Quick viewing(Text Mode)

07 Define Ring, Subrings, Modules, and Submodules. Example 1. Rings

07 Define Ring, Subrings, Modules, and Submodules. Example 1. Rings

RINGS AND FIELDS ‘07

RINGS AND FIELDS ‘07

Define , , modules, and submodules. Example 1. Rings: (a) We have the well-known rings; Z, Q, R, and C. (b) Inside C we find rings as follows; select an α that is n n−1 n−2 the root to some x +an−1x +an−2x +···+a0 0 where the ajs are , and R = Z[α] = {t ∈ C | t = n−1 n−2 bn−1α + bn−2α + ··· + b0, where each bi ∈ Z}. (c) Given any ring R, we can construct a ring from the collection of n × n matrices whose entries lie in R;

S = Matn(R), where and are the standard op- erations. (d) The upper, 2×2 matrices over a R with equal diagonal entries forms a commutative of Mat2(R). Example 2. Modules Over a Ring R:

(a) For an index set I, ΠI R and ⊕I R are modules over R. (b) An R- F is called free on a X if for every f : X → A such that f(rx + sy) = rf(x) + sf(y) for x, y ∈ X and r, s ∈ R, where A is an R-module, there is a unique h : M → A such that h|X = f. (c) An R-module M is called projective if, for every f : A → B and homomorphism g : M → B, there is a homo- ∼ Theorem 1. An R-module M is free if and only if M = ⊕I R. Theorem 2. An R-module M is projective if and only if M ⊕ N = F for some F .

Let us use the projective and free modules as a foundation for our course. 1 2 RINGS AND FIELDS ‘07

Theorem 3. Every R-module is free if and only if R is a ring.

When is it true that every R-module is projective? Artin-Wedderburn Theorem . Every R-module is projective if and only if R is a finite of matrix rings over division rings; i.e.,

R = Matn1 (D1) × Matn2 (D2) × · · · × Matnk (Dk), for some natural n1, n2, . . . , nk and division rings D1,D2,...,Dk. Example 3. Let H denote the (rational) Hamiltonian ; H = {a + bi + cj + dk | a, b, c, d ∈ Q}, where addition is component-wise and multiplication extends linearly 2 2 2 from i = j = k = −1. Then, the left of R = Mat2(H) consist- ing of the matrices with second column zero is a projective (left) ideal of R that is not free. √ √ √Example 4. Let R = Z[ −5] = {a + b −5 | a, b ∈ Z and I = (3, 1 + −5). We claim that I is projective but not free. First we will argue that I 6= R. Notice 3R√⊆ I. Modulo√ 3R, the general element√ of I can be represented as (1+ −5)·√(a+b −5) = (a−5b)+(a+b) −5, which is congruent to (a + b)[1 + −5] mod 3R. Therefore, I/3R ∼= Z/3Z, and since R/3R has 9, I is properly contained in R√. If I were free,√ then I would be principal; say I = (a + b −5). Since I contains 3, 1+ −5, by norms, a2 +5b2 must divide 9 and 6. Hence a2 + 5b2 = 1 or 3. We rule out 1 since I 6= R. However, a2 + 5b2 = 3 is not possible. Therefore, I is not free. Set 2 √ α = (1 + −5), and β = −1. 3 Note that α · I ⊆ R and β · I ⊆ R. Define√ the homomorphism θ : F = R ⊕ R → I by θ(r, s) = r3 + s(1 + −5). The epimorphism θ is split by the map δ : I → F where δ(x) = (βx, αx). (Check: √ 2 θ(δ(x)) = θ((βx, αx)) = 3βx + (1 + −5)αx = x(−3 + 3 6) = x. Therefore, I is projective. √ Example 5. The ring R = Z[ −5] of the last example has the property that every nonzero nonunit can be expressed as a product of irreducible elements (see homework), yet R is not a UFD (see the next example). RINGS AND FIELDS ‘07 3

Example 6. Let d be a -free 6= 1, and define √ R = Z[ d] provided d ≡ 2, 3 (mod 4), or √ 1 + d R = Z[ ] when d ≡ 1(mod 4). 2 Then every ideal of R is projective. It is not known whether or not there are infinitely many d0s for which R is a PID. However, for such an R, R is a PID if and only if R is a UFD. 4 RINGS AND FIELDS ‘07

