g∗(¯α)=α ¯ g which are natural in the following sense. Given any R-module homomorphisms φ : M◦ M ′ and ψ : N N ′, the following diagram commutes: → → g∗ HomR(M N,P ) / Bilinear(M N,P ) O ⊗ O × (φ⊗ψ)∗ (φ×ψ)∗

′ ′ ′ ′ HomR(M N ,P ) / Bilinear(M N ,P ). ⊗ (g′)∗ × Hint: 2.6. 2.11. Lemma (Hom- adjunction). Let M,N,P be R-modules.There are isomorphisms ⊗ γ : Hom (M N,P ) Hom (M, Hom (N,P )). M R ⊗R ≃ R R which are natural in the following sense. Given any R-module homomorphism φ : M M ′, the following diagram commutes: → γM HomR(M R N,P ) / HomR(M, HomR(N,P )) O ⊗ O ∗ ∗ (φ⊗idN ) φ

′ ′ HomR(M R N,P ) / HomR(M , HomR(N,P )) ⊗ γM′ 3 Proof. Use the commutative diagrams in 2.3 and 2.10 with ψ = idN . f g 2.12. Lemma (Exactness property of ). Let M ′ M M ′′ 0 be an exact sequence of R-modules. Then the sequence ⊗ −→ −→ → f⊗id g⊗id M ′ N N M N N M ′′ N 0 ⊗ −−−−→ ⊗ −−−→ ⊗ → is exact for all R-module N. Proof. Let N,P be any two R-modules. Then the implication (a) = (b) in 2.1 tells us that ⇒ g∗ f ∗ 0 Hom (M ′′, Hom (N,P )) Hom (M, Hom (N,P )) Hom (M ′, Hom (N,P )) → R R −→ R R −→ R R is exact. Now the 2.11 tells us that the induced sequence (g⊗id )∗ (f⊗id )∗ 0 Hom (M ′′ N,P ) N Hom (M N,P ) N Hom (M ′′ N,P ) → R ⊗ −−−−−→ R ⊗ −−−−−→ R ⊗ is exact. Note that this uses the naturality in 2.11. Now the implication (b) = (a) in 2.1 f⊗id g⊗id ⇒ tells us that M ′ N N M N N M ′′ N 0 is exact too. ⊗ −−−−→ ⊗ −−−→ ⊗ → .Lemma. Let M be an R-module and I be an ideal in R. Let i : I ֒ R be the inclusion .2.13 i⊗idM → (a) Show that I M IM and I M R M is simply the inclusion of IM ⊗R ≃ ⊗R −−−→ ⊗R in M once we identify R R M with M. (b) Show that ⊗ (R/I) M M/IM ⊗R ≃ as R modules. The ideal I is in the annihilator of both (R/I) R M and M/IM, so they are both also (R/I)-modules and the isomorphism above is an isomorphism⊗ of R/I-modules. Proof. Part (a) is easy verification. b) We have the exact sequence I ֒ R R/I 0. So 2.12 implies that) → → → I M R M (R/I) M 0 ⊗R → ⊗R → ⊗R → is exact. By part (a) this is equivalent to saying that IM ֒ M (R/I) R M 0 is exact. → → ⊗ → 2.14. Lemma. (a) If f : Rn Rm is a surjection, then n m. (b) If Rn Rm, then n = →m. ≥ ≃ f Proof. (a) Let I be a maximal ideal in R and let k = R/I. Since Rn Rm 0 is exact, id⊗f −→ → 2.12 implies that (R/I) Rn (R/I) Rm 0 is exact. But (R/I) Rn Rn/IRn (R/I)n = kn, where the⊗ first−−−→ isomorphism⊗ is from→ 2.13 and the second⊗ was≃ a homework≃ exercise. So we have a surjection kn km 0. Since k is a field, we get n m from linear algebra. This proves part (a). Part (b)→ follows→ from part (a). ≥

4 3. Vector Spaces Let F be a field. An F -module is called a F -vector space. Fix a field F . In the section below, vector space means F -vector space and the letters V , W , denote vector spaces. 3.1. Definition. The definition of linear independece, span, basis, homomorphism (linear map), isomorphism, are as in the case of modules. 3.2. Lemma. Let V be a vector space and B V . TFAE: (a) B is a maximal linearly independent set.⊂ (b) B is a minimal spanning set. (c) B is a basis of V . 3.3. Remark. Note that the implication (a) = (c) above uses the fact that every non-zero element of F is invertible. So this lemma fails⇒ for arbitrary R-modules. 3.4. Lemma. Let φ : V W be a linear map and B V . (a) If φ(B) is linearly→ indep then so is B. ⊆ (b) If B spans V and φ is onto, then φ(B) spans W . 3.5. Lemma. Let φ : V W is an iso and B V . Then (a) B spans V if and→ only if φ(B) spans V . ⊆ (a) B is linearly indep if and only if φ(B) is linearly indep. (c) B is a basis of V if and only if φ(B) is a basis of W . 3.6. Theorem. Every non-zero vector space has a basis. So V is free as a F -module. Proof. By Zorn’s lemma there is a maximal linearly independent set. Apply 3.2. 3.7. Remark. An argument similar to the above proof show that any linear independent set in V can be extended to a basis of V . 3.8. Definition. Say that V is finite dimensional if V has a finite basis. 3.9. Example. Let F n be the rank n free module over F consisting of all row vectors of length n with entries from F . Let ej be the vector whose j-th entry is 1 and all other entries n are zero. Then verify that e1, , en is a basis of F . This is called the standard basis of F n. { ··· } 3.10. Lemma. V is finite dimensional if and only if V can be spanned by finitely many elements.

Proof. Suppose S = v1, , vn span V . Among all subsets of S that span V , pick one that has the smallest size;{ call··· it B.} Then B is a minimal spanning set. Apply 3.2.

3.11. Lemma. Let b1, , bn be a sequence of elements from V . Then there exists a linear n ··· map φ : F V given by φ(ei)= bi for all i (universal property of free module). The map φ is an iso if→ and only if B is a basis of V . In particular, if V has a basis consisting of n elements then V F n. ≃ 3.12. Lemma. Let B= (b1, , bn) be an ordered basis of V and w1, ,wn W . Then there is a unique linear map···φ : V W such that φ(b )= w for{ all j.··· } ⊆ → j j 3.13. Lemma. Suppose B = (b , , b ) is a basis of V . Let v = (λ b + + λ v ) with 1 ··· n 1 1 ··· n n λj =0. Then (B bj ) v is also a basis of V . 6 \{ } ∪{ } 5 Proof. Deﬁne φ : V V by φ(bj)= v and φ(bi)= bi for then i = j. Verify that ψ : V V given by → 6 → ψ(b )= λ−1(λ b + + λ b b + λ b + + λ b ) j j 1 1 ··· j−1 j−1 − j j+1 j+1 ··· n n and ψ(b )= b for all i = j is the inverse of φ. So φ and ψ are mutually inverse iso. i i 6 3.14. Lemma. Let v , , v , , v be a linearly independent set and Suppose B = { 1 ··· s ··· k} u1, ,ur, v1, , vs be a basis of V . Then there exists 1 j r such that (B uj ) {v ···is also a··· basis} of V . ≤ ≤ \{ } ∪ { s+1} Proof. Write vs+1 = (λ1u1 + + λrur)+(µ1v1 + + µsvs). Since the vi’s are linearly independent, λ = 0 for some ···j. Pick just a j. Then··· apply 3.13. j 6 3.15. Lemma. Let v , , v be a linearly independent set and Suppose B = u , ,u { 1 ··· k} { 1 ··· m} be a basis of V . Then there is a k element subset B0 of B such that (B B0) v1 , vk is a basis of V . \ ∪{ ··· } Proof. Apply 3.14 k times to replace k of the u ’s with v , , v successively in the basis. i 1 ··· k 3.16. Lemma. Any linearly independent set in V can be extended to a basis of V . Proof. Follows from the 3.15 since every vector space has a basis. 3.17. Theorem. Suppose V if ﬁnite dimensional. Then there is a number n such that any basis of V has n elements. This number n is called the dimension of V . We write dim(V )= n.

Proof. Suppose B = (b1, , bn) and C = (v1, , vk) be two bases of V . Without loss, ··· ′ ··· suppose k n. Then 3.15 implies that B = (B B0) C is a basis of V where B0 = k. But C already≤ a basis, so a maximal linearly independent\ ∪ set, so if B′ properly contains| | C, it cannot be a basis. Ir follows that B B = , so k = n. \ 0 ∅ 3.18. Definition. Define external and internal direct sum, quotients. Get isomorphism theorems, same as in modules. 3.19. Lemma. Let U and V be vector spaces. Then dim(U V ) = dim(U) + dim(V ). ⊕ f g 3.20. Theorem. Let 0 U V W 0 be an exact sequence of vector spaces. Then: → −→ −→′ → ′ (a) there exists some subspace W V such that g W ′ : W W is an isomorphism and V is the internal direct sum of f(U) ⊆and W ′. In other| words,→ every short exact sequence in the category of vector spaces splits. In particular V U W . (b) dim(U) + dim(W ) = dim(V ). ≃ ⊕ 3.21. Definition. Let φ : M N be a linear map. Define rank(φ) = dim(Im(φ)). → 3.22. Theorem. Suppose φ : V W is a linear map. Then we get an exact sequence → φ .ker(φ) ֒ V Im(φ) 0 0 → → −→ → So V ker(φ) Im(φ). In particular, dim(V ) = dim(ker(φ)) + rank(φ). ≃ ⊕ 3.23. Theorem. Let φ : V W be a linear map between finite dimensional vector spaces and dim(V ) = dim(W ). TFAE:→ (a) φ is one-to-one. (b) φ is onto. (c) φ is an iso. 6 Proof. Suppose φ is one to one. Then the exact sequence 0 V W V/φ(W ) 0 implies dim(W ) = dim(V ) + dim(V/φ(W )). So dim(V/φ(W ))→ =, that→ is φ(→W )= V . so→φ is onto. Conversely, if φ is onto, then the exact sequence 0 ker φ V W 0 implies that dim(V ) = dim(W ) + dim(ker(φ)). So ker(φ) = 0 and →φ is injective.→ → →

7 4. Matrices, Hom spaces between free modules, duals. Let R be a commutative ring with 1. Let Rn be the free R-module consisting of all column vectors of length n with entries from R. Let e , , e be the standard basis of Rn. { 1 ··· n} 4.1. Deﬁnition. Let V be a free R module of rank n. A choice of an ordered basis B = (b , , b ) gives us an isomorphism [ ] : V Rn, given by [b ] = e . 1 ··· n B → i B i 4.2. Deﬁnition. Let M (R) be the set of all m n matrices with eneties in R. Write m,n × Mn(R) = Mn,n(R). With the usual matrix operations, Mm,n(R) is a R-module and Mn(R) is a R-algebra with the inclusion R Mn(R) given by r rIn where In denotes the n n identity matrix. → 7→ × Let A M (R). Then A determines a linear map l : Rn Rm given by l (v)= Av. ∈ m,n A → A 4.3. Matrix of a linear map: Let V, W be free R-modules with basis B = v1, , vn and C = w , ,w . Let φ Hom (V, W ). Then there exists α R such that{ ··· } { 1 ··· m} ∈ R ij ∈ m φ(v )= α w . j X ij i i=1

The matrix m n matrix ((αij)) is called the matrix representing (or the matrix of) the linear map φ with respect× to the bases B and C. We write C [φ]B = ((αij)). One verifies that C [φ]B[v]B = [φ(v)]C . C The correspondence φ [φ]B gives an R-module isomorphism HomR(V, W ) Mm,n(R). One verifies that 7→ ≃ C [φ]B[v]B = [φ(v)]C . n m In pariticular we have the isomorphism HomR(R , R ) Mm,n(R) given by φ [φ] where [φ] is the matrix of φ with respect to the standard bases≃ of Rn and Rm. One has7→ [[ ] φ []−1] = [φ]C . C ◦ ◦ B B 4.4. Lemma. Let V1, V2, V3 be free R-modules. Let Bj be an ordered basis of Vj for j =1, 2, 3. φ ψ Let V V V be linear maps. Then 1 −→ 2 −→ 3 [ψ]B3 [φ]B2 = [ψ φ]B3 . B2 B1 ◦ B1 n n In particular we have an R-algebra isomorphism HomR(R , R ) Mn(R) given by φ [φ] whose inverse is given by A l . → 7→ 7→ A 4.5. Definition. Let GLn(R) denote the set of all units in the ring Mn(R). So A GLn(R) if and only if the linear map l : Rn Rn is an isomorphism of R-modules. ∈ A → 4.6. Change of basis. Similarity. Let F be a field. Two matrices A, B Mn(F ) are similar if there exists an invertible matrix P such that P −1AP = B. Verify that∈ A and B are similar if and only if they represent the same linear transformation with respect two two choices of bases of F n.

8 5. Duality 5.1. Definition. Let R be a commutative ring with 1 and M be an R-module. An R-valued linear functional on M is an R-linear map from M to R. Define the dual module M ∗ to be the set of all R-valued linear functionals on M, that is, ∗ M = HomR(M, R). Let T : M N be a linear map, then we get a linear map T ∗ : N ∗ M ∗ given by → → T ∗(α)= α T for all φ N ∗. ◦ ∈ The linear map T ∗ is called the adjoint of T . Verify that for all M T1 M T2 M one has 1 −→ 2 −→ 3 (T T )∗ = T ∗ T ∗. 2 ◦ 1 1 ◦ 2 In other words, (M M ∗), and (T T ∗) is a functor from the category R-Mod to R-Modop. Given v M and7→α M ∗, define7→ ∈ ∈ α, v = α(v). h i Then , : M ∗ M R is a R-bilinear map (also called pairing). The pairing gives a “natural”h i map ev× : M→ M ∗∗ given by M → ev (v)= , v . M h• i In other words ev (v)(α)= α, v = α(v). M h i 5.2. Exercise: Note that if T : V W is a R-linear map, then there is the unique linear map S : W ∗ V ∗ such that → → w,Tv = Sw,v for all v V,w W ∗. h i h i ∈ ∈ Infact, S = T ∗. 5.3. Exercise: Verify that for all R-linear map T : M N, the following diagram commutes: → evM M / M ∗∗

T T ∗∗

evN N / N ∗∗ In other words the collection of maps evM defines a natural transformation from the identity functor of R-Mod to the functor (M M ∗∗). 7→ 5.4. Definition. Now assume M be a free module of finite rank. Choose a basis v1, , vn. ∗ ··· Define αi M by ∈ i i α (vj)= δj. ∗ Exercise: Show that α1, ,αn are linearly independent and any α M can be written n i { 1··· }n ∗ ∈ as α = i=1 α(vi)α . So α , ,α is a basis of M . It is called the dual basis to v1, , vn . In particularP rank(M ∗)= n ···= rank(M). { ··· } 5.5. For the rest of the section, let F be a field. Let V be an F -vector space. Given a non-zero vector v V , we can extend it to a basis v = v1, , vn of V . So there exists a ∈ ∗ { ··· } ∗∗ linear functional α V such that α(v1) = 0 (Why?). This shows that if 0 = , v V , ∈ 6 ∗∗ h• i ∈ then v = 0. In other words, the map evV : V V is injective. 9→ So if V is a finite dimensional F -vector space, then evV is an isomorphism, since V and V ∗∗ have the same dimension. Thus we obtain the following result:

5.6. Theorem. Let FdvsF be the category of finite dimensional vector spaces over F . The collection of maps evV defines a natural isomorphism from the identity functor of FdvsF to the functor (V V ∗∗). For finite dimensional vector spaces, one often identifies V with V ∗∗ 7→ 1 n using the natural isomorphisms evV . In particular, verify that if α , ,α is a basis of ∗ i j { ··· } V , then the is basis v1, , vn of V such that α , vj = δi . One says that v1, , vn is the dual basis to α ,{ ,α··· . } h i { ··· } { 1 ··· n} 5.7. Definition. Let W be a F -vector space and A be a subset of W . Define A⊥ W ∗ by ⊆ A⊥ = α W ∗ : α,w = 0 for all w A . { ∈ h i ∈ } Verify that A⊥ is a subspace of W ∗. 5.8. Lemma. If U is a subspace of a finite dimensional space V , then dim(U) + dim(U ⊥)= dim(V ). 1 k ⊥ Proof. Let α , ,α be a basis of U . The linear functionals αj’s can be put together { ··· } k k i to define a linear map α : V F given by α(v)= i=1 α (v)ei where ei’s are the standard basis of F k. Verify that ker(α→)= U. P 1 k 1 n ∗ Now extend α , ,α to a basis α , ,α of V . Let v1, , vn be the dual basis ··· i { ···j } { ··· } k of V , that is, vj W such that α , vj = δi . Verify that α(vj)= ej, so α : V F is onto. ∈ αh i → So the sequence 0 U ⊥ V F k 0 is exact → → −→ → 5.9. Theorem. Let T : V W be a linear map between finite dimensional vector spaces. Then rank(T ∗) = rank(T ). → ∗∗ ∗ ⊥ Proof. Identify V with V using evV . Then Im(T ) is a subspace of V . One has v Im(T ∗)⊥, if and only if v, T ∗(α) = 0 for all α W ∗, that is, Tv,α = 0 for all α W ∈∗, that is, T v = 0, i.e, v hker(T ). Soi ∈ h i ∈ ∈ Im(T ∗)⊥ = ker(T ). Using 5.8, we obtain, dim(Im(T ∗)) = dim(V ) dim(Im(T ∗)⊥) = dim(V ) dim(ker(T )) = dim(Im(T )). − −

n 5.10. Remark. Let Mn,1(F ) = F be the standard n dimensional F vector space of column n vectors. Identify F with its dual such that the dual basis to e1, , en is itself. Then the canonical pairing (F n)∗ F n F is given by the usual dot product{ ··· on F}n: v,w = vTr w. Using 5.2, one veriﬁes that× the→ adjoint of the linear map (v Av) is givenh by (vi ATr·v). So 5.9 says: The row rank and the column rank of a matrix are7→ equal. 7→