1. EXERCISE SET 1 In this homework set, R is an integral . A nonzero, nonunit r of R is said to be irreducible provided r = st for s, t ∈ R can only happen when one of s or t is a . (Recall, u ∈ R is a unit if uv = 1 for some v ∈ R). A nonzero, nonunit r of R is said to be prime if r | ab for a, b ∈ R, implies r | a or r | b. R is said to have the ascending chain condition on principal ideals, if any increasing chain of ideals

(a1) ⊆ (a2) ⊆ (a3) ⊆ · · · stablizes (i.e., there is an index n such that (am) = (an) for all m ≥ n). 1. Show that if R has the ascending chain condition on principal ideals, then every nonzero nonunit r of R is a product of ir- reducible elements. Hint: To argue that r has an irreducible factor, either r is irreducible, or r = r1s1 with r1, s1 nonzero nonunits. Note (r) is properly contained in (r1) in this case. Re- peat for r1 and use the hypothesis to eventually obtain r = s1t1 with s1 irreducible. Now write r = r1v1 = r1r2v2 = ... with r1, r2,... irreducible, and use the hypothesis to show that even- tually some vj is a unit. 2. Show that if R is a UFD then every irreducible element is prime and R has the ascending chain condition on principal ideals. 3. Show that if irreducible elements are prime in R and R has the ascending chain condition on principal ideals, then R is a UFD. 4. Show that R is a UFD if and only if every nonzero nonunit is a product of primes. 5. Show any PID is a UFD. RINGS AND FIELDS ‘07 5

2. Integral Domains

Theorem 4. Let R be an . Every submodule of ⊕nR for any n is free if and only if R is a PID. Proof. An ideal of R is a submodule of R. Hence if the condition is in force, R is necessarily a PID. Conversely, assume that R is a PID. We will show, by induction on n, that every submodule of ⊕nR is free. When n = 1, the nonzero submodules of R are ideals and are principal (hence isomorphic to R). Inductively, let L be a nonzero submodule of F = ⊕nR. For πj : F → th R equal to the projection map onto the j coordinate, I = πj(L) 6= 0 for some j. Since I = aR for some 0 6= a ∈ R, we can define a splitting map f : aR → L by f(ar) = rx where x ∈ L is any element such that πj(x) = a. By the standard argument, ∼ L = Im f ⊕ Ker πj|L = aR ⊕ Ker πj|L. th Note that Ker πj|L = {y ∈ L | y has a zero j component} ≤ ⊕i6=jR, where the index i runs over 1, 2, . . . , j − 1, j + 1, . . . , n. By induction, ∼ ∼ Ker πj|L = ⊕mR is free, and therefore, L = ⊕m+1R. ¤ Given an ideal I 6= 0 of R, define I−1 = {t ∈ Q | tIsubseteqR}. The ideal I is said to be invertible, if I · I−1 = R. Analogously, we have

Theorem 5. Let R be an integral domain. Every submodule of ⊕nR for any n is projective if and only if every ideal of R is invertible (pro- jective). In order to see this we need a few observations. Bear Injective Test Lemma . The module U is injective if and only if U is injective relative to any 0 → I → R where I is an ideal of R. Proof. Assume that U is injective with respect to any sequence 0 → I → R, and suppose that 0 → A → B is exact and f : A → U. Consider the set C of all 2-tuples (g, C) where A ≤ C ≤ B and g|A = f. There is a maximal element (g0,C0). If C0 6= B let x ∈ B \ C0, and let I = {r ∈ R | rx ∈ C0}. Set 0 0 C = C0 + hxi. If I = 0, then hxi ∩ C0 = 0 and so C = C0 ⊕ hxi ≤ B; 0 0 0 we then have the over-module C of C0 and can define g : C → U by 6 RINGS AND FIELDS ‘07

0 g (c0, rx) = g(c0), contradicting the maximality of (g, C0). Therefore I 6= 0. By hypothesis, the map h : I → U given by h(r) = g(rx) can be lifted to a map h0 : R → U. We then define g0 : C0 → U by 0 0 g (c0 + rx) = g(c0) + h (r). If c0 + rx = b0 + sx with c0, b0 ∈ C0 and r, s ∈ R, then (r − s)x = b0 − c0 ∈ C0 and so r − s ∈ I. But then 0 0 0 g(b0 − c0) = g((r − s)x) = h(r − s) = h (r − s) = h (r) − h (s) and 0 0 0 0 so g(b0) − g(c0) = h (r) − h (s) implying g(b0) + h (s) = g(c0) + h (r) making g0 well-defined. It is easy to check that g0 is a homomorphism that extends g which contradicts the maximality of (g, C0). Thus, B = C0 as desired. ¤ Corollary 6. The quotient field Q of R is injective. Proof. Let I be an ideal of R and 0 6= a ∈ I. If 0 6= f : I → Q, define h : R → Q to be multiplication by (1/a)f(a). If b ∈ I, then h(b) = (1/a)f(a) · b = (1/a)f(ab) = (1/a)af(b) = f(b). By Baer’s Injective Test Lemma, Q is injective. ¤ Corollary 7. If I,J are ideals of R, then Hom(I,J) is naturally iden- tifiable with {t ∈ Q | tI ⊆ J}. Proof. It is well-known that Hom(I,Q) can be identified with Q via multiplication the elements of Q, and that Hom(I,J) is naturally a submodule of Hom(I,Q). The assertion follows. ¤ Lemma 8. Let I 6= 0 be an ideal of R. Then I is invertible if and only if I is projective. −1 Proof. Suppose II + R. Then 1 = r1s1 + r2s2 + ... + rnsn with −1 −1 rj ∈ I and sj ∈ I for each j by the definition of II . The map θ : F = ⊕nR → I given by θ(x1, x2, . . . , xn) = Σjrjxj is split by the map f : I → F given by f(a) = (as1, as2, . . . , asn) ∈ F . Therefore, I ⊕ Ker θ ∼= F and I is projective. Conversely, if I is projective, there is an epimorphism θ : F → I that is split by some map f : I → F . Let θ(ei) = xi ∈ I where ei is the standard element of F having a 1 in the ith-component and 0 0 s elsewhere. Write f = (f1, f2, . . . , fn) where fj : I → R. By the last −1 corollary, regard fj ∈ Hom(I,R) = {t ∈ Q | tI ⊆ R} = I . It follows that Σjxjfja = a for every a ∈ I, and consequently that Σjxjfj = 1. That is, I · I−1 = R. ¤ The proof of the last theorem now follows easily from the proof at the beginning this section in light of the last lemma. RINGS AND FIELDS ‘07 7

3. Dedekind Domains A domain such that every nonzero ideal is invertible is called a . An R-submodule J of Q is called a of R if rJ ⊆ R for some 0 6= r ∈ R. The domain R is said to be integrally closed if the only over-ring of R inside Q that is fractional over R is R itself. An ideal P of R is said to be prime, if ab ∈ P implies a ∈ P or b ∈ P for any a, b ∈ R. Equivalently, P is prime if R/P is an integral domain. Recall that a UFD is a domain such that every nonzero nonunit is a product of primes. Theorem 9. TFAE: (a) R is Dedekind. 1 (b) Every nonzero ideal is 1 2 -generated (i.e., can be generated by two nonzero elements with the first generator chosen arbitrar- ily). (c) Every proper ideal of R is (uniquely) a product of prime ideals of R. (d) R is integrally closed, each nonzero is a , and every ideal of R is finitely generated. (e) R is and for every prime ideal P of R, RP is is a PID. 8 RINGS AND FIELDS ‘07

4. A Fundamental Theorem A module is free if it is isomorphic to a of copies of R. The of copies of R is called the of the free module. Note ∼ ⊕nR = ⊕mR if and only if n = m. (One way to justify this is to note that ⊕nR has at most n linearly independent elements). If M is a module over an integral domain, the submodule T of M is T = {x ∈ M | rx = 0 for some 0 6= r ∈ R}. A module whose torsion submodule is zero is said to be torsion-free. For example, the torsion submodule of M/T is zero, and so M/T is torsion-free. We can rephrase Theorem 4. Proposition 10. If R is an integral domain, then any finitely generated torsion-free module is isomorphic of a submodule of ⊕nR for some n.