5.11. Remark. Let Mn,1(F ) be the standard n dimensional vector space of column vectors ∗ with standard basis e1, , en . Sometimes it is convenient to identify Mn,1(F ) with the n-dimensional vector{ ··· space} of row vectors M (F ), so that eTr, , eTr is the dual 1,n { 1 ··· n } basis to e1, , en . Then the canonical pairing M1,n(F ) Mn,1(F ) F is given by v,w = v{ w···. An m} n matrix A corresponds to the linear× map l : M →(F ) M (F ) h i · × A 1,n → 1,m given by lA(v) = Av. Using 5.2, one verifies that the adjoint of this linear map is given by r : M (F ) M (F ) given by r (v)= vA. A 1,m → 1,n A 10 6. multilinear algebra For this section, let R be a commutative ring with 1. Assume 2 is invertible in R. Let M,N,P be R-modules. Let m , m , denote elements of M. 1 2 ··· 6.1. Exercise: Let f : M P is a R-linear map of R modules. Suppose there are x , , x V such that x ’s span→ M and f(x ), , f(x ) are linearly independent in P . 1 ··· n ∈ j 1 ··· n Then x1, , xn is a basis of M. In particular M is free and f is an isomorphism onto its image. ··· 6.2. Theorem. Let M and N be free modules over a commutative ring R with bases v1, , vk and w1, ,wl . Then vi wj : 1 i k, 1 j l form a basis for {M ···N. In} particular,{ ···M } N is free{ and⊗ rank (≤M ≤ N) =≤ rank≤(M}). rank (N). ⊗R ⊗R R ⊗R R R Proof. Consider a free R module P of rank kl with basis given by e where 1 i k, ij ≤ ≤ 1 j l. Define f : M N P by f( i civi, j djwj) = ij cidjeij. Verify that f is bilinear.≤ ≤ So by the universal× property→ of theP tensorP product, thereP exists a unique linear map f¯ : M N P such that f¯(vi wj)= eij. Since the vectors vi wj clearly span M N, it follows⊗ that→ f¯ is an iso (by 6.1).⊗ ⊗ ⊗ 6.3. Definition. Let M k = M M M (k-times). A map f : M k N is called × ×···× → multilinear if f is linear in each arguement. More precisely, for each fixed m1, , mk M, the functions ··· ∈ m f(m1, , mj−1, m, mj+1, , mk) k7→ ··· ··· are linear. Let MultiR(M , N) be the set of all such multilinear maps. Given f Multi (M k, N) and a permutation σ S , define f σ Multi (M k, N) by ∈ R ∈ k ∈ R (f σ)(m , , m )= f(m , , m ). 1 ··· k σ(1) ··· σ(k) k Verify that this defines a right action of the permutation group Sk on MultiR(M , N). We σ say that f is symmetric if f = f for all σ Sk. We say that f is skew symmetric if f σ = sign(σ)f for all σ S . ∈ ∈ k k 6.4. Exercise: Show that f MultiR(M , N) is skew symmetric (resp. symmetric) if and only if the value of f changes by∈ a negative sign (does not change) when we interchange two of the arguments. Hint: The permutation group is generated by transpositions. Let T k(M) = M M denote the k-th tensor power of M. Let i : M k T k(M) ⊗···⊗ → k be the canonical multilinear map ι(m1, , mk)= m1 mk. The tensor power T (M) has the following universal property: ··· ⊗···⊗ 6.5. Lemma. Given any R-module N, and a R-multilinear map f : M k N, there exists a unique R-module homomorphism f¯ : T k(M) N such that f = f¯ ι. In→ other words, there are natural isomorphisms, Multi (M k, N) →Hom (T k(M), N). ◦ R ≃ R 6.6. Lemma. Given an R-module M, there exists an R-module S together with a multilinear map i : M k S with the following universal property: Given any R-module N, and a symmetric R-multilinear→ map f : M k N, there exists a unique R-module homomorphism f˜ : S N such that f = f˜ i.→ Such a pair (S, i) is necessarily unique upto → k ◦ unique iso. We write S = Sym (M) and call it the k-th symmetric power of M. We write i(m1, m2, , mk)= m1m2 mk. (The proof is similar to the proof of 6.7). ··· ··· 11 6.7. Lemma. Given an R-module M, there exists an R-module A together with a multilinear map i : M k A with the following universal property: Given any R-module N, and a skew symmetric→R-multilinear map f : M k N, there exists a unique R-module homo- → morphism f˜ : A N such that f = f˜ i. Such a pair (A, i) is necessarily unique upto → k ◦ unique iso. We write A = Alt (M) and call it the k-th alternating power of M. We write i(m , m , , m )= m m m . 1 2 ··· k 1 ∧ 2 ∧···∧ k ⊗n Proof. Let J M be the linear span of the tensors of the form m1 mk such that ⊆ k n ⊗···⊗ mi = mj for some i = j. Define Alt (M)= T (M)/J and let i be ι followd by the quotient map. A multilinear6 map f : M k N induces a R-module map f¯ : M ⊗k N such that f = f¯ ι. If f is skew symmetric,→ then f vanishes on J, so it induces f˜ : M ⊗→n/J N such ◦ → that f = f˜ i. Since image(i) spans Altk(M), the map f˜ is determined by f. ◦ 6.8. Lemma. (a) Let f : M N be an R-linear map. Then there exists a unique linear k k k → k map T (f): T (M) T (N) such that T (f)(m1 mk)= f(m1) f(mk). f g → ⊗···⊗ ⊗···⊗ (b) If M N P are R-linear maps, then T k(g f)= T k(g) T k(f). k −→ −→ ◦ ◦ (c) T (idM ) = idT k(M). In other words, T k is a functor on the category of R-modules. k k Proof. Define F : M T (N) by F (m1, , mk)= f(m1) f(mk). Then F is multilinear, so by universal→ property of T n(M)··· we get a unique R⊗···⊗-linear map T n(f): T k(M) T k(N) such that → T n(f)(m m )= F (m , , m )= f(m ) f(m ). 1 ⊗···⊗ k 1 ··· k 1 ⊗··· k This proves part (a). Verify that the equation φ(m m )=(g f)(m ) (g f)(m ) 1 ⊗···⊗ k ◦ 1 ⊗···⊗ ◦ k holds if φ = T k(g) T k(f). But the uniqueness in part (a) says that T k(g f) is the only map that satisfy this◦ equation. So T k(g) T k(f)= T k(g f). Thir proves part◦ (b). Part (c) also follows from the uniqueness condition◦ in part (a). ◦ 6.9. Exercise: Show the the lemma 6.8 holds, word for word, if T k is replaced by Symk or Altk, and the tensor product replaced with the symmetric or the wedge product. The proofs are identical.

6.10. Remark. Let m1, , mk M. Elements of the form x = m1 mk or its image k k ··· ∈ ⊗···⊗ in Sym (M) or Alt (M) are called monomials of degree k and the mj are called the entries k k of the monomial. Observe that the permutation group Sk acts on T (M) and Sym (M) and k σ Alt (M) by permuting the factors in the monomials: x = mσ(1) mσ(k) etc. (Warning: this is a right action, not a left action). ⊗···⊗ 6.11. Lemma. (a) One has m m =0 if m = m for some i = j. 1 ∧···∧ k i j 6 (b) One has m1 mj mj+1 mk = m1 mj+1 mj mk, that is, when two consecutive terms∧···∧ are interchanged,∧ ···∧ the monomial− ∧···∧ changes by∧ a minus···∧ sign. (c) One has m m = sign(σ)m m . σ(1) ∧···∧ σ(k) 1 ∧···∧ k In other words, when we rearrange the entries of a monomial, the monomial changes by the sign of that permutation. (d) If some mj is a linear combination of the rest of the mi’s, then m1 mk =0. 12 ∧···∧ k k Proof. Recall the construction of Alt (M)= T (M)/J. Since m1 mk J if mi = mj for some i = j, part (a) holds. Part (a) implies that ⊗···⊗ ∈ 6 0=(m + m ) (m + m )= m m + m m j j+1 ∧ j j+1 j ∧ j+1 j+1 ∧ j since mj mj = mj+1 mj+1 = 0. Part (b) follows by a similar calculation. Part (b) veriﬁes (c) in the∧ special case,∧ when σ is a transposition of the form (j, j + 1). Since x xσ is a 7→ group action and the the adjacent transpositions generate Sk, the general part (c) follows. Part (d) follows from part (c).

6.12. Theorem. Let M be a free R-module with basis v1, v2, , vn. k ··· (a) T (M) has a basis given by vi1 vik where i1, , ik run from 1 to n. So dim(T k(M)) = nk. ⊗···⊗ ··· k r1 r2 rn (b) Sym (M) has a basis given by v v v where rj 0 and rj = k. So 1 2 ··· n ≥ Pj dim(Symk(V )) = n+k−1 . k (c) If k n, then Altk(M) has a basis given by v v where 1 i < i < ≤ i1 ∧···∧ ik ≤ 1 2 ≤··· i n. So dim(Altk(M)) = n . If k > n, then Altk(M)=0. k ≤ k Proof. (a) follows from 6.2. We prove (c) and leave (b) as an exercise. (c) Let u u be any monomial of degree n + 1. We can write each u as a linear 1 ∧···∧ n+1 j combination of v1, , vn and then multiply out to get a sum of monomials of degree (n +1) whose terms are v···, , v . In each of these monomials some v appears atleast twice, so 1 ··· n j the monomial is zero. So u1 un+1, hence any higher degree monomial is also zero. So Altk(M)=0 for k > n. ∧···∧ Let = (i , , i ) Zk : 1 i < i < i n . If I =(i , , i ) , we write J { 1 ··· k ∈ ≤ 1 2 ≤··· k ≤ } 1 ··· k ∈J mi1 mik = mI . Since the monomial mi1 mik changes sign when two of the entries ∧···∧ ∧···∧ k are interchanged, it is clear that the monomials mI : I span Alt (M). To show that k { ∈J} they are linearly independent, we embed Alt (M) inside T k(M) using an averaging process. To be precise, deﬁne µ : M k T k(M) by 1 → µ (m , , m )= sign(σ)m m m . 1 1 ··· k X σ(1) ⊗ σ(2) ⊗···⊗ σ(k) σ∈Sk Check that µ is skew-symmetric. So we get a linear map µ : Altk(M) T k(M) satisfying → µ(m m )= sign(σ)m m m . 1 ∧···∧ k X σ(1) ⊗ σ(2) ⊗···⊗ σ(k) σ∈Sk

Suppose cImI = 0 for some scalars cI . Then applying the map µ, we get PI∈J sign(σ)c m m =0 X X I iσ(1) ⊗···⊗ iσ(k) I=(i1,··· ,ik)∈J σ∈Sn

Observe that as I and σ varies over and Sn, the ordered tuples (iσ(1), , iσ(k)) are distinct, J k ··· so miσ(1) miσ(k) run through distinct basis elements of T (M). So all the cI ’s must be zero. ⊗···⊗ 6.13. Corollary. Let M be a free module of rank n. Then Altn(M) is a free module of rank 1. If m , , m is any basis of M, then (m m ) is a basis of Altn(M). 1 ··· n 1 ∧···∧ n

13 7. determinant 7.1. Deﬁnition. let M be a rank n free module over R and φ : M M be a linear map. Then φ determines a linear map Altn(φ) : Altn(M) Altn(M) satisfying→ → φ(m ) φ(m ) = Altn(φ)(m m ). 1 ∧··· n 1 ∧···∧ n Since Altn(M) is one dimensional, there is a scalar det(φ) such that Altn(φ)(x) = det(φ)x for all x Altn(M). The scalar det(φ) is called the determinant of the linear transformation φ. So we∈ have φ(m1) φ(mn) = det(φ)(m1 mn). ∧··· n ∧···∧ Let e1, , en be the standard basis of R = Mn,1(R). Let A = ((aij)) Mn(R). Then A determines··· the linear map l (m) = Am from Rn to Rn. Write A = (a∈, , a ) where A 1 ··· n aj = Aej is the j-th column of A. We write det(a , , a ) = det(A) = det(l ). 1 ··· n A So a a =(Ae ) (Ae ) = det(A)(e e ). 1 ∧···∧ n 1 ∧···∧ n 1 ∧···∧ n n 7.2. Theorem. Let a1, , an R . Then (a1, , an) det(a1, , an) is a skew symmetric multilinear function··· from∈ Rn Rn ··· R. In7→ other words,··· the determinant is linear in each column of the matrix×···× and it changes→ sign when two of the columns of the matrix are interchanged. One has det(In)=1. The determinant is the only function from Rn Rn to R satisfying the above properties. ×···× Proof. Follows from the identity det(a , , a )(e e )= a a (5) 1 ··· n 1 ∧···∧ n 1 ∧···∧ n n n and 6.11(c), using the fact that e1 en is a nonzero element of Alt (R ). The uniqueness is left as an exercise. ∧···∧ 7.3. Theorem. If φ, ψ Hom (Rn, Rn), then det(φ ψ) = det(φ) det(ψ). So if A, B ∈ R ◦ ∈ Mn(R), then det(AB) = det(A) det(B). Proof. Follows from the identity Altn(φ) Altn(ψ) = Altn(φ ψ) (see 6.9), since the Altn(φ) is simply multiplication by the scalar det(◦φ) and so on. ◦

7.4. Theorem. Let A = ((aij)) Mn(R). (a) If A is upper triangular or∈ lower triangular then det(A) is the product of the diagonal entries of A. (b) More generally det(A) = sign(σ)a a a . Since sign(σ) = σ∈Sn 1,σ(1) 2,σ(2) ··· n,σ(n) sign(σ−1), from this formula one seesP that det(A) = det(Atr).

Proof. Proof of (b): Substitute aj = i aijei in the right hand side of (5), multiply out and rearrange terms using 6.11. Proof ofP (a) is similar and easier.

7.5. Deﬁnition. Let A Mn(R). Let Aij Mn−1(R) to be the matrix obtained from A by deleting the i-th row and∈j-th column. Let∈c =( 1)i+j det(A ) and deﬁne adj(A) = ((c )). ij − ij ji 7.6. Exercise: Writing a , , a in terms of e ’s and expanding, show that 2 ··· n j e a a =( 1)i−1 det(A )e e i ∧ 2 ∧···∧ n − 1i 1 ∧···∧ n

14 7.7. Theorem. (a) One has A. adj(A) = det(A)In = adj(A).A. Equating the i-th diagonal entry in the identity adj(A)A = det(A)In give us the “cofactor expansion” of an n by n determinant in terms of its i-th column. (b) Let A Mn(R). Then A GLn(R) if and only if det(A) is a unit in R and in that case A−1 = det(∈ A)−1 adj(A). ∈ Proof. (a) Let η = e e . Using (5) and 7.6, we have 1 ∧···∧ n det(A)η = a e a a =(a det(A ) a det(A )+ )η. X i1 i ∧ 2 ∧···∧ n 11 11 − 21 21 ··· i This checks the cofactor expansion of the determinant along the ﬁrst column. The cofactor expansion along the rest of the columns follow by appealing to 7.2. Let i = j. Let A be 6 (ij) the matrix obtained from A by replacing the j-th column by the i-th column. Since A(ij) has two identical columns, det(A(ij)) = 0. The cofactor expansion of det(A(ij)) along its j-th column gives a det(A ) a det(A )+ = det(A )=0 1i 1j − 2i 2j ··· (ij) This veriﬁes the (i, j)-th entry of the equation Adj(A).A = det(A)I for i = j. The other n 6 equality A. adj(A) = det(A)In. interchanging the role of rows and columns since det(A) = det(Atr). −1 (b) If A GLn(R), then 7.3 implies det(A) det(A ) = 1, so det(A) is a unit in R. The converse follows∈ from part (a).

A 0 7.8. Theorem. One has det B D = det(A) det(D) for A Mn(R), D Mk(R) and B M (R). ∈ ∈ ∈ k,n

15 8. Noetherian modules and rings 8.1. Definition. Let R be a commutative ring with 1 and let M be an R-module. Let M 1 ⊆ M2 be an ascending chain of submodules of M. Say that the chain M1 M2 is stationary⊆··· if there exists k such that M = M = . ⊆ ⊆··· k k+1 ··· 8.2. Theorem. Let R be a commutative ring with 1 and let M be an R-module. TFAE: (a) Any increasing sequence of submodules in M is stationary. (b) Any non-empty collection of submodules of M has a maximal element. (c) Every submodule of M is finitely generated. 8.3. Definition. Say that M is a Noetherian R-module if it satisfies the equivalent properties above. Say that R is a Noetherian ring if it is Noetherian as a R-modules. g Theorem. Let 0 M ′ ֒ M M ′′ 0 be an exact sequence of R modules. Then M .8.4 is Noetherian if and only→ if M→′ and−→M ′′ are→ Noetherian. ′ ′′ Proof. Assume M and M are Noetherian. Let M1 M2 M3 be a ascending chain of ⊆ ⊆ ··· ′ ′ submodules of M. Then the sequences g(M1) g(M2) and M M1 M M2 ⊆ ⊆··· ∩ ⊆ ∩ ′ ⊆··· must be stationary, so there exists k such that g(Mk) = g(Mk+1) = and M Mk = ′ ··· ∩ M Mk+1 = . Suppose a Mk+1. Since g(Mk) = g(Mk+1), there exists b Mk such ∩ ··· ∈ ′ ′ ∈ that g(a)= g(b), so (a b) ker(g Mk+1 )= Mk+1 M = Mk M , so a Mk too. It follows that M = M = −. So ∈M is Noetherian.| The∩ other implication∩ is easy∈ exercise. k k+1 ··· 8.5. Corollary. (a) A finite direct sum of Noetherian modules is Noetherian. (b) Finitely generated modules over Noetherian rings are Noetherian modules. (c) If A is Noetherian ring and I is an ideal, then A/I is noetherian ring, in particular, homomorphic images of Noetherian rings are Noetherian. Proof. (c) Note that A/I is a Noetherian as a A-module, hence also as a A/I-module! 8.6. Theorem (Hilbert basis theorem). If R is a Noetherian ring then the polynomial ring R[x] is Noetherian. Proof. If f(x) R[x], we write lc(f) to denote the leading coefficient of f. Let I be an ideal in R[x]. One verifies∈ that I = lc(f): f I is an ideal in R. Since R is Noetherian, there lc { ∈ } exists a1, , an R such that Ilc =(a1, , an). Pick fi(x) I such that lc(fi)= ai. Let ′ ··· ∈ ··· ∈ I be the ideal generated by f1, , fn and let r = max deg(f1) , deg(fn) . Claim: One has I = (I M···)+ I′, where M is the{R-submodule··· of R[x}], generated by 1, , xr−1. ∩ ···Suppose f is the polynomial of the least degree in I ((I M)+ I′). Then deg(f) r. n \ ∩ ≥ Since lc(f) Ilc, we can write lc(f) = i=1 riai for some rj R. Let g(x) = f(x) n deg(f∈)−deg(fi) P ∈ − i=1 rix fi(x). Then g I and g(x) has degree less than f(x). By minimality of Pf, we get g (I M)+ I′, but then∈ so does f. This contradiction proves the claim. Since M is∈ finitely∩ generated R-module, it is Noetherian. So I M is finitely generated as a R-module, hence also as a R[x] module. So I = (I M)+ I′∩is also finitely generated as a R[x] module. ∩ 8.7. Corollary. If R is a Noetherian ring and A is a finitely generated R-algebra, then A is Noetherian. 16 9. Finitely generated module over a PID Throughout this section, let R be a (commutative) PID with 1. 9.1. Theorem. A PID is Noetherian. 9.2. Remark. Recall that for every finitely generated non-zero free R-module M, there is a well defined natural number rankR(M), the rank of the module. This integer is the number of elements in a basis of M. The rank of the zero module is defined to be 0. If M and N are two free R-modules then so is M N and rank (M N) = rank (M) + rank (N). ⊕ R ⊕ R R 9.3. Theorem. Let F be a free R-module and M be a submodule. Then M is free and rank(M) rank(F ). ≤ Proof. We only give a proof where F is finitely generated. We induct on n = rankR(F ). For n = 1, F R, so a submodule is just an ideal in R which is principal, hence either zero or free of rank≃ 1. So assume F has a basis x1, , xn and that the theorem holds when the free module has rank less than n. Consider{ the··· projection} π : F R given by π(x ) = 1 and π(x )=0 → 1 j for all j > 1. Then ker(π) is the submodule of F generated by x2, , xn, hence it is free of rank (n 1). Restricting π to M, we get an exact sequence ··· − π| 0 ker(π ) M M π(M) 0. → |M −→ −−→ → Now π(M) is a submodule of R, hence is free of rank atmost 1; and ker(π M ) is a submodule of the rank (n 1) free module ker(π), so ker(π ) is free of rank atmost| (n 1). The exact − |M − sequence above splits since π(M) is free, so M ker(π M ) π(M) is free of rank atmost n. ≃ | ⊕ 9.4. Theorem. If M is a finitely generated R-module, then every submodule of M is finitely generated. Proof. Since M is finitely generated, it is a Noetherian R-module. 9.5. Definition. Let M be a R-module. If K M define ⊆ ann(K)= r R: rK =0 { ∈ } One verifies that ann(K) is an ideal in K, it is called the annihilator of K. In paricular, if x M, then ann(x) = r R: rx = 0 . An element x M is called a torsion element if there∈ exists a non-zero r{ ∈R such that}rx = 0, that is, ann(∈ x) = 0. Let M be the set of ∈ 6 tor all torsion elements of M. One verifies that Mtor is a submodule of M, called the torsion submodule of M. Say that M is a torsion module if M = Mtor. Say that M is torsion-free module if Mtor = 0. Say that m M is a primitive element if m = rm′ for some r R and m′ M implies r is a unit in R. Verify∈ that m =(n , , n ) Rk is primitive if and∈ only if gcd(∈n , , n ) = 1. 1 ··· k ∈ 1 ··· k 9.6. Lemma. One has (a) ann(R/cR)= cR. (b) ann(M M ) = ann(M ) ann(M ). 1 ⊕ 2 1 ∩ 2 9.7. Theorem. A finitely generated torsion free R-module is free. Proof. Let M be a finitely generated torsion free R-module. Since M is Noetherian R- module, we can pick a maximal free submodule H of M. Let x , , x be a basis for H. 1 ··· n The maximality of H implies that given any v M 0 , there exists nv R 0 such 17 ∈ \{ } ∈ \{ } that n v + m x + + m x = 0 (why?). So for each v M, there exists a non-zero integer v 1 1 ··· k k ∈ nv such that nvv M. Since M is finitely generated, we can find a non-zero integer n such that nM H. Since∈ H is free, 9.3 implies that nM is free. But nM M, so M is free. ⊆ ≃ k 9.8. Corollary. Let v = (n1, , nk) R such that gcd(n1, , nk)=1. Then v can be extended to a basis for Rk. ··· ∈ ··· p Proof. Consider the short exact sequence 0 Rv Rk Rk/Rv 0. Suppose x Rk/Rv such that nx = 0 for some non-zero integer→ n. Let→ x ˜ −→Rk be such→ that p(˜x) = ∈x. Then ∈ nx˜ = mv for some m R, this means n diviedes mni for all i. Since gcd(n1, , nk) = 1, it follows that n divides∈ m. Sox ˜ = n−1mv Rv, so x = 0. Thus Rk/Rv···is a finitely generated and torsion free. Theorem 9.7 implies∈ that Rk/Rv is free, so the exact sequence above splits. In other words, there is a free submodule H Rk isomorphic to Rk/Rv such that Rk = Rv H. If B is any basis of H then v B is⊆ a basis of Rk. ⊕ { } ∪ 9.9. Theorem. Let M be a finitely generated module over R. Then one can write M = Mtor F for some finitely generated free module F . The rank of F is determined by M and is called⊕ the free rank of M.

Proof. Verify that M/Mtor is torsion free, hence it is free, by 9.7. So the exact sequence 0 M M M/M 0 → tor → → tor → splits. 9.10. Lemma. (a) Let N be a submodule of a R-module M. If λ : M R is a linear functional, then λ(N) is a ideal in R. → (b) If M is a free module and N is a non-zero submodule of N, then there is a linear functional λ such that λ(N) =0. 6 Proof. (a) is a easy verification. (b) Let n be a non-zero element of N. Then choose m M 1 ∈ such that n = rm1 and m1 is primitive in M. By 9.8, we may extend m1 to a basis m1, , mk of M. Then we may define a linear functional λ : M R by λ(m1) = 1 and λ{(m )=0··· for} j > 1. Then λ(n)= r = 0, so λ(N) = 0. → j 6 6 9.11. Theorem. Let N be a submodule of a finitely generated free R-module M such that M/N is torsion. Then there exists a basis m , , m of M and r , ,r R such that 1 ··· k 1 ··· k ∈ r1R r2R rkR and N = r1m1R + r2m2R + + rkmkR. ⊇ ⊇···⊇′ ′ ···′ ′ Suppose m1, , mk were another basis for M and r1, ,rk were another set of scalars such that r′ R ··· r′ R and N = r′ m′ R + + r′ m′ R···. Then r R = r′ R for all j. 1 ⊇···⊇ k 1 1 ··· k k j j 9.12. Definition. The scalars r1, ,rk are uniquely determined upto scalar by the submodule N. These scalars are called··· the elementary divisors of N.

Proof of the existence part of 9.11. Let rankR(M)= k. By induction assume the existence part of the theorem is true when rank of M is less than k. (The case rankR(M) = 1 just comes down to saying R is a PID). Let e1, , ek be a basis of M and µ1, ,µk be the dual basis of HomR(M, R) so that ···j ··· µ (e ) = δ . Verify that if λ : M R is a linear functional on M, then λ(N) is an ideal i j i → in R. Choose λ HomR(M, R) such that λ(N) = r1R is the maximal in the collection of ideals λ(N): λ ∈ Hom (M, R) . Lemma 9.10(b) implies that λ(N) is not the zero ideal, { ∈ R } so r1 = 0. Pick n1 N such that λ(n1)= r1. 6 ∈ 18 Claim: If µ HomR(M, R), then µ(n1) r1R. ∈ ∈ ′ ′′ proof of claim: Let d be the gcd of µ(n1) and λ(n1). Then there exists r ,r R such that d = r′µ(n )+ r′′λ(n )=(r′µ + r′′λ)(n ), so (r′µ + r′′λ)(N) dR r R. So by∈ maximality 1 1 1 ⊇ ⊇ 1 of the ideal r1R, we must have dR = r1R. This proves the claim. Write n = c e + + c e . By the claim above, c = µ (n ) is a multiple of r for all j. 1 1 1 ··· k k j j 1 1 So we can write n1 = r1m1 for some m1 M, so λ(m1) = 1. Now we have a commutative diagram ∈ λ 0 / ker(λ) / M / R / 0 O O O

λ 0 / ker(λ ) / N / r R / 0 |N 1 where the vertical arrows are inclusions and the rows are exact. The exact rows split since R and r1R are free. Verify that it now follows: M = ker(λ) m R and N = ker(λ ) r m R. ⊕ 1 |N ⊕ 1 1 Now ker(λ) is a free module of rank (k 1) and ker(λ ) is a submodule. Since M/N is − |N torsion, so is ker(λ)/ ker(λ N ). So by induction, we can ﬁnd a basis m2, , mk of ker(λ) and r , ,r R such that| r R r R and ker(λ )= r m R + ···+ r m R. 2 ··· k ∈ 2 ⊇···⊇ k |N 2 2 ··· k k To complete the proof, It remains to show r1R r2R. Let ν1, , νk HomR(M, R) be the dual basis to m , , m . Let d = ar + br be⊇ the gcd of r and··· r .∈ Then 1 ··· k 1 2 1 2 (aν1 + bν2)(r1m1 + r2m2)= d, so (aν + bν )(N)= dR r R. So by maximality of r R, we must have r R = dR. 1 2 ⊇ 1 1 1 Before proving the uniqueness, we need a lemma:

9.13. Lemma. Let F = b1R + + bkR be a free R-module with basis b1, , bk and let S = c b R + + c b R be a submodule··· with c R and c c for all j.Then ···ann(F/S)= c R. 1 1 ··· k k j ∈ j | k k Proof. Note that F/S R/c R R/c R. So ann(F/S)= c R c R = c R. ≃ 1 ⊕···⊕ k 1 ∩···∩ k k proof of uniqueness part of 9.11. Suppose M = m1R + + mkR and N = r1m1R + + ···s ··· rkmkR as in 9.11. Since m1, , mk is a basis for M, Alt (M) is a free module with basis ··· s mi1 mi2 mis with 1 i1 < i2 < < is k. In other words, Alt (M) is the internal direct∧ sum:∧···∧ ≤ ··· ≤ Alts(M)= (m m )R. M i1 ∧···∧ is 1≤i1<···

Note that r1 rk implies that the product (rk−s+1 rk−1rk) is divisible by all the | ··· | ··· s coefficient (ri1 ris ) that occur in the above sum. So 9.13, applied with F = Alt (M) and S = Alts(N) implies··· that ann(Alts(M)/ Alts(N)) = r r r R. k−s+1 ··· k−1 k So the scalars r , r r , r r r , are determined by the pair (M, N) upto units. So k k k−1 k k−1 k−2 ··· the scalars rk,rk−1,rk−2, are determined upto units too. ··· 19 9.14. Theorem (Structure theorem for finitely generated modules over PID). Let M be a finitely generated R-module. Then there exists a non-negative integer k and non-units r , ,r R such that M Rk (R/r R) (R/r R) and r r r . The 1 ··· k ∈ ≃ ⊕ 1 ⊕···⊕ n 1 | 2 | ···| n integer k is determined by M and the sequence of scalars (r1, ,rn) is determined by M upto units. Two finitely generated modules M and M ′ are isomorphic··· if and only if they determine the same (k,r , ,r ). 1 ··· n Proof. By 9.9 we can write M =(M/M ) M where the first summand is free and the tor ⊕ tor second summand is a torsion module. The integer k = rankR(M/Mtor) is determined by M. Let n be the minimal number of generators of Mtor. Let x1, , xn be a minimal set n ··· of generators of Mtor. Let R be the standard free module with standard basis e1, , en. n ··· consider the surjection π : R Mtor given by π(ej) = xj. Let N = ker(π). Then n → n Mtor R /N. By theorem 9.11, we know that there is a basis m1, , mn of R and r , ≃,r R such that r r and N = r m R + + r m R. So··· 1 ··· n ∈ 1 |···| n 1 1 ··· n n n n M m R/r m R R/r R. tor ≃ M j j j ≃ M j j=1 j=1 Now π(m ) = 0 since otherwise the (n 1) elements π(m ), , π(m ) would generate M . 1 6 − 2 ··· n tor On the other hand r1m1 N, so π(r1m1) = 0, so r1 is not a unit. So the rj’s are all non-units. ∈ Note that the integer n is determined by M as the minimal number of generators of Mtor. Hence so is the map π and the submodule N Rn. So the uniqueness part of 9.11 shows that the scalars (r , ,r ) are determined upto⊆ units by M. 1 ··· n 9.15. Theorem. Let M be a finitely generated torsion R-module. If p is a prime in R, define M(p)= m M : pnm =0 for some n 0 . { ∈ ≥ } There is a finite set of primes S (considered upto units) in R such that M(p) = 0 and M = M(p). For each p S, one has 6 ⊕p∈S ∈ M(p) R/pn1,p R R/pn2,p R R/pnkp,p R ≃ ⊕ ⊕ with 0 n1,p nk,p and ordered sequence of intergers (n1,p, , nk,p) is uniquely deter- ≤ ≤···≤ n ··· mined by M. The list of prime powers (pn1,p , ,p kp,p ): p S , are called the elementary divisors of M. Two torsion R-modules{ are isomorphic··· if and∈ only} if they have the same list of elementary divisors. Proof. If a R, let M = m M : am = 0 . Suppose a = bc where b and c are relatively ∈ a { ∈ } prime. Then one verifies that Ma = Mb Mc. Since M is finitely generated and torsion, ⊕ np there exists a R 0 such that aM = 0, so M = Ma. Let a = p be the prime ∈ \{ } Qp∈S factor decomposition of a. Then by induction it follows that M = M = M np . If q is a ⊕p∈S p any prime, then verify that M(q) M np =0. So M(q) = 0 if q / S. Also, if q S, ∩ ⊕p∈S\{q} p ∈ ∈ then M(q) Mqnq . It follows that M(q)= Mqnq . Since each⊇ M(p) is finitely generated, there exists n N such that pnM(p) = 0. By ∈ theorem 9.14, we can write M(p) R/r1R R/rkp R where r1 r2 rkp are uniquely ≃ ⊕···⊕ | ···|n determined upto units. Then pn(R/r R) =0, so r pn, that is, r = p kp,p for some n . k 1 | k kp,p Since r r r , it follows that r = pnj,p , with n n n and the 1 | 2 | ···| kp j 1,p ≤ 2,p ≤ ···≤ kp,p integers nj,p are determined by M. 20 9.16. Definition. Let spec(R) be the set of prime ideals in R. For each prime ideal I R, ⊆ choose a prime pI such that pI R = I. Let S = pI : I spec(R) . So each prime element of R is a unit multiple of exactly one element of{S. Let∈M be a R} -module. If p is a prime element of R, then define M(p)= m M : pnm = 0 for some n 0 . { ∈ ≥ } Since M(p) is finitely generated and every element of M(p) is killed by some power of p, there exists n N such that pnM(p) = 0. ∈ 9.17. Lemma. Let N be a R module. Let x N 0 such that ann(x) is a maximal element in the collection of ideals ann(y): y N ∈0 \{. Then} ann(x) is a prime. { ∈ \{ }} Proof. Homework. 9.18. Theorem. Let M be a torsion R-module. then M = M(p). ⊕p∈S Proof. Let M ′ = M(p) and suppose N = M/M ′ = 0. Pick x N 0 such that ann(x) Pp∈S 6 ∈ \{ } is maximal among annihilators of non-zero elements of M. Then 9.17 implies ann(x)=(p0) is a prime. Pickx ˜ M such that x =x ˜ mod M ′. Then p x˜ M ′. So we may write ∈ 0 ∈ p0x˜ = m0 + m1 + + mk where mj M(pj) where pj’s are distinct elements of S. Let ··· ∈ n q = p1 pk. Then some power of q kills m1, , mk. So q p0x˜ M(p0), which implies n ··· n ··· n ∈ q x˜ M(p0), that is q x = 0 in N. Also px = 0. Since gcd(p, q ) = 1, it follows that x = 0. This∈ contradiction proves M = M ′ = M(p). Pp∈S Let x M(p0) (M(p1)+ + M(pk)) where p0R,p1R, ,pkR are distinct prime ideals. ∈ ∩ nj··· ··· n There exists nj such that pj M(pj) = 0. Let n = max n0, n1, , nk . Then pj M(pj)=0 {n ··· } for all j = 0, 1, , n. Since x M(p0), we have p0 x = 0. Let q = p1 pn. Since ··· ∈ n n n ··· x M(p1)+ + M(pk), we have q x = 0. Now gcd(p0 , q ) = 1, so there exists a, b R ∈ n ··· n n n ∈ such that ap0 + bq =1. So x = ap0 x + bq x =0. So M(p0) (M(p1)+ + M(pk)) = (0). So the sum M(p) is a direct sum. ∩ ··· Pp∈S