Proof. Let C be finitely generated with generating set {x1, x2, . . . , xn} of nonzero elements. Select x1; if x2 is independent of x1, select x2; otherwise reject x2. Having selected a subset S of {x1, x2, . . . , xm} in this fashion, select xm+1 if S ∪ {xm+1} is an independent set; otherwise reject xm+1. In this way we find a linearly independent subset U of {x1, x2, . . . , xn} such that for any xi ∈/ U, U∪{xi} is linearly dependent. Define F = ⊕x∈U hxi, a free module contained in C. Given any index j, there is a nonzero ring element rj such that rjxj ∈ F (this is because U is a linearly independent set but U∪{xj} is linearly dependent. Thus, if c = Σjaixi ∈ C with ai ∈ R, multiply by r = Πjrj 6= 0 to obtain ∼ rc = Σjairxi ∈ F . Therefore, C = rC ≤ F and C is isomorphic to a submodule of the free module F (the sends c 7→ rc which is 1 − 1 since C is torsion-free). ¤ A domain is said to be noetherian if each ideal of R is finitely gen- erated. Theorem 11. The following are equivalent for an integral domain R: (a) R is noetherian. (b) The finitely generated torsion-free modules are precisely the mod- ules that are isomorphic to a submodule of ⊕nR for some n. (c) If M is a finitely generated module, then any submodule of M is finitely generated. (d) If K is a finitely generated module and K1 ⊆ K2 ⊆ K3 ⊆ · · · are submodules of K, then there is an index m such that Kn = Km for all n ≥ m. RINGS AND FIELDS ‘07 9

Proof. (a) → (b). Clearly the submodules of ⊕nR are torsion-free. Induction on n shows that the submodules are finitely generated as well. Conversely, the case n = 1 shows that (b) → (c). Let N be a submodule of M. As M is finitely generated, there is a free module F = ⊕nR and an epimorphism φ : F → M → 0. By the Correspondence Theorem, φ−1(N) = K is a submodule of F and by (b) is finitely generated. Therefore N = φ(φ−1(N)) is finitely generated. (c) → (d). Set N = ∪nKn. Since K is a submodule of M, K is finitely generated; generated by x1, x2, . . . , xm say. For each i, there is an index ni such that xi ∈ Kni . With m = max{n1, n2, . . . , xn}, each xi ∈ Km and so N ⊆ Km ⊆ N. (d) → (a). Given an ideal I of R, let x1 ∈ I and I1 = Rx1. If I 6= I1, there is an x2 ∈ I \ I1. Set I2 = Rx1 + Rx2. If I 6= I2 there is an x3 ∈ I \ I2. Set I3 = ΣRxi. Continuing in this manner, by (d), some Ik must coincide with I. ¤ Furthermore we have a fundamental splitting result: Corollary 12. If M is a finitely generated module over a PID, then M = T ⊕ N where T is the torsion submodule of M, and N is free, of finite rank. Proof. The map M → M/T splits because M/T is finitely generated, torsion-free (hence free) by the last theorem. ¤ Finitely Generated Modules Over a PID . If R is a PID and M is a finitely generated module, then M is a (finite) direct sum of cyclic modules. Proof. It remains to show that any finitely generated torsion mod- ule T is a direct sum of cyclic modules. Given a prime p ∈ R, let k Tp = {t ∈ T | p t = 0 for some k ≥ 1}. The submodule Tp is called the p-primary submodule of T . We claim that T = ⊕p primeTp where the index includes exactly one p from each associate of a given prime. (Note that Tp = Tq when p and q are associated primes.) This reduces the problem to finitely generated primary modules (i.e., a finitely gen- erated module K such that for some prime p, every element in K can be annihilated by a power of p). Let 0 6= t ∈ T . Then rt = 0 for some 0 6= r ∈ R. Express r as r = e1 e2 em ej p1 p2 ··· pm with p1, p2, . . . , pm non-associate primes. Set rj = r/pj and note that the of r1, r2, . . . , rm is 1. This ej is because, any prime dividing rj must divide rjpj = r and some must 0 be one of the pis; but pi does not divide ri. 10 RINGS AND FIELDS ‘07