21 10. Canonical forms of matrices In this section F will be be a field and V will be a F -vector space. We let R = F [x]. Sometimes we will denote the identity element of a ring A by IA. 10.1. Definition. Say that a field E is an field extension of F if E F . Say that an extension E of F is algebraic if element of E is solution to some polynomial⊇ equation with coefficients in F . A field F is called algebraically closed if every polynomial in F [x] has a root in F , hence splits into linear factors over F . We note a couple of facts: Given any field F , there exists an algebraic extension F¯ such that F¯ is algebraically ◦ closed. We say that F¯ is an algebraic closure of F . The field of complex numbers, denoted C, is algebraically closed. ◦ 10.2. Definition. Let T End(V ). Say that λ F is an eigenvalue of T in F , if ker(T λI) = 0 . In this case, the∈ non-zero vectors in ker(∈ T λI) are called the eigenvectors of −T corresponding6 { } to the eigenvalue λ. We shall let spec(T−) denote the set of eigenvalues of T . Assume dimF (V ) is finite. Then define the characteristic polynomial of T by c (x) = det(xI T ). T − Note that λ F is an eigenvalue of T if and only if cT (λ) = 0. In other words, the eigenvalues of∈T in F are exactly the solutions of the characteristic polynomial in F . If f(x)= a + a x + + a xk F [x], then we write f(T )= a I + a T + + a T k. 0 1 ··· k ∈ 0 1 ··· k 10.3. Theorem (Cayley-Hamilton). Let V be a finite dimensional F -vector space. Let T ∈ EndF (V ) and let cT (x) be the characteristic polynomial of T . Then cT (T )=0. Proof. Let x , , x be a basis of V . Let E = End (V ). Let V n = V V (n-times). 1 ··· n F ⊕···⊕ Write T xj = i aijxi. Consider the matrix S Mn(E) given by S = ((aijIE δijT )). Note P ∈ − n that by definition cT (T ) = det(S). The matrix S defines an element of End(V ) and the tr n equations T xj = i aijxi translate into Sx = 0 where x =(x1, , xn) V . Multiplying P adj ··· ∈ by the adjugate of S, we get 0= S Sx = det(S)IMn(E)x. This means det(S)xj = 0 for all j, that is det(S) = 0. 10.4. Extension of scalars: Let F E be a field extension. Then E V is a E-vector ⊆ ⊗F space. If v1, v2, is a F -basis of V , then verify that the same set froms an E-basis of E V . We{ say···} that E V is the vector space obtained from V by extension of scalars. ⊗F ⊗F Let T EndF (V ). Then T naturally extends to a E-linear map from E F V to E F V . We shall∈ denote this linear transformation again by the same symbol T . Let⊗ be a F⊗-basis of V of size n. Then T is represented by some matrix A M (F ) with respectB to . The ∈ n B extended linear map on V F E is then represented by the same matrix A with respect to the E-basis of (E V ).⊗ We are now simply thinking of A as an element of M (E). B ⊗F n Suppose dimF (V )is finite. Sometimes, given a linear map T EndF (V ), it is convenient to extend scalars and consider T as a linear map on the vector∈ space F¯ V over the ⊗F algebraically closed field F¯. We can always solve the equation cT (x) = 0 in F¯ and then find bases of the vector spaces ker(λI T ) for each λ spec(T ). This way we find all the − ∈ eigenvalues of T in F¯ and all corresponding eigenvectors in F¯ F V . Then we can check which of these eigenvalues (resp. eigenvectors) actually lie in F (resp.⊗ V ). 10.5. Notation: Let R be a ring. Recall that an R-module is a pair (M, ρ) where M is an abelian group and ρ : R EndZ(M) is a ring homomorphism. Suppose F R, that is, → 22 ⊆ R is an F -algebra. Then M becomes an F -module, that is, a F -vector space. Think of M with this F -vector space structure. In this case, verify that for all r R, the map ρ(r) is F -linear, so ρ : R End (M) and further that ρ is F -linear. ∈ → F 10.6. Lemma. (a) Given an F -vector space V and T End (V ), there exists an unique ∈ F F [x]-module structure (V, ρT ) on V that extends the F -module structure on V and that satisfies ρT (x)= T . Conversely, any F [x]-module (V, ρ) is determined by the F -vector space structure on V and the F -linear map ρ(x) End (V ). ∈ F (b) The modules (V, ρT ) and (V, ρS) are isomorphic if and only if T and S are similar. Proof. Easy verification. 10.7. Observation: Recall that F [x] is a PID. The above lemma tells us that:

specifying an F [x]-modules structure on an F -vector space V is the equivalent to specifying an element T End (V ). ∈ F This is the key observation that lets us apply the structure theorem for finitely generated modules over PID’s to deduce results in linear algebra. 10.8. Definition. From now on, let V be a finite dimensional F -vector space. Let ρ : F [x] End (V ) T → F be the F [x]-module structure on V defined by ρT (f(x)) = f(T ). The map ρT is in particular a F -linear map from the infinite dimensional F vector space F [x] to the finite dimensional F -vector space EndF (V ). So ker(ρT ) is a non-zero ideal in F [x]. So there is a unique monic polynomial mT (x) such that

annF [x](V ) = ker(ρT )=(mT (x)).

Exercise: Verify that mT (x) is the unique monic polynomial of least degree such that mT (T ) = 0. We say that mT (x) is the minimal polynomial of T .

10.9. Theorem. The minimal polynomial mT (x) divides the characteristic polynomial cT (x). The characteristic polynomial divides some power of the minimal polynomial. So the solution set of both polynomials is the exactly the set of eigenvalues of T .

Proof. By the Cayley-Hamilton theorem the characteristic polynomial cT (x) ker(ρT ). So m (x) c (x). So each roots of m (x) in F¯ is an the eigenvalues of T . Conversely,∈ suppose T | T T λ is an eigenvalue of T in F¯. If(x λ) does not divide mT (x), then gcd((x λ), mT (x)) = 1, so we can write a(x)(x λ)+ b(x−)m (x) = 1 for some a(x), b(x) F¯[x]. But− then − T ∈ I ¯ = ρ (a(x)(x λ)+ b(x)m (x)) = a(T )(T λI)+ b(T )m (T )= a(T )(T λI) End(F ⊗F V ) T − T − T − which contradicts the fact that (T λI) is not invertible. So (x λ) divides mT (x) in F¯[x]. − αλ − So in F¯[x] we have a factorization mT (x)= (x λ) with each αλ > 0. So cT (x) Qλ∈spec(T ) − divides some power of mT (x) in F¯[x] and hence in F [x]. 23 10.10. Theorem. Assume the setup in 10.8. Let R = F [x]. (a) Then there exists monic polynomials a1(x), , ad(x) R, (the invariant factors of the R-module (V, ρT )) such that a (x) a (x) ···a (x) and∈ we have an R-module isomorphism 1 | 2 |···| d V R/a R R/a R. ≃ 1 ⊕···⊕ d The polynomials aj(x) called the invariant factors of T and they are uniquely determined by T . (b) Two linear maps are similar if and only if they have the same invariant factors. (c) One has ad(x)= mT (x). Proof. Since R is inﬁnite dimensional as F -vector space, the R-module V cannot have a free component. So V is a torsion R-module. Now (a) follows from 9.14. We know two endomorphisms S and T are similar if and only if (V, ρS) and (V, ρT ) are isomorphic R- modules. Since the invariant factors of a torsion R-module determines and is completely determined by its invariant factors, part (b) now follows. Finally, Observe that ak(x) is the annihilator of the R-module V , so ak(x) and mT (x) generate the same ideal and both are monic. This proves (c). k−1 k 10.11. Notation. Let a(x)= b0 + b1x + + bm−1x + x F [x]. Then deﬁne C to be the k k matrix whose last column is ( b ···, b , , b )tr,∈C = C = = C = 1 × − 0 − 1 ··· − k−1 2,1 3,2 ··· k,k−1 and all other entries of C are zero. Write C = Ca(x). 10.12. Exercise: Show that we have a R module isomorphism (F k, ρ ) R/a(x)R. Ca(x) ≃ 10.13. Theorem. Assume the setup in 10.8. Let a1(x), , ad(x) be as in 10.10. (a) Then there exists a basis of V with respect to··· which the matrix of T is block diagonal with the diagonal blocks being C , ,C . We denote this matrix by a1(x) ··· ad(x) Block[Ca1(x), ,Cad(x)] and call it the rational canonical form of T . (b) The rational··· canonical form of T is uniquely determined by T (upto permutation of the blocks). (c) Two linear maps are similar if and only if they have the same rational canonical form.

Proof. (a) Identify V with d R/a R as R-modules using 10.10. The linear map T ⊕j=1 j ∈ End (V ) corresponds to the map L End ( d R/a R) given by L (p)= x.p. The space F x ∈ F ⊕j=1 j x R/a R has a F -basis 1, x, , xdeg(aj )−1 and one veriﬁes that the matrix of L with j { ··· } x|R/aj R respect to this basis is Caj (x). This proves (a).

(b) Let A = Block[Ca1(x), ,Cad(x)] be a rational canonical form of T with respect to ··· ′ some basis of V where a1(x) ad(x) and deg(a1) 1. Let A = Block[C ′ , ,C ′ ] a1(x) ad′ (x) |···| ≥ ··· ′ be another rational canonical form of T with respect to some other basis of V where a1(x) ′ ′ | ad′ (x) and deg(a1) 1. Suppose V has dimension n. ···| ≥ n Let dimF (V )= n. Let LA be the linear map on F determined by A: LA(v)= Av. Then n LA determines a R-module structure on F . Using 10.12, one veriﬁes that we have an R- n d module isomorphism F j=1R/ajR. So the polynomials a1(x), , ad(x) are just the ≃ ⊕ ′ {′ ··· } invariant factors of the linear map L . Similarly a (x), , a ′ (x) are the invariant factors A { 1 ··· d } of the linear map LA′ . Since LA is similar to LA′ , 10.10(b) implies that they determine the ′ ′ ′ same invariant factors. So d = d and the lists a1(x), , ad(x) and a1(x), , ad′ (x) are identical. { ··· } { ··· } 24 (c) Note that specifying the rational canonical form of T is equivalent to specifying the invariant factors of the module (V, ρT ). Recall that S and T are similar if and only if the F [x]-modules (V, ρT ) and (V, ρS) are isomorphic. Also (V, ρT ) and (V, ρS) are isomorphic if and only if they have the same invariant factors.

10.14. Notation: Let Nk be the k k matrix whose (i, j)-th entry is equal to 1 if j = i +1 and zero otherwise. Let J = λI ×+ N . Then (J λI)k = N k = 0, so λ is the only k,λ k k k,λ − k eigenvalue of Jk,λ. We say that Jk,λ is a Jordan block of length k and type λ. A matrix is said to be in the Jordan form if it is block diagonal with Jordan blocks along the diagonal. Exercise: Show that we have a R-module isomorphism (F k, ρ ) R/(x λ)kR. Jk,λ ≃ − 10.15. Theorem. Let V be a ﬁnite dimensional F -vector space. Let T EndF (V ). Suppose the minimal polynomial breaks into linear factors over F . ∈ (a) Then there is a basis of V with respect to which the matrix of T is in the Jordan form. (b) The Jordan form of a linear map is unique (upto permutation of Jordan blocks). (c) Two linear maps are similar if and only if they have the same Jordan from (upto permutation of Jordan blocks).

Proof. Let ρT : F [x] End(V ) be the R = F [x]-module structure on V determined by T . By 9.15, there exists →p (x), ,p (x) F [x] (not necessarily distinct) such that 1 ··· k ∈ V k R/p (x)nj R. ≃ ⊕j=1 j nj Since mT (x) acts as zero on V , it follows that mT (x) acts as zero on R/pj(x) R. So nj pj(x) must divide mT (x) for all j. So each pj(x) is a prime factor of mT (x). Since mT (x) decomposes into linear factors in F [x], each pj(x) has the form (x λ) for some nj −n eigenvalue λ of T . So each summand R/pj(x) R has the form R/(x λ) R for some λ spec(T ) and n 1. The operator T on V corresponds to the “multiplication-by-− x” ∈ ≥ operator on k R/p (x)nj R. Call this map L . Verify that R/(x λ)nR has a F -basis ⊕j=1 j x − given by 1, (x λ), (x λ)2, , (x λ)n−1 and the matrix of L with respect to this basis { − − ··· − } x is just Jn,λ. This proves part (a). Proofs of part (b) and (c) are similar to those in 10.13, once we note that spcifying the Jordan form of T is equivalent to specifying the elementary divisors of the module (V, ρT ). 10.16. Corollary. Let V be an n-dimensional vector space and T End(V ). Consider the Jordan form of T . Then ∈ (a) Multiplicity of λ in cT (x) is is the sum of the lengths of the Jordan blocks of type λ. (b) Multiplicity of λ in mT (x) is the length of the longest Jordan block of type λ. (c) The dimension of the space of eigenvectors corresponding to a eigenvalue λ, that is, dim(ker(T λI)) (also called the geometric multiplicity of the eigenvalue λ) is equal to the number of Jordan− blocks of type λ in A. (d) T is diagonalizable (i.e. similar to a diagonal matrix) if and only if all the Jordan blocks are of length 1. Proof. Proof of (c): Let A be a matrix in Jordan form and let λ be an eigenvalue of A. If

S end(V ) and V = jVj such that SVj Vj for all j, then ker(S)= j ker(S Vj ). So the null∈ space of (A λI)⊕ is equal to the direct⊆ sum of the null spaces of the⊕ Jordan| blocks in (A λI). Let J − be a jordan block in A. If µ = λ, then (J λI) is invertible. If µ = λ, − µ,k 6 µ,k − then (Jµ,k λI)= Nk which clearly has rank (k 1), so it has a one dimensional null space. The other− parts are left as exercise. − 25 10.17. A method to ﬁnd a basis in which a matrix has Jordan form: Let A ∈ MN (C). Find the eigenvalues of A by solving the characteristic equation. For each eigenvalue λ of A, follow the following procedure: i Let Ni be the null space of (A λI) . For i = 1, 2, ﬁnd a basis i for Ni and ◦ calculate n = dim(N ), n = dim(− N ) dim(N ), ···, n = dim(N )B dim(N ), 1 1 2 2 − 1 ··· k k − k−1 until dim(Nk) = dim(Nk+1). Extend to a basis of N . ◦ Bk−1 Bk−1 ∪ Sk k Verify that k−2 (A λI)( k) is a linearly independent subset of Nk−1. Extend this ◦ linearly independentB ∪ set− to aS basis of N . Bk−2 ∪ Sk−1 k−1 Verify that k−3 (A λI)( k−1) is a linearly independent subset of Nk−2. Extend ◦ this linearlyB independent∪ − setS to a basis of N . Bk−3 ∪ Sk−2 k−2 Repeat this process k times to get 1, , k. By induction on r, we verify that ◦ the disjoint union of , , formsS a··· basisS for N . It follows that = n . By S1 ··· Sr r |Sj| j construction (A λI) j j−1. So we can write Sj = vj,1, , vj,nj such that (A λI)v = v − . ThusS ⊆ we S end up with the following list{ of··· vectors: } − j,r j−1,r 00 0 0 00 : v v v v v v S1 1,1 1,2 ··· 1,nk ··· 1,nk−1 ··· 1,n2 ··· 1,n1 2 : v2,1 v2,2 v2,nk v2,nk−1 v2,n2 S . . ··· . ··· . ··· . . . . ··· ··· k−1 : vk−1,1 vk−1,2 vk−1,nk vk−1,nk−1 S : v v ··· v ··· Sk k,1 k,2 ··· k,nk and multiplication by the matrix (A λI) moves the vectors vertically up one step. k − Let Vλ = Null(A λI) be the subspace spanned by vj,k’s; this is the “generalized eigenspace − N of A corresponding to eigenvalue λ”, Then C is the direct sum of the Vλ’s. Each Vλ is

A-stable. The vj,k’s form a basis for Vλ and the matrix of A Vλ with respect to this basis gives the collection of Jordan blocks of type λ. We get a Jordan| block of type λ for each column in the above diagram. The length of the column is the size of the corresponding Jordan block. Note that the Jordan form of A can be found without actually computing the vectors vj,k’s. All we need to compute are the numbers (n1, , nk) for each eigenvalue λ. In the following, denotes any matrix norm. ··· kk 10.18. Deﬁnition. Let A Mn(C). The number ρ(A) = sup λ : λ spec(A) is called the spectral radius of A. ∈ {| | ∈ } A map : Mn(C) R≥0 is called a matrix norm if (i) A = 0 if and only if A = 0, (ii) rA k= kr A , (iii)→ A + B A + B and (iv) ABk k A B . For example k k | |k k k k≤k k k k k k≤k kk k 2 ((aij)) 2 = aij is a matrix norm. k k qPij| | n 10.19. Lemma. Let J = λIk + Nk as in 10.14 with λ C. Then limn→∞ J =0 if and only if λ < 1. If λ > 1, then J n : n 1 is unbounded.∈ | | | | {k k ≥ } Proof. Since N k = 0, for all n>k we have J k = λkI+ n λn−1N + + n λn−k+1N k−1. k 1 ··· k−1 n 10.20. Lemma. Let A Mn(C). Then limn→∞ A =0 if and only if ρ(A) < 1. If ρ(A) > 1, then An : n 1 is∈ unbounded. {k k ≥ } 26 Proof. Write A = SBS−1 where B is in Jordan form. Then An = SBnS−1. So An 0 if and only if Bn 0. Now the lemma follows from 10.19. → → n 1/n 10.21. Theorem (Gelfand’s spectral radius formula). One has ρ(A) = limn→∞ A for all A M (C). k k ∈ n Proof. Fix r > ρ(A) >s> 0. Then ρ(r−1A) < 1, so (r−1A)n 0, that is, for large n we have 1 > (r−1A)n = r−n An , so r > An 1/n. On the other→ hand ρ(s−1A) > 1, so for large n, wek have 1 k< (s−1kA)nk = s−n Akn ,k so s < An 1/n. So s < An 1/n < r for all large n. k k k k k k k k

27 11. Field Extensions Notation: The letters F,K,L,M would usually denote fields. 11.1. Definition. Let F be a field. The characteristic of a field, denoted ch(F ) is the smallest natural number n such that n.1F = 0. If no such n exists, then one says that F has characteristic zero, otherwise one says that F has finite characteristic. If F is finite characteristic, then it is easy to see that ch(F ) is a prime number. Clearly the intersection of a collection of subfields of F is again a subfield. If F is a field and S is some subset of F , then the subfield of F generated by S is defined to be the intersection of all subfields of F that contain S. So the subfield of F generated by S is the smallest subfield of F that contains S. The prime subfield of F is defined to be the subfield generated by 1F . If F has characteristic zero, then the prime subfield is isomorphic to Q. If F has characteristic p, then the prime subfield is isomorphic to Fp. 11.2. Field extensions: If K is a field containing a field F , then we say that K is an extension field of F . We use the shorthand K/F to denote a field extension. Two extensions K/F and K′/F are called isomorphic if there is a field isomorphism φ : K K′ such that the restriction of φ to F is the identity. → If F K M is a “tower” of extensions, we say that K/F is a subextension of M/F . If K/F⊆ is⊆ an field extension, then K is a F -vector space. The dimension of this vector space is called the degree of the extension and denoted by [K : F ] = dimF (K). We say that K/F is a finite extension if [K : F ] is finite. An element a K called algebraic over F if there exists p(x) F [x] such that p(a) = 0. Call K/F an algebraic∈ extension if each element of K is algebraic∈ over F . 11.3. Theorem. Suppose F K M is a tower of extensions. Then M/F is finite if and only if M/K and K/F are both⊆ finite.⊆ Suppose a1, , am be a basis of K/F and let b1, , bn be a basis of M/F . Then the a b : 1 i ···m, 1 j n forms a basis of M/K.··· So [M : F ] = [M : K][K : F ]. { i j ≤ ≤ ≤ ≤ } Proof. Exercise. 11.4. Theorem. Finite extensions are algebraic. Proof. Let K/F be a finite extension and a K. The elements 1, a, a2, cannot be all linearly independent over F , so some nontrivial∈ linear combination of these··· must be zero. In other words, a must satisfy of some polynomial with coefficients in F . 11.5. Minimal polynomials: Suppose K/F is an algebraic extension and a K. Let F [a] and F (a) be respectively the subring and subfield of K generated by F and∈ a. Let φa : F [x] F [a] be the surjective homomorphism obtained by sending x to a. Then ker(φ ) = →f F [x]: f(a)=0 is an ideal in F [a]. Since a is algebraic over F , the ideal a { ∈ } ker(φa) is not zero. So there exists a unique monic polynomial pa(x) such that ker(φa) = (pa(x)). Verify that pa(x) is the unique monic polynomial in F [x] of minimal degree such that pa(a) = 0. We call pa(x) the minimal polynomial of a (over F ). Thus we have

φa 0 (pa(x)) F [x] F [a] 0, where φa(x)= a. → → −→ 28 → Recall that in a PID every prime ideal is maximal. Since F [a] F [x]/(pa) is an integral domain, (p ) is a prime ideal, hence maximal. So F [a] F [x]/(p≃) is actually a field. So a ≃ a F [x]/(p ) F [a]= F (a). a ≃ 2 d−1 If the polynomial pa has degree d then one verifies that 1, a, a , , a forms a basis of F (a) as a K vector space. Hence [F (a): F ]= d. Conversely we have··· the following: 11.6. Theorem. Let p be an irreducible monic polynomial of degree d in F [x]. Then there exists a extension K/F of degree d and a K such that p is the minimal polynomial of a and K = F [a] F [x]/(p(x)). ∈ Suppose K and≃ K′ are two extensions of F and a K, a′ K′ such that p(x) is the minimal polynomial of a and a′ over F . Then the extensions∈ K/F∈ and K′/F are isomorphic via an isomorphism that sends a to a′. Proof. Since p(x) is irreducible, it is a prime. In a PID every prime ideal is maximal, so K = F [x]/(p) is a field. Let φ : F [x] F [x]/(p) be the surjection. The composition F F [x] K = F [x]/(p) is injective. Identify→ F with the image of of this injection. Then L/K→ is a field→ extension. Let a = φ(x). Then K = F [a] and p(a) = 0 holds in K. Since p(x) is irreducible, it must be the minimal polynomial of a. Finally, 1, a, a2, , ad−1 is a basis of K as a F -vector space. ··· 11.7. Generators for a field extension: Let F [x , , x ] denote the ring of polynomials 1 ··· n in n variables x1, , xn, with coefficients in the field F . The fraction field of F [x1, , xn] is the field of all rational··· functions, denoted by F (x , , x ). ··· 1 ··· n Let K/F be a field extension and a1, , an K. Then one verifies that the smallest subfield of K generated by F and a , ···, a consists∈ of all rational expressions of the form 1 ··· n f(a1, , an)/g(a1, , an) where f,g F [x1, , xn] and g(a1, , an) = 0. This field is denoted··· by F (a , ···, a ). Say that the∈ extension··· K/F is generated··· by a6 , , a K if 1 ··· n 1 ··· n ∈ K = F (a1, , an). Say that an extension K/F is finitely generated if K = F (a1, , an) for some a ···, , a . Observe that ··· 1 ··· n F (a , , a )= F (a , , a )(a ). 1 ··· n 1 ··· n−1 n Say that K/F is a simple extension if K = F (a) for some a K. ∈ 11.8. Theorem. An extension K/F if finite if and only if there exists finitely many elements a , , a K such that each a is algebraic over F and K = F (a , , a ). 1 ··· n ∈ j 1 ··· n Proof. Suppose a1, , an K such that each aj is algebraic over F and K = F (a , , a ). Let···m be the∈ degree of the minimal polynomial of a over F . A for- 1 ··· n j j tiori, aj satisfies a polynomial of degree mj which has coefficients in F (a1, , aj1 ). So [F (a , , a )(a ), F (a , , a ] m . So every step in the tower ··· 1 ··· j−1 j 1 ··· j−1 ≤ j F F (a ) F (a , a ) F (a , , a ) ⊆ 1 ⊆ 1 2 ⊆···⊆ 1 ··· n is finite and 11.3 implies that F (a1, , an)/F is finite too; infact [F (a1, , an): F ] m m m . ··· ··· ≤ 1 2 ··· n Conversely, suppose K/F is finite. Let a1, , an be a basis of K as a F -vector space. Since finite extensions are algebraic, each a is··· algebraic over F and a , , a generate K j 1 ··· n as a F -vector space, hence also as a field, that is, F (a1, , an)= K. 29 ··· 11.9. Corollary. (a) Let F K M be fields. If K/F and M/K are algebraic then so is M/F . ⊆ ⊆ (b) Let K/F be an extension and a, b K are algebraic over F . Then (a + b) and ab are also algebraic over F . So the set of∈ elements in K that are algebraic over F form a subextension of K/F . n i Proof. (a) Let a M. Then there is a polynomial f(x) = aix K[x] such that ∈ Pi=0 ∈ f(a)=0. So a is algebraic over L = F (a0, , an), so L(a)/L is finite. Since each aj K is algebraic over F , 11.8 implies that the extension··· L/F is finite. So L(a)/F is also finite∈ and hence a is algebraic over F . (b) Since a and b are algebraic over F , the extension F (a, b)/F is finite. Since (a + b) belong to F (a, b), the extensions F (a + b)/F is also finite. Hence a + b is also algebraic over F . Similar argument for ab. 11.10. Definition. Let K,K ,K , F be fields such that F K K for j =1, 2. Then we 1 2 ⊆ j ⊆ define K1K2 to be the subfield generated by K1 and K2, that is, the smallest subfield of K that contains both K1 and K2. We say that K1K2 is the compositum of K1 and K2. Exercise: If [K , F ] and [K : F ] are finite, then [K K : F ] [K : F ][K : F ]. 1 2 1 2 ≤ 1 2 11.11. Definition. An injective homomorphism of fields is called an embedding of fields. Let σ : F L be a homomorphism of fields. Since F does not have any nonzero ideal, ker(σ) is either→ 0 or is equal to F . So either σ = 0 or σ is an embedding. Let K/F be an extension and σ : F L be an embedding. We say that σ extends to K if there exists a →′ ′ field homomorphism σ : K L such that σ F = σ. Let R,S be rings. Any ring→ homomorphism| σ : R S induces a ring homomorphism i i → R[x] S[x] by defining σ( i aix ) = i σ(ai)x . This homomorphism R[x] S[x] will again→ be denoted by σ. P P → 11.12. Lemma. Let σ : F L be an embedding of a field F into a field L. Let p(x) F [x] be an irreducible polynomial→ and let K = F [x]/(p(x)) be the extension of of F obtained∈ by adjoining a root of p. The embedding σ extends to K if and only if σ(p)(x) has a root in L. Proof. Let a be the image of x in K. So K = F [a] = F (a) F [x]/(p(x)). Ifσ ˜ is an extension of σ to K thenσ ˜(a) is a root of σ(p) in F . Conversely,≃ suppose α be a root of σ(p) in L. Then define a map σ : F [x] L such that σ = σ and σ (x) = α. Then 1 → 1|F 1 σ1(p(x)) = σ(p)(α) = 0, so ker(σ1) contains (p(x)) which is a maximal ideal in F [x]. Since σ1 = 0, it follows that ker(σ1)=(p(x)). So σ1 induces an embeddingσ ˜ : F [x]/(p(x)) L which6 extends σ. →

30 12. splitting field, Algebraic closure 12.1. Definition. Let f(x) F [x]. Say that f(x) splits in F [x] if it can be decomposed into linear factors in F [x]. An∈ extension K/F is called a splitting field of some non-constant polynomial f(x) if f(x) splits in K[x] but f(x) does not split in M[x] for any any K ) M F . If an extension K of F is the splitting field of a collection of polynomials in F [x], then⊇ we say that K/F is a normal extension. 12.2. Lemma. Let p(x) be a non-constant polynomial in F [x]. Then there is a finite extension M of F such that p(x) splits in M. Proof. Induct on deg(p); the case deg(p) = 1 is trivial. If p is already split over F , then F is already a splitting field. Otherwise, choose an irreducible factor p1(x) of p(x) such that deg(p1(x)) 1. By, there is a finite extension M1/F such that p1 has a root α in M1. So we can write p≥(x)=(x α)q(x) in M [x]. Since deg(q) < deg(p), by induction, there is a finite − 1 extension M/M1 such that q splits in M. Then M/F is also finite and p splits in M. 12.3. Theorem. (a) Let p(x) be a non-constant polynomial in F [x]. Then there is a finite splitting field K for p(x). (b) Any such K is a finite extension of F . (c) The splitting field for p(x) is unique in the sense that if K and K′ are two splitting fields for p(x), then K/F and K′/F are isomorphic extensions. Proof. (a) By 12.2 that there is a finite extension M of F such that p(x) splits in M. Let K be the intersection of all subextensions of M/F in which p(x) is split. Verify that K is a splitting field for p(x). n (b) let K be a splitting field of p(x). Suppose p(x) factors in K as p(x)= c i=1(x ai). Since p(x) already splits in F (a , , a ) K, we must have K = F (a , Q, a ).− Since 1 ··· n ⊆ 1 ··· n each aj is algebraic over F , the extension F (a1, , an)/F is finite. (c) Let i′ : F K′ be the inclusion map. Consider··· the collection of all fields M such that F M K and→ there exists an embedding σ : M K′ that extends i′. Among these, choose⊆ an⊆M such that [M : F ] is maximal. Let σ : M → K′ be an extension of i′. We claim that M = K. → Proof of claim: If M = K, then there exists a root a K of p(x) such that a / M. Let q(x) be the minimal polynomial6 of a in M[x]. Then q(x)∈ is a factor of p(x) in M∈[x], hence σ(q(x)) σ(p(x)) = i′(p(x)) = p(x) in K′[x]. But p(x) splits in K′[x]. So σ(q(x)) has a root in K′. By| lemma 11.12, the embedding σ can be extended to M(a) M[x]/(q(x)). This contradicts the maximality of M and proves the claim. ≃ Identify F inside K and K′. Then, by the claim, we have an embedding σ : K K′ such → that σ F = idF . Since σ(K)/F and K/F are isomorphic extensions, the polynomial p(x) splits already| in σ(K) K′. Since K′ is a minimal extension in which p splits, we must have σ(K)= K′. ⊆ 12.4. Definition. Say that a field K is algebraically closed, if every polynomial in K[x] splits in K. For example, the fundamental theorem of algebra states that C is algebraically closed. An extension K of F is called an algebraic closure of F , if K/F is algebraic extension and K is algebraically closed. 12.5. Theorem. Let F be a field. Then there exists an algebraically closed field containing F . 31 Proof. (Due to E. Artin) Let Σ be the set of all irreducible monic polynomials in F [x]. For each f Σ we choose an indeterminate x . Let A be the polynomial ring generated by the ∈ f indeterminates xf with coefficients in F . Let I be the ideal in A generated by f(xf ): f Σ . We claim that I = A. { ∈ } proof of claim:6 Suppose I = A. Then there exists f , , f Σ such that 1 ··· n ∈ g f (x )+ g f (x )=1 1 1 f1 ··· n n fn with g A. For simplicity, let us write x = x . The polynomials g ’s and f ’s together j ∈ i fi j j involve only finitely many variables. Let us call them x1, , xN (with N n). So our identity now reads ··· ≥ g (x , , x )f (x )+ g (x , , x )f (x )=1. 1 1 ··· N 1 1 ··· n 1 ··· N n n Let M be a finite extension of F such that each f has a root a M. Then substituting j j ∈ xj = aj yields 0 = 1; this contradiction proves the claim. Let m be a maximal ideal of A containing I and let F = A/m. Then each f Σ has a 1 ∈ root in the field F1. Repeat the construction with F1 instead of F to inductively obtain a tower of extensions F F1 F2 such that each monic irreducible polynomial in Fj[x] ⊆ ⊆ ∞ ⊆··· has a root in Fj+1. Let K = n=1Fn. Then verify that K is a field (with operations induced from F ’s). If f K[x] is any∪ non-constant polynomial, then f has coefficient in some F n ∈ n hence it has a root in Fn+1, hence f has a root in K. So every non-constant polynomial in K[x] splits. 12.6. Theorem. Let L/F be an extension with L algebraically closed. Let K/F be an algebraic extension. Then there exists an embedding σ : K L fixing F . → Proof. Let i : F L be the inclusion map. Consider the collection → = (M, σ): F M K, σ : M K′, σ = i . F { ⊆ ⊆ → |F } Define a partial order on by defining (M, σ) (M ′, σ′) if and only if M M ′ and ′ F ≤ ⊆ σ M = σ. One verifies that every chain in has an upper bound. So by Zorn’s lemma, there| exists a maximal element (M, σ) in . WeF claim that M = K. The proof of this claim is similar to the proof of the claim in 12.3(c).F 12.7. Theorem. Every field has an algebraic closure. The algebraic closure is unique in the sense that if K and K′ are two algebraic closures of F , then the extensions K/F and K′/F are isomorphic. Proof. Let F be a field. Theorem 12.5 implies that there exists an extension K/F such that K is algebraically closed. Let L be the set of all elements of K that are algebraic over F . From 11.9(b), we know that L is a field; so L/F is an algebraic sub-extension of K/F . Let f(x) L[x] L. Then f(x) has a root a in K. Since L(a)/L is algebraic and L/F is algebraic,∈ corollary\ 11.9(a) implies that a is algebraic over F . But then a L. So each non-constant polynomial in L[x] has a root in L. Consequently, L is algebraically∈ closed, so it is an algebraic closure of F . Identify F inside K and K′. By 12.6, there exists an embedding σ : K K′ such that → σ F = idF . Since σ(K)/F and K/F are isomorphic extensions, σ(K) is algebraically closed. But| K′/σ(K) is an algebraic extension; so we must have K′ = σ(K). 32 12.8. Definition. Let K/F be an extension and f : i I be a collection of polynomials { i ∈ } in F [x]. We say that K is a splitting field for the family fi : i I , if each fi splits in K and K is a minimal extension with this property. In other{ words,∈ K} contains all the roots of all the polynomials in f : i I and K is generated by these roots over F . This field K { i ∈ } contains a splitting field Ki for each fi and K is the compositum of the Ki’s. Let M be any splitting field of fi : i I . Fix an algebraic closure M¯ of M. Then M is the subfield of M¯ generated over{ F by∈ all} the roots of all the polynomials in f : i I . { i ∈ } One verifies that M is the unique subfield of M¯ that is a splitting field for fi : i I . Let K be any splitting field for f : i I . Then 12.6 implies that we have{ an embedding∈ } { i ∈ } σ : K F¯ fixing F . The image σ(K) is a subfield of F¯ and a splitting field of fi : i I . So σ(K→)= M. So splitting field of f : i I is an unique extension of F upto isomorphism.{ ∈ } { i ∈ } 12.9. Lemma. Let K/F be an algebraic extension. Let σ : K K be an embedding fixing F . Then σ is an automorphism of K. → Proof. Let a K and let p(x) F [x] be the minimal polynomial of a. Consider the subfield ∈ ∈ M of K generated by all the roots of p that lie in K. Then restriction σ M takes M to itself since a root of p maps to a root of p under σ. But [M : F ] is a finite| extension and any injective map of finite dimensional vector spaces is an isomorphism. So σ M is an isomorphism. So a is in the image of σ, hence σ is an onto. | 12.10. Theorem. Let K be an algebraic extension of F . Fix an algebraic closure F¯ of F containing K. Then the following conditions are equivalent: (a) Every embedding of K into F¯ fixing F induces an automorphism of K. (b) K is a splitting field of a family of polynomials fi : i I in F [x]. (c) Every irreducible polynomial in F [x] which has a{ root∈ in K} splits in K. Proof. Assume (a). Let a K and p F [x] is the minimal polynomial of a over F . ∈ a ∈ Let b be a root of pa in F¯. Then there is an isomorphism F (a) F (b) mapping a to b whose restriction to F is the identity. Extend this to an embeddding→ σ : K F¯. By our assumption this embedding is an automorphism of K, so σ(a)= b belongs to K→. Thus every root of pa belongs to K. So K is the splitting field of the family pa(x): a K F . Thus (a) implies (b) and (c). { ∈ \ } Assume (b). Then verify that any embedding σ : K F¯ maps K to itself. Then 12.9 shows that σ is an automorphism. So (b) implies (a). → Assume (c). Let a K and p(x) is the irreducible polynomial of a over F . Let σ : K F¯ be an embedding fixing∈ F . Then σ(a) is another root of p(x), which also belongs to K→, so σ(K) K. Now, 12.9 implies σ is an automorphism of K. ⊆

33 13. finite fields 13.1. Definition. Let F be a field. Fix an algebraic closure F¯ of F . A polynomial f(x) F [x] is called separable if all the roots of f(x) in F¯ are distinct. ∈ 13.2. Lemma. A non-constant irreducible polynomial f(x) F [x] is separable if and only if f ′(x) =0. So any polynomial with coefficients in a characteristic∈ zero field is separable. Suppose6 char(F )= p. Let f(x) F [x] be a non-constant irreducible polynomial. Then f is not separable if and only if there∈ exists g(x) F [x] such that g(xp)= f(x). ∈ Proof. A polynomial f(x) has multiple roots if and only if f ′(x) and f(x) has a common factor in F¯[x], that is gcd(f, f ′) = 1. But since f is irreducible, gcd(f, f ′) = 1 if and only if f(x) divides f ′(x). But deg(f ′) <6 deg(f), so we must have f ′(x) = 0. In characteristic6 zero the derivative of any non-constant polynomial is non-zero. So every irreducible polynomial is separable. Now suppose F has characteristic p. If f(x) = g(xp), then g′(x) = pxp−1g′(xp) = 0. k p Conversely, let f(x)= akx be a polynomial which is not of the form g(x ). Then there Pk exists a k 0 not divisble by p such that ak = 0. Pick the largest k with this property. Then deg(f≥′)= k 1, so f ′ = 0. 6 − 6 13.3. Definition. If d a natural number, let Φ(d) be the set of integers x such that 1 x d and x is relatively prime to d. Let φ(d) be the number of elements in Φ(d). Observe≤ that≤ d.Φ(n/d) is exactly the set of elements of 1, 2, , n whose gcd with n is equal to d. So { ··· } 1, 2, , n = d.Φ(n/d). (6) { ··· } a d|n Counting both sides, we get n = φ(d). (7) X d|n 13.4. Lemma. Let H be a finite group of order n. Suppose, for each divisor d of n, the set x H : xd =1 has atmost d elements. Then H is cyclic. { ∈ } Proof. Let d be a divisor of n. Let Hd be the set of elements of H of order d. If a Hd, then (a)= 1, a, , ad−1 already gives d elements in H satisfying the equation xd =1.∈ So (a) must be{ the set··· of all} elements satisfying xd = 1, so H (a), so H = φ(d). It follows d ⊆ | d| that Hd =0 or Hd = φ(d). But Hd = n = φ(d). So we must have Hd = φ(d) | | | | Pd|n| | Pd|n | | for all d. In particular H = , so H has an element of order n. n 6 ∅ 13.5. Theorem. The multiplicative group of a finite field is cyclic. Proof. Let H be the multiplicative group of a finite field. Then H satisfies the hypothesis of 13.4 since a polynomial of degree d with coefficients in a field has atmost d roots. 13.6. Definition. Let F be a field of characteristic p. Then σ : F F , defined by σ(x)= xp, is a non-zero endmomorphism of F , since (x + y)p = xp + yp. So→σ is an isomorphism from F to some subfield σ(F ) of F . The map σ is called the Frobenius endomomorphism.

13.7. Existence and uniqueness of finite fields: Fix an algebraic closure F¯p of Fp. Let n 1. Observe that σ is an automorphism of F¯ since the equation xp = a has a solution ≥ p for every a F¯p. ∈ 34 pn ′ Let Fpn be the set of all roots of the polynomial f(x)= x x in F¯p. Since f (x)= 1 = 0, n n− − 6 the polynomial f has p disctinct roots in F¯p; so Fpn has p elements. Observe that n F n = x F¯ : σ (x)= x . p { ∈ p } n n Since σ is an automorphism of F¯p, it follows that Fpn is a field of order p . So Fpn it is n an extension of Fp of degree n. On the other hand, if F is any subfield of F¯p of order p , then 13.5 implies that the multiplicative group of F is a cyclic group of order pn−1. So every pn a F satisfies a = a, so F = F n . Thus: ∈ p n Fpn is the unique subfield of order p inside the algebraic closure F¯p.

Suppose F be any finite field. Then char(F ) = p is some rational prime, so F is a fin- tie extension of Fp. By 12.6, there is an embedding σ : F F¯p fixing Fp. The image σ(F ) n → is then a subfield of F¯p of order p , so σ(F )= Fpn .

n Every ﬁnite ﬁeld is isomorphic to Fpn for some prime power p .

35 14. Seperability 14.1. Definition. Let F be a field. Fix an algebraic closure F¯ of F . Let K/F be an extension. Let a K and let p(x) F [x] be a minimal polynomial of a. A root of p in F¯ is called a conjugate∈ of a. So a has∈ atmost d = deg(p) many conjugates. Say that a is an separable element (over F ) if a has d conjugates, or equivalently, if p(x) has d distinct roots, or equivalently, if p(x) is separable. Call an algebraic extension K/F separable if each a K is separable. ∈ 14.2. Lemma. Any algebraic extension of a field of characteristic zero or any algebraic extension of a finite field is separable. Proof. Let K be any algebraic extension of a field F of characteristic zero. Let a K. Then the minimal polynomial p(x) of a is irreducible, so 13.2 implies that a is separable.∈ So K/F is a separable. Let K be any algebraic extension of Fp. Let a K, then Fp(a) is a finite extension of Fp, ∈ pn so F (a) F n for some n 1. So a satisfies the polynomial f(x)= x x So the minimal p ≃ p ≥ − polynomial p(x) of a over Fp is a factor of f(x). Since f(x) does not have any repeated root in F¯p, the minimal polynomial p(x) does not have any repeated roots either. So a is separable an separable element. Thus K/Fp is separable. 14.3. Example. A standard example of nonseparable extension is F (t) F (t1/p). p ⊆ p 14.4. Lemma. Let K/F be an extension and a K. Then a is separable if and only if a satisfies a polynomial f(x) F [x] that has all distinct∈ roots, i.e. that splits into distinct linear factors in F¯[x]. ∈ Proof. Let p(x) F [x] be the minimal polynomial of a. If f(a) = 0, then p(x) divides f(x). So if f(x) has distinct∈ roots, then so does p(x).

14.5. Definition. Let K F be an extension with [K : F ]= n. Let [K : F ]s be the number of embeddings σ : K F¯⊇fixing F pointwise (i.e. σ restricted to F is identity). This number is called the separable→ degree of the extension K/F . 14.6. Theorem. Let K = F (a) be a finite simple extension of a. Let p(x) F [x] be the minimal polynomial of a. Then there is a bijection between the set of roots of∈p(x) and the set of embedding of K in F¯ fixing F . So the separable degree [K : F ]s is equal to the number of distinct conjugates of a in K. One has [K : F ] deg(p) = [K : F ]. The equality s ≤ [K : F ]s = [K : F ] holds if and only if a is a separable element.

Proof. Let a1, a2, , ar be the distinct conjugates of a, i.e. the roots of the minimal polynomial p. An embedding··· σ : K F¯ sends a to a conjugate of a since p(a) = 0 implies → i p(σ(a)) = 0. Conversely, verify that, for each conjugate ai one gets an embedding of K F¯ by sending a to a . So the separable degree of F (a)/F is equal to r deg(p) = [K : F ].→ i ≤ The equality [K : F ] = [K : F ]s holds if and only if p has deg(p) many distinct roots, in other words, all distinct roots; that is, a is separable. 14.7. Theorem. Let K/F be an finite extension. Let M be a field such that K M F . ⊇ ⊇ (a) Then one has [K : F ]s = [K : M]s[M : F ]s. (b) One has [K : F ]s [K : F ] and equality holds if and only if K/F is separable. (c) K/F is separable≤ if and only if K/M and M/F are separable. 36 Proof. (a) Fix an algebraic closure E M. Then E is also an algebraic closure of F ; we write. We have a map ⊇ res : Embeddings of K in E fixing F Embeddings of M in E fixing F M { }→{ } given by resM (λ) = λ M . Any embedding µ : M E can be extended to an embedding λ : K E (by 12.6) and| in [K : M] many different→ ways. So the map res is onto and the → s M preimage of every element has size [K : M]s. Part (a) follows. (b) The inequality [K : F ] [K : F ] is proved by induction on [K : F ]. Pick a K F s ≤ ∈ \ and let M = F (a). Theoerm 14.6 implies [M : F ]s [M : F ]. Since [K : M] < [K : F ], by induction we may assume [K : M] [K : M]. Using≤ part (a) we now get s ≤ [K : F ] = [K : M] [M : F ] [K : M][M : F ] = [K : F ]. s s s ≤ If K/F is separable, then [K : F ] = [K : F ]s follows by similar induction. Conversely, suppose [K : F ] = [K : F ] . We have to show that, for each a K, the s ∈ minimal polynomial p(x) of a has deg(p) many distinct roots. From [K : F ] = [K : F ]s, we get, [K : F (a)][F (a): F ] = [K : F (a)]s[F (a): F ]s. Since the separable degree is always less than or equal to the degree, one must have [F (a): F ]s = [F (a): F ]. Now 14.6 implies that a is separable. (c) Suppose K/F is separable. Since each x M is also an element of K the minimal polynomial of x must have distinct roots, so M/F∈is separable. Now let y K. The minimal polynomial of y over M is a factor of its minimal polynomial over F , so∈ must have distinct roots, whence K/M is separable too. Conversely, If K/M and M/F are separable, then [K : M]s = [M : F ] and [M : F ]s = [M : F ], so [L : K]s = [L : K] which, now implies L/K is separable. 14.8. Theorem (Primitive element theorem). If K/F be a finite separable extension. Then there is an γ K such that K = F (γ). ∈ Proof. If F is finite then K = F (γ) where γ is any generator for the multiplicative group of K. So assume that F is infinite. Let F (α, β) be a separable extension of degree n. Then there are n distinct embeddings σ1, , σn of F (α, β) in F¯. Consider the nonzero polynomial P (x)= (σ α σ α + x{(σ β···σ β)).} Since F is an infinite field, there exists c F such i6=j i − j i − j ∈ that P (cQ) = 0 which implies that γ =(α + cβ) has n distict images under the σ , i.e. γ has 6 i n distinct conjugates, so [F (γ): F ]s = n. So [F (γ): F ]= n and F (γ)= F (α, β). 14.9. Corollary. If K is an separable extension of F such that for every α K, we have [F (α): F ] n. Then [K : F ] n. ∈ ≤ ≤ Proof. Pick α so that [F (α): F ]= m is maximal. If β is K but not in F (α) then the extension F (α, β)/F has degree strictly larger than m and is simple by the primitive element theorem. This contradicts the maximality of m; so K = F (α).

37 15. Cyclotomic extensions

15.1. Roots of unity: Let ζn = exp(2πi/n). Let µn be the group of n-th roots of unity over Q. So µn is the multiplicative group generated by ζn; it is a cyclic group of order n. Note that µd µn if and only if d divides n. We say that ζ is a primitive n-th root of unity, ⊆ ∗ if ζ µn and ζ / µd for any d < n. Let µn denote the set of primitive n-th roots of unity. So ∈ ∈ ∗ r µn = ζn : 1 r n, gcd(r, n)=1 ∗ { ≤ ≤ } and µn = φ(n). Deﬁne the n-th cyclotomic polynomial Φn(x) to be the polynomial whose roots| are| the primitive n-th roots of unity:

Φn(x)= (x ζ). Y∗ − ζ∈µn Using (6), one obtains, n xn 1= (x ζr)= (x ζsr). − Y − n Y Y − n r=1 s|n r∈Φ(n/s) sr r Now (x ζ )= (x ζ ) = Φn/s(x). It follows that Qr∈Φ(n/s) − n Qr∈Φ(n/s) − n/s xn 1= Φ (x). (8) − Y d d|n

2 4 2 15.2. Example. Φ3(x)= x + x +1, Φ12(x)= x x + 1, 48 47 46 43 42 41 40 39 36 35− 34 33 32 31 28 26 24 22 20 Φ105(x) = x + x + x − x − x − 2x − x − x + x + x + x + x + x + x − x − x − x − x − x + x17 + x16 + x15 + x14 + x13 + x12 − x9 − x8 − 2x7 − x6 − x5 + x2 + x + 1. 15.3. Lemma. Let K/F be an extension. Suppose f(x), h(x) F [x] and f(x) divides h(x) in K[x]. Then f(x) divides h(x) in F [x]. ∈ Proof. By division algorithm, we can write h(x) = f(x)g(x)+ r(x) with g(x),r(x) F [x], deg(r) < deg(f). But the identity holds in K[x] as well. Since f(x) divides h(x) in∈ K[x], we must have r(x) = 0.

15.4. Lemma. The cyclotomic polynomial Φd(x) belongs to Z[x], is monic of degree φ(d).

Proof. By deﬁnition, the polynomials Φd(x) is monic of degree φ(d). By induction, assume Φd(x) Z[x] for all d < n. Let f(x)= d|n,d6=n Φd(x). Then f(x) is a monic polynomial in ∈ n Q n Z[x]. Since f(x) divides (x 1) in Q[ζn][x], lemma 15.3 implies that f(x) divides (x 1) in Q[x]. Since both f(x) and− (xn 1) are monic, Gauss’ lemma implies that f(x) divides− (xn 1) in Z[x]. So (8) implies that− Φ (x)=(xn 1)/f(x) Z[x]. − n − ∈ 15.5. Remark. Using equation (8), one can inductively calculate the cyclotomic polynomials n as Φn(x)=(x 1)/( Φd(x)). − Qd|n,d

Proof. If Φn(x) reducible in Q[x], then it is reducible in Z[x] and we can write Φn(x) = f(x)g(x) with f(x),g(x) Z[x] and f(x) irreducible. Since Φn(x) is monic, we may take f(x) and g(x) to be monic∈ as well. Let ζ be a primitive root of unity that is a root of f(x).

38 Claim: If p is a prime such that p ∤ n, then ζp is also a root of f(x).

Note that ζp is also a primitive n-th root of unity. So ζp is a root of f(x) or g(x). Suppose, p p p if possible, ζn is a root of g(x). Then ζn is a root of g(x ). So f(x) divides g(x ) and we can write g(xp) = f(x)h(x) for some monic polynomial h(x) Z[x]. Reducing the coeﬃcients modulo p we get ∈ p p g¯(x) =g ¯(x )= f¯(x)h¯(x), in Fp[x], where f¯(x) denotes the reduction of f(x) modulo p. This implies that f¯(x) andg ¯(x) have a common factor which implies Φ¯ n(x) has a repeated root in F¯p[x]. But since p does not divide n, (xn 1)′ = nxn−1 = 0, so (xn 1) has all distinct roots in F¯ and hence so does − 6 − p Φn(x). This contradiction proves the claim. By repeatedly applying the claim we ﬁnd that ζa is a root of f(x) whenever a is relatively prime to n. But every primitive n-th root of unity can be written in the form ζa for some a relatively prime to n. So every root of Φn(x) is a root of f(x), hence f(x) = Φn(x) is irreducible. 15.7. Corollary. One has Q(ζ ) Q[x]/(Φ (x)), so it is an extension of Q of degree φ(n). n ≃ n The conjugates of ζn are precisely the primitive roots of unity.

39 16. Galois theory 16.1. Definition. Say that K/F is a normal extension if K is a splitting field of a collection of polynomials in F [x], in other words, if every embedding of K in F¯ fixing F , is an automorphism of F . 16.2. Lemma. Let K/F be an algebraic extension. The following are equivalent: (a) K/F is normal. (b) every embedding of K in F¯ fixing F is an automorphism of K. (c) If a K, each conjugate of a also belong to K. (d) Every∈ a K satisfies a polynomial in F [x] that splits into linear factors in K[x]. ∈ Proof. The equivalence of (a) and (b) is contained in 12.10. An embedding of K F¯ takes each a K to some conjugate of a, hence the equivalence of (b) and (c). ∈ Let a∈ K and let p(x) F [x] be the minimal polynomial of a. Recall that the conjugates of a are∈ precisely the roots∈ of p(x). So if all the conjugates of a are in K, then p(x) splits into linear factor in K[x], hence (c) implies (d). Conversely assume (d). Let f(x) F [x] such that f(a) = 0 and f splits into linear factors in K[x]. Now p(x) f(x), so p(x∈) also splits into linear factors in K[x], i.e, all the roots of p(x) are in K. Thus| (d) implies (c). 16.3. Lemma. (a) A compositum of normal extensions is normal. (b) An Intersection of normal extensions is normal. (c) Suppose F M K be fields. If K/F is a normal, then so is K/M. ⊆ ⊆ Proof. Let K : i I be a collection of fields F K K such that K is the compositum { i ∈ } ⊆ i ⊆ of Ki : i I . Suppose each Ki/F is normal. Let σ : K F¯ be any embedding fixing F . Then{ σ ∈is} an embedding of K in F¯ so 12.10 implies that→σ is an automorphism of K , |Ki i |Ki i that is, σ(Ki)= Ki for each i. Since K is generated by the Ki’s it follows that σ(K)= K. This proves (a). Part (b) and (c) are easier exercises. 16.4. Definition. Let K/F be an extension. We let Aut(K/F ) be the set of field automorphisms of K fixing F pointwise. Clearly, Aut(K/F ) is a group. Let H be a subset of Aut(K/F ). Then we let KH denote the set of elements of K that are fixed by each element of H. One verifies that KH is a subfield of K containing F . We say that KH is the fixed field of H. 16.5. Lemma. Let K/F be an finite extension. The following are equivalent: (a) K/F is normal and separable. (b) Aut(K/F ) = [K : F ]s = [K : F ]. (c)| Each a K| satisfies a polynomial in F [x] that splits into distinct linear factors in K[x]. ∈ Proof. If τ is any embedding of K in F¯ fixing F , then τ σ : σ Aut(K/F ) gives Aut(K/F ) many distinct embeddings. Thus, one always has{ ◦ ∈ } | | Aut(K/F ) [K : F ] [K : F ]. | | ≤ s ≤ The equality [K : F ]s = [K : F ] holds if and only if K/F is separable (see 14.7) and the equality Aut(K/F ) = [K : F ]s holds if and only if K/F is normal (see 12.10). This proves the equivalence| of (a)| and (b). Equivalence of (a) and (c) follows from 16.2 and 14.4. 40 16.6. Definition. A finite extension K/F is called Galois, if it satisfies the equivalent conditions of 16.5. If K/F is a Galois extension, then the group Aut(K/F ) called the Galois group of K/F and is denoted by Gal(K/F ). 16.7. Lemma. Let K/F be a Galois extension. Let M be a field such that F M K. Then K/M is Galois and M/F is separable. ⊆ ⊆ Proof. Exercise. 16.8. Lemma. Let K/F be a Galois extension with Galois group G. Then KG = F . Proof. Let α KG and σ be any embedding of F (α) in F¯ fixing F . Then σ extends to an embedding σ ∈: K F¯. Any such embedding σ must be an automorphism of K fixing F , 1 → 1 so σ1 must fix α. In other words σ fixes α. Thus [F (α): F ]s = 1, so F (α)= F . 16.9. Lemma. Let K/F be a Galois extension with galois group G. Fix an algebraic closure F¯ of F containing K. Let a K. Then the orbit G.a of a under G is equal to the set of all ∈ the conjugates of a in F¯. Let G.a = a1, , ar and { ··· r } p(x)= (x a ). Y − j j=1 Then p(x) F [x] and is the minimal polynomial of a over F . ∈ Proof. If σ G, then σ(a) and a satisfies the same polynomials in F [x], so σ(a) is a conjuate of a. Conversely,∈ let b F¯ be a conjugate of a , then there exists an embedding σ′ : F (a) F¯ such that σ′(a)= b. Lemma∈ 12.9 implies that σ′ extends to an embedding σ : K F¯ fixing→ F . Since K/F is normal, σ G. So b = σ(a) G.a. → ∈ ∈ Each g G permutes the set G.a = a1, , ar . So g(p) = p, hence the coefficients of p are in K∈G. But KG = F by 16.8 So p{(x)··· F [x}] and p(a)=0. So p(x) is a multiple of ∈ the minimal polynomial of a. On the other hand each aj is a conjugate of a, so is a root of the minimal polynomial of a. Since aj’s are distinct, j(x aj) is a factor of the minimal polynomial of a. Q − 16.10. Theorem (Artin’s theorem). Let G is a finite group of automorphisms of K of order n and F = KG be the fixed field. Then K/F is Galois of degree n with Galois group G.

Proof. Let a in K. Let G.a = a1, , ar be the set of conjugates of a. By 16.9, p(x) = r (x a ) F [x] is the minimal{ ··· polynomial} of a. Since p(x) splits in K[x] into distinct j=1 − j ∈ Qlinear factors, 16.5 implies that K/F is Galois. One has [F (a): F ] = deg(p) n, so 14.9 implies that Gal(K/F ) = [K : F ] n. On the other hand, clearly, G Gal(K/F≤ ) which already has order| n. So G| = Gal(K/F≤ ). ⊆ 16.11. Theorem (Fundamental theorem of Galois theory). Let L/F be a finite Galois extension with Galois group G. Then there is an inclusion reversing bijection between the subgroups of G and the subfields of L that contain F , given by M Gal(L/M), where M is some subfield of L containing F. 7→ This is called the Galois correspondence. The inverse correspondence is given by by H LH . (b) Let H be a subgroup of G. Then L/LH is Galois with galois group H. 7→ (c) Let F M L. The extension M/F is normal (Galois) if and only if H = Gal(L/M) is normal in⊆G and⊆ in that case Gal(M/F )= G/H. 41 Proof. (a) Let M be some field between L and F . We already know L/M is Galois. If H = Gal(L/M), then M = LH by 16.8. Hence the injectivity of the corresopndence M Gal(L/M). (note that this part of the theorem holds even for infinite extensions). The7→ surjectivity of the correspondence as well as part (b) follows from Artin’s theorem. (c) Suppose M/F is normal. One verifies that there is a homomorphism res : G = Gal(L/F ) Gal(M/F ) given by res (σ)= σ M → M |M with kernel H = Gal(L/M). So H normal subgroup of G. Furthermore any automorphism of M fixing F extends to an embedding and hence an automorphism of L. So resM is onto and Gal(M/F )= G/H. Conversely, if M is not a normal extension then there is an embedding λ of M into L fixing F such that λM = M. Verify that Gal(L/λM)= λ Gal(L/M)λ−1. So Gal(L/λM) and Gal(L/M) are conjugates6 and belong to distinct subfields M and λM, so they are not equal, showing that Gal(L/M) is not normal. 16.12. Theorem (The normal basis theorem). If L/K is a finite Galois extension and then there is an element w in L such that its images under Gal(L/K) form a basis of L/K. −1 Proof. For a infinite field K look at the polynomial det(σi σj) as a polynomial function of the automorphisms σ1, , σn in Gal(L/K). This polynomial is nonzero, so find an w in L −1 ··· with det(σi σj(w)) = 0. Now a relation a1σ1(w)+ + anσn(w) = 0 with ai in K implies 6 −1 ···−1 n linear equations by applying σi to it. Since ((σi σj(w))) is invertible ai must all be zero. 16.13. Theorem (base change). Suppose K/F be a Galois extension and F ′/F is any extension. Then KF ′/F ′ is also Galois and Gal(KF ′/F ′) = Gal(K/K F ′). It follows that ∩ [KF ′ : F ] = [K : F ][F ′ : F ]/[K F ′ : F ]. ∩ 16.14. Theorem (Compositum and intersection of Galois is Galois). Suppose K1 and K2 are Galois extensions of F with Galois group G and G respectively. Then K K and 1 2 1 ∩ 2 K1K2 is a Galois extension of F . One has Gal(K K /F )= (σ, τ) G G : σ = τ . 1 2 { ∈ 1 × 2 |K1∩K2 |K1∩K2 } In particular, if K K = F , then Gal(K K /F ) G G . 1 ∩ 2 1 2 ≃ 1 × 2 16.15. Theorem (Existence of Galois closure). Let K/F be any separable extension. Fix an algebraic closure F¯ containing K. Then there is a smallest subfield L of F¯ such that L contains K and L/F is Galois. We say that L is the Galois closure of K. 16.16. Definition. Let G be a monoid and K be a field. A character of G in K is a monoid homomorphism from G to K∗. 16.17. Theorem (Linear independence of Characters). Let G be a monoid and K be a field. Let χ1, , χn be distinct characters of G in K. Then χ1, , χn are linearly independent in the space··· of K valued functions on G. ··· Proof. Suppose, if possible, there exists a , , a K∗ such that 1 ··· n ∈ a χ + + a χ =0 (9) 1 1 ··· n n We may assume that (9) is a dependence relation with n as small as possible. Then we must have n 2. Since χ = χ , there exists g G such that χ (g) = χ (g). Now ≥ 1 6 2 ∈ 1 6 2 a1χ1(gx)+ + anχn(gx) = 0 for all x G. ··· 42 ∈ Since χ is multiplicative, we get the linear dependence relation a χ (g)χ + + a χ (g)χ =0. (10) 1 1 1 ··· n n n By taking a linear combination of (9) and (10), we get a dependence relation among the χj’s with less than n terms, which contradicts minimality of n. 16.18. Definition. Let K/F be a Galois extension. Define the relative norm map N K : K F by N K (a)= σ(a) F → F Y σ∈Gal(K/F ) Clearly an element and its galois conjugates have the same norm and the norm map is K K K multiplicative, NF (ab) = NF (a)NF (b). An extension is called cyclic, if it is Galois with cyclic Galois group. 16.19. Theorem (Hilbert’s theorem 90). Let K/F be a cyclic extension. Let σ be a generator K ∗ of G and a K. Then N(a) = NF (a)=1 if and only if there exists b K such that a = b/σ(b). ∈ ∈ K Proof. Suppose Gal(K/F ) Z/nZ. Write N(a) = NF (a). If a = b/σ(b), then clearly ≃ 2 j−1 N(a) = 1. Conversely, suppose N(a) = 1. Let a0 = 1 and aj = aσ(a)σ (a) σ (a) for ··· ∗ j > 0. Note that aσ(aj)= aj+1 for all j and an = N(a) = 1. Consider f : G K defined by → n−1 f = a σj. X j j=0 By linear independence of characters, f is not the zero function. Choose x K such that f(x) = 0. Observe that aσ(f(x)) = f(x). ∈ 6 The next theorem describes cyclic extensions, when the base field contains enough roots of unity. 16.20. Theorem. Let F be a field and n Z. Assume that n is relatively prime to Char(F ) and F contains an n-th root of unity. ∈ (a) Let K/F be a cyclic Z/nZ extension. Then there exists α K such that K = F (α) and α satisfies the equation xn a =0 for some a K. ∈ (b) Conversely, suppose a F− and let α be a root∈ of xn a =0. Then F (α)/F is cyclic of degree d for some d n such∈ that αd F . − | ∈ Proof. Let ζ be a primitive n-th root of unity in F . (a) Assume Gal(K/F ) is a cyclic group of order n generated by σ. Since N(ζ−1) = 1, by Hilbert’s theorem 90, there exists α K such that σα = ζα. So σja = ζja for j = 0, 1, , n 1 are n distinct conjugates∈ of α in K, so [F (α): F ] n. It follows that F (α)=··· K.− So the minimal polynomial of α over F has degree n. Now≥a = αn is fixed by σ, hence a F . So α satisfies xn a = 0. (b) Conversely,∈ suppose a −F and let α be a root of xn a = 0. Then xn a has n distinct roots ζja: j =0, 1 ∈ , n 1 and they all belong to −F (α). So F (α) is the− splitting field of the separable{ polynomial··· x−n } a, hence F (α)/F is Galois. Let G = Gal(F (α)/F ). Let g G. Then g sends α to one of− its conjugates, so there exists n Z/nZ such that ∈ g ∈ gα = ζng α. One verifies that g n gives an injective homomorphism from G Z/nZ. 7→ g → 43 17. Integrality and the Nullstellensatz In this section R and S will denote rings with R S. ⊆ 17.1. Definition. Let R S be an extension of rings. An element a S is called integral over R if a satisfies a monic⊆ polynomial with coefficients in R. ∈ 17.2. Exercise: Let R S be rings. Let M be a S-module. If S is a finitely generated R-module and M is a finitely⊆ generated S-module, then M is a finitely generated R-module.

17.3. Lemma. Let R S be rings. Let a S. The following are equivalent: (a) a is integral over⊆R. ∈ (b) R[a] is a ﬁnitely generated R-module. (c) R[a] is contained in a subring M of S such that M is a ﬁnitely generated R-module.

Proof. (c) implies (a): Suppose m1, , mk generate M. Since R[a] is a subring of M, we have aM M, that is, M is a R[a]-module.··· So ⊆ am = r m for some r R. j X ij j ij ∈ ij Consider G = ((aδ r )) M (R[a]), and m = (m , , m )tr M n. Note that M n is a ij − ij ∈ k 1 ··· n ∈ module over Mk(R[a]). The equations above combines to give Gm = 0. Multiplying by the adjoint of G, we get, det(G)Ikm = 0, so det(G)mj = 0 for all j. Hence det(G) = 0. Writing k k−1 out the formula for determinant, we get an equation of the form a + t1a + + tk = 0, with t R. ··· j ∈ 17.4. Corollary. Let x1, , xn be elements of S, each integral over R. Then R[x1, , xn] is a finitely generated R-module.··· In other words, if S is a finitely generated R-algebra··· and is an integral extension of R, then S is a R-module of finite rank. Proof. Let A = R[x , , x ]. Induct on n. The case n = 1 follows from 17.3. By induction r 1 ··· r on n, we may assume that An−1 is finitely generated R-module. Since xn is integral over An−1, 17.3 implies that An = An−1[xn] is finitely generated An−1-module. using 17.2, it follows that An is finitely generated R-module. 17.5. Lemma. The set of elements of S that are integral over R forms a subring of S. Proof. Suppose a, b S are integral over S. Then 17.4 implies that R[a, b] is finitely generated R-module. If c∈= ab or c = a + b, then Since R[c] R[a, b]. So 17.3 implies that ab and a + b is integral over R too. ⊆ 17.6. Definition. Let A be the set of elements of S that are integral over R. If A = R, then we say R is integrally closed in S. If A = S, then we say S is an integral extension of R, or that S is integral over R. 17.7. Lemma. Suppose R S A be rings. If S is integral over R and A is integral over S, then A is integral over R⊆. ⊆ Proof. Let a A. Since A is integral over S, there exists an identity ∈ n n−1 a + s1a + + sn =0 44 ··· with s in S. Let S = R[s , ,s ]. Then a satisfies a monic polynomial with coefficients j 1 1 ··· n in S1, so S1[a] is a finitely generated S1 module. Since sj are integral over R, 17.4 implies that S1 is a finitely generated R-module. So 17.2 implies that S1[a] is a finitely generated R-module. Now 17.3 tells us that a is integral over R. 17.8. Lemma. Let R be a UFD and let S be the fraction field of R. Then R is integrally closed in S. In particular, Z is integrally closed in Q. ◦ For any field k, the polynomial ring K[x1, , xn] is integrally closed in the quotient ◦ field of rational functions K(x , , x ). ··· 1 ··· n Proof. Suppose c S is integral over R. Write c = a/b with a, b R such that they have ∈ n n−1 ∈ no common factor. Suppose c satisfies the equation x + r1x + + rn = 0, with rj R. n n−1 n n ··· ∈ Then a + r1a b + + rnb = 0, so b divides a , since a and b has no common factor, it follows that b must be··· a unit in R, so a/b R. ∈ 17.9. Definition. Let R S be a ring extension and a1, , an S. We say that a1, , an are algebraically independent⊆ over R if for any nonzro polynomial··· ∈ f R[x , , x ] one··· has ∈ 1 ··· n f(a1, , an) = 0. In other words, the map R[x1, , xn] R[a1, , an] given by xj aj is an··· isomorphism.6 ··· → ··· 7→ 17.10. Lemma. Let f =0 be polynomial in k[x]. Then k[x, f −1] is integrally closed in k(x). 6

Proof. Suppose g k(x) be integral over k[x, f −1]. Then we have an equation of the form ∈ gn + f −r1 a gn−1 + f −r2 a gn−2 + =0, 1 2 ··· nr where ai k[x]. Let r = max r1,r2, . Multiplying the identity by f clears the denominators∈ and we get an equation{ of···} integral dependence of (f rg) over k[x]. Since k[x] is integrally closed in the fraction field k(x), we have f rg k[x], So g k[x, f −1]. ∈ ∈ 17.11. Theorem. Let k be a field and S be a finitely generated k algebra. Suppose S is a field. Then S is a finite extension of k. Proof. Let x , , x generate S as a k-algebra. Induct on n. The case n = 1, is clear. 1 ··· n So assume n > 1. Let R = k[x1]. By induction, S is a finite extension of k(x1). So x , , x satisfies some monic polynomials with coefficients in k(x ). Let a /b be all these 2 ··· n 1 j j coefficients, aj, bj k[x1]. Let f be the product of all the bj’s. Then x2, , xn are integral −1 ∈ −1 ··· −1 over k[x1, f ]. So S is integral over k[x1, f ], so k(x1) is integral over k[x1, f ]. If x1 is not algbraic over k, then k[x1] is isomorphic to the polynomial ring and 17.10 implies that −1 −1 k[x1, f ] is integrally closed in k(x1). This forces k[x1, f ] = k(x1), which is impossible (why?). So x1 must be algebraic over k. Thus k(x1) is a finite extension of k. So S is a finite extension of k. 17.12. Theorem (weak Nullstellensatz). Let k be an algebraically closed field and R = k[x1, , xn]. Then every maximal ideal in R is the form (x1 a1, , xn an) for some (a , ···, a ) kn. − ··· − 1 ··· n ∈ Proof. Given a =(a , , a ) kn one has a surjection ev : R k given by ev (f)= f(a). 1 ··· n ∈ a → a One verifies that ker(eva)=(x1 a1, , xn an). So these are maximal ideals in R. Let m − ··· −45 be any maximal ideal in R. Then the composition k ֒ R R/m is a non-zero map since 1 / m, hence is an injection. So R/m is a finitely generated→ →k-algebra and a field, So 17.11 implies∈ R/m is a finite extension of k. Since k is algebraically closed, we get k R/m. Let ≃ e : R K be a surjection with kernel m and let e(xj)= aj. Then (xj aj) m. It follows that m→=(x a , , x a ). − ∈ 1 − 1 ··· n − n 17.13. Definition. Given S k[x , , x] define ⊆ 1 ··· (S)= a kn : f(a) = 0 for all f I . Z { ∈ ∈ } Subsets of this form in kn are called algebraic sets. Given A kn, define ⊆ (A)= f k[x , , x ]: f(a) = 0 for all a A . I { ∈ 1 ··· n ∈ } Say that an ideal I k[x , ,k ] is a radical ideal if I = Rad(I). ∈ 1 ··· n 17.14. Theorem (strong Nullstellensatz). Let k be an algebraically closed field and and I be an ideal in k[x , , x ]. Then (I) = Rad(I). 1 ··· n IZ Proof. Suppose f1, , fm generate I. Suppose f (I). Then f1, , fm, (1 x0f) k[x , , x ] do not··· have any common zeros, so there∈ IZ exists g k[x , ···, x ] such− that ∈ 0 ··· n j ∈ 0 ··· n g (1 x f)+ g f + + g f =1. 0 − 0 1 1 ··· n n Now substitute x = f −1 and clear denominators to see that f M I for some M N. 0 ∈ ∈ 17.15. Corollary (Ideal-Variety correspondence). Let k be an algebraically closed field. Then: (a) Every algebraic set in kn has the form (J) for some radical ideal J. (b) If A is an algebraic set in kn, then (AZ) Is a radical ideal. I (c) There is a inclusion reversing bijection between the radical ideals in K[x1, ,kn] and the algebraic subsets of kn, given by The maps I (I) and A (A). ··· 7→ Z 7→ I Proof. (a) Let S k[x1, , xn]. Let I be the ideal generated by S and J = Rad(I). Then one verifies that ⊆(S)=···(I)= (J). Note that J is a radical ideal (why?). (b) Let A be anyZ algebraicZ setZ in kn. Write A = (J) for some radical ideal J. Then using Nullstellensatz, we get (A)= (J) = Rad(J)=Z J. (c) Now suppose J be a radicalI ideal.IZ Then (J) = Rad(J)= J by Nullstellensatz. On the other hand, if A is an algebraic set, then writeIZ A = (J) for some radical ideal. We get (A)= (J)= (J)= A. Z ZI ZIZ Z

46 18. Some applictions Symmetric polynomials 18.1. Definition. Let R be a commutative ring and A = R[x1, , xn]. The permutation σ ··· group Sn acts on the right on A by (f )(x1, , xn) = f(xσ(1), , xσ(n)). A polynomial σ··· ··· Sn f A is called a symmetric polynomial if f = f for all σ Sn. Let A be the ring of ∈ n ∈ symmetric polynomials. Note that the polynomial i=1(t xi) A[t] is fixed by Sn, so its Sn Sn Q − ∈ coefficients belong to A . Define s1, ,sn A by Define n ··· ∈ (t x )= tn s tn−1 + s tn−2 +( 1)ns (11) Y − i − 1 2 −··· − n i=1 So n s = x , s = x x , , s = x x x . 1 X i 2 X i j ··· n 1 2 ··· n i=1 i

18.3. Theorem. Let F be a field. The permutation group Sn acts on L = F (x1, , xn) by Sn ··· permuting the variables. Let M = K(x1, , xn) be the fixed field. The extension L/M is Galois with Galois group S . One has M···= F (s , ,s ). n 1 ··· n Proof. By 16.10, the extension L/M is Galois with Galois group Sn and x1, , xn are the Galois conjugates of x . Clealy M = F (s , ,s ) M. Let p(x)= n (t ···x ). Equation 1 1 1 ··· n ⊆ j−1 − j (11) implies that p(t) M [t] and x ’s are the roots of this equation.Q So L is the splitting ∈ 1 j field of p(t) over M1. Since p(t) has degree n, we have [L : M1] n!. But M1 M and we already have [L : M]= S = n!. So M = M. ≤ ⊆ | n| 1 18.4. Theorem (Fundamental theorem of symmetric polynomials). Let K be a field. A symmetric polynomial in n-variables over K is a polynomial in the elementary symmetric Sn polynomials, that is, K[x1, , xn] = k[s1, ,sn]. ··· ···47 Proof. Let f K[x , , x ]Sn be a symmetric polynomial. Let R = K[s , ,s ] and ∈ 1 ··· n 1 ··· n S = K[x1, , xn]. Note that (11) tells us that xj’s satisfy monic polynomial with coefficients in S. So by··· 17.5, the extension R S is an integral extension. In particular f is integral over ⊆ R. On the other hand, from 18.3, we know that f K[x , , x ]Sn K(x , , x )Sn = ∈ 1 ··· n ⊆ 1 ··· n K(s1, ,sn). But 18.2 shows that K[s1, ,sn] is isomorphic to the polynomial ring, so 17.8 tells··· us that K[s , ,s ] is integrally··· closed in K(s , ,s ), so f K[s , ,s ]. 1 ··· n 1 ··· n ∈ 1 ··· n Cubic oplynomials n 18.5. Definition. Let K be a field. Let R = K[x1, , xn]. Let f(x)= j=1(x xj ) R[x]. Define ··· Q − ∈ δ(f)= (x x ) and ∆ = δ(f)2. Y j − k f j

Observe that ∆f is a symmetric polynomial, so ∆f is a polynomial in the elementary symmetric functions in x1, , xn. Now suppose f(x)=···xn s xn−1 +s xn−2 +( 1)ns K[x] be a polynomial. Suppose − 1 2 −··· − n ∈ f(x) splits as f(x) = (x xj) over the algebraic closure of K. Then we define δ(f) and Qj − ∆f as above. Since s1, ,sn are the elementary symmetric polynomials in x1, , xn, it follows that ∆ is a polynomial··· in s , ,s , so disc(f) K. The polynomial f···(x) has a f 1 ··· n ∈ repeated root if and only if ∆f = 0. 18.6. Theorem. Assume char(K) = 2. Let f(x) = x2 + px + q K[x], so ∆ = p2 4q. 6 ∈ f − Then f(x) is irreducible in K if and only if ∆ is not a square in K and in that case K[ ∆ ] f p f is the splitting field of f. The generator for the Galois group Gal(K[ ∆ ]/K) Z/2Z takes p f ≃ ∆ to ∆ . p f −p f 18.7. Cubic polynomials: Let K be a field. Assume char(K) is not 2 or 3. Consider a cubic polynomial y3 + αy2 + βy + γ with coefficients in K. On substituting y =(x α/3), the polynomial takes the form − f(x)= x3 + px + q (12) with p, q K. So we can restrict ourselves to studying cubic polynomials of this form. If f is reducible,∈ then it has a root in K and solving f amounts to solving a quadratic equation. So we shall assume that f is irreducible in K[x], in particular q = 0. Since char(K) = 3, we ′ 6 6 have f (x) = 0, so f is a separable polynomial. Let a = a1, a2, a3 be the three roots of f. So f(x) splits6 in K¯ as f(x)=(x a )(x a )(x a ) and − 1 − 2 − 3 a + a + a =0, a a + a a + a a = p, a a a = q. (13) 1 2 3 1 2 1 3 2 3 1 2 3 − 18.8. Theorem. One has ∆ =(a a )2(a a )2(a a )2 = 4p3 27q2. f 1 − 2 2 − 3 1 − 3 − − Proof. By the fundamental theorem of symmetric polynomials, ∆f is a polynomial in p and q i j (since a1 + a2 + a3 = 0). So we have an identity of the form ∆f = aijp q . Note that the Pi,j left hand side if homogeneous polynomial of degree 6 in a1, a2, a3. Since p is homogeneous of degree 2 and q is homogeneous in degree 3, the monomial piqj is homogeneous of degree i j (2i +3j) in a1, a2, a3. p q is homogeneous of degree 6 in aj’s if and only if (2i +3j) = 6. 3 2 3 2 So only such monomials are p and q . So we must have ∆f = a3,0p + a0,2q (why?). The constants a3,0 and a0,2 can now be found by considering two specific cubic polynomials (say f(x)= x3 1 and f(x)= x3 x). − − 48 18.9. Lemma. Let a = a1 be a root of f(x)=0. Then the other two roots can be expressed as ( a + f ′(a)−1 ∆ )/2 where ∆ is any square root of ∆ in the splitting field of f. − p f p f f Proof. Let f(x)=(x a)g(x) be the factorization of f in K[a][x]. Then − g(x)=(x a )(x a )= x2 (a + a )x + a a = x2 + ax qa−1. − 2 − 3 − 2 3 2 3 − Note that ∆ =(a a )2(a a )2(a a )2 = g(a )2∆ . f 1 − 2 1 − 3 2 − 3 1 h −1 The two roots of g(x) are ( a ∆g)/2=( a g(a) ∆f )/2. Finally, note that g(a)= f ′(a). − ± p − ± p 18.10. Theorem. Let L be the splitting field of f and let a be a root of f in K[x]. Then L = K[a, ∆f ]. There are two possibilities for the Galois group: (a) If ∆p is a square in K, then L = K[a] and Gal(L/K) Z/3Z. f ≃ (b) If ∆ is not a square in K, then L = K[a, ∆ ] and Gal(L/K) S . f p f ≃ 3 Proof. Since ∆ has a square root δ =(a a )(a a )(a a ) in L, we have K[a, ∆ ] f f 1 − 2 1 − 3 2 − 3 p f ⊆ L. On the other hand 18.9 implies that K[a, ∆ ] contains all the roots of f. So L = p f K[a, ∆f ]. Let G = Gal(L/K). Since f is irreducible and separable, it has three distinct roots.p Using the action of G on the three roots of f, we identify G S . Since f(x) is ⊆ 3 irreducible [K[a1]: K] = 3. If ∆f is a square in K, then 18.9 implies that L = K[a1], So G has order 3, hence G Z/3Z. Now, suppose ∆ is not a square in K. Since δ L is a square root of ∆ , ≃ f f ∈ f we must have [K[δf ]: K] = 2. So L/K has a subextension K[δf ]/K of degree 2 and a subextension K[a1]/K of degree 3, so [L : K] = 6 and G = S3. 18.11. Elementary method for solving a cubic by radicals: One solves a quadratic equation x2+αx+β = 0 by using a linear substitution of the form y = x+u and then choosing u so as to complete square, that is, transforming the equation to the form y2 + C = 0. The most direct generalization that works for a cubic x3 + αx2 + βx + γ = 0 is to use a quadratic substitution y = x2 + ux + v. One can work out the cubic polynomial satisfied by y and then choose u, v so as to complete a cube, so as to transform the equation to the form y3 + C = 0. One can make the computation more managable, by using a variation as follows. First use the linear substitution to make the quadratic term vanish. Then, substituting x =(u p/3u) 3 3 p3 − in f(x) = x + px + q = 0, one gets, after some simplification u 27u3 + q = 0, or, 6 3 p3 3 − 3 u + qu 27 = 0. This is a quadratic equation in u , so we can solve for u and then take a cube root− to find an expression for a root in terms of radicals. 18.12. Solving a cubic by radicals (Lagranges method): Assume that the base field K contains a cube root of unity ω. Let σ be the 3-cycle that takes (a1, a2, a3) to (a3, a1, a2). The proof of Hilbert theorem 90 suggests considering the element 2 y = a1 + ωa2 + ω a3. Then σy = ωy. So y3 belongs to LA3 = K[ ∆ ]. The cyclic Z/3Z extension L/K[ ∆ ] is p f p f obtained by adding a cube root of y3. We find an formula for y3 K[ ∆ ]. One has ∈ p f 3 3 3 3 2 2 2 2 2 2 2 y =(a1 + a2 + a3)+3ω(a1a2 + a2a3 + a3a1)+3ω (a1a2 + a2a3 + a3a1)+6a1a2a3. 49 3 2 2 2 Note that j aj =3a1a2a3 = 3q since j aj = 0. Let u+ =(a1a2 +a2a3 +a3a1) and u− = 2 P2 2 − P (a1a2 + a2a3 + a3a1). Then y3 = 9q +3ωu +3ω2u . − + − Note that u and v are solutions of the quadratic equation g(x)= x2 (u + u )x + u u , − + − + − whose coefficients are fixed by all permutations of a1, a2, a3, so g(x) K[x]. One verifies 3q ∈ that (u + u )=3q and ∆ =(u u )2 = ∆ . So u = 1 ∆ . It follows that + − g + − − f ± 2 ± 2 p f y3 =(a + a ω + a ω2)3 = 27 q3 + 3 (ω ω2) ∆ . 1 2 3 − 2 2 − p f The transposition τ interchaning a and a is not in A , so it does not fix K[ ∆ ], so 2 3 3 p f τ( ∆ )= ∆ . Let y = y and y = τ(y )= a + a ω2 + a ω, then p f −p f + − + 1 2 3 y = 27 q3 3 (ω ω2) ∆ 1/3 ± − 2 ± 2 − p f Now we can get a1, a2, a3 by solving the system of linear equations 2 2 a1 + a2 + a3 =0, a1 + a2ω + a3ω = y+, a1 + a2ω + a3ω = y−. 18.13. Quartic equation: Consider a separable quartic polynomial in k[x]. By a linear substitution of the variable any quartic polynomial in one variable can be brought to the 4 2 4 form f(x)= x + px + qx + r. Let L be the splitting field of f. Let f(x)= i=1(x ai) be the factorzation of f in L[x]. Consider the elements Q −

θ1 =(a1 + a2)(a3 + a4), θ2 =(a1 + a3)(a2 + a4), θ3 =(a1 + a4)(a2 + a3).

Let S4 be the group of permutation of the four roots. Each such permutation permutes the θi’s, so the coefficients of g(x)=(x θ1)(x θ2)(x θ3) are fixed by all permutations in the roots, so g(x) k[x]. Elementary computation− − with− symmetric functions show that ∈ g(x)= x3 2px +(p2 4r)x + q2. − − The polymomial g(x) is called the resolvent cubic of the quartic polynomial f(x). One can solve the cubic g(x) to express θj’s in terms of radicals in p,q,r. Let M = k[θ1, θ2, θ3]. Then One verifies that M LH where H = 1, (12)(34), (13)(24), (14)(23) is a subgroup 2 ⊆ { } isomorphic to (Z/2Z) . Since i ai = 0, we see that the elements (ai + aj) are the solutions 2P of the quadratic equations x θk = 0. It follows that the extension L/M is atmost a biquadratic extension, so it has− degree atmost 4. So LH = M. We get that for each i = j, 6 there is some k such that (ai + aj)= √θk. The actual roots can be found by enumerating these finitely many possibilities. ±