In general, it is easy to see that the gcd of a1, a2, . . . , an is a where aR = a1R + a2R + ··· + amR. Write b1r1 + b2r2 + ··· + bmrm = 1 for some b1, b2, . . . , bm ∈ R. Then t = (b1r1t) + (b2r2t) + ··· + (bmrmt) and ej with tj = bjrjt, pj tj = 0 so tj ∈ Tpj . Thus t ∈ ΣjTpj . n If x ∈ Tp1 ∩ Σj=2Tpj with p1, p2, . . . , pm non-associate primes, then f1 fj p1 x = 0 and x = x2 + x3 + ··· + xn with pj xj = 0 for j ≥ 2. f1 f2 f3 fn The elements p1 and p2 · p3 ··· pn are relatively prime (i.e., have f1 f2 f3 fn gcd 1) and so ap1 + bp2 · p3 ··· pn = 1 for some a, b ∈ R. Then f1 f2 f3 fn f1 f2 f3 fn x = (ap1 + bp2 · p3 ··· pn )x = ap1 x + bp2 · p3 ··· pn x = 0. From the next home set, we obtain that any finitely generated nonzero primary module M has a M = hxi ⊕ K, where x 6= 0. Since K is either 0 or can be likewise factored

K = hx2i ⊕ K2, we obtain progressive decompositions

M = hxi ⊕ hx2i ⊕ · · · hxji ⊕ Kj, which must stabilize since the chain of submodules hxi ≤ hxi ⊕ hx2i ≤ hxi ⊕ hx2i ⊕ · · · hxji stabilizes (i.e., terminates with some Kj = 0). Therefore, M is a direct sum of cyclics and the proof is complete. ¤ Analogously, we have Finitely Generated Modules Over a Dedekind Domain . If R is Dedekind and M is a finitely generated module, then M = P ⊕ C where P is a and C is a finite direct sum of cyclic modules. RINGS AND FIELDS ‘07 11

5. EXERCISE SET 2 In this set R is a PID. A module M is said to be p-primary where p is a prime element of R, if for every x ∈ M there exists a k such that pkx = 0.

1. Let M be a finitely generated p-primary module. Show there exists a natural number n such that pnM = 0 (i.e., for every x ∈ M, pnx = 0). 2. Let M be a finitely generated p-primary module. Argue that chains of submodules K1 ⊆ K2 ⊆ · · · ⊆ Kn ⊆ · · · must stablize (i.e., there exists an m such that for all n ≥ m, Kn = Km). 3. Let M be a finitely generated p-primary module. (a) The order of an element 0 6= y ∈ M is the least positive integer ` such that p`y = 0 (this is justified by Problem 1). Use Problem 1 to show there is an element x ∈ M that has maximal order k (i.e., pkx = 0, while pk−1x 6= 0 and if 0 6= y ∈ M, then pky = 0). (b) With the x in part (a), show there is a submodule K of M that is maximal with respect to hxi ∩ K = 0. (Hint: Use Problem 2) 4. Let M be a finitely generated p-primary module. This problem shows that for x and K from Problem 3, M = hxi ⊕ K. Let X = hxi. (a) Given r ∈ R a nonzero, nonunit, write r = pis such that i ≥ 0 and p does not divide s (R is a UFD). Show that ry ∈ X ⊕ K if and only if piy ∈ X ⊕ K. Hint: For one direction, pn, s relatively prime, where pny = 0, implies apn + bs = 1 for some a, b ∈ R. Multiply by pi. (b) To show that M = hxi ⊕ K, it suffices to show that py ∈ X ⊕K implies y ∈ X ⊕K, for every y ∈ M, using induction and part (a). (c) We will now assume that py ∈ X ⊕ K and show that y ∈ X ⊕K. Show there is an index j < k (where k is given in Problem 1) such that pjpy = 0. (Take j = 0 if py = 0 and otherwise compare the orders of y and x). (d) Write py = rx + z with r ∈ R and z ∈ K. Using the j from part (c), conclude that p | r. Write r = ps. (e) With K0 = K + hy − sxi, either K = K0 or X ∩ K0 6= 0. In the latter case, write 0 6= tx = z0 + a(y − sx) for some t, a ∈ R and z0 ∈ K. Argue that p does not divide a since otherwise, tx ∈ K. 12 RINGS AND FIELDS ‘07

(f) Continuing part (e), write up + va = 1, and conclude that y = upy + vay ∈ X ⊕ K. RINGS AND FIELDS ‘07 13

6. EXERCISE SET 3 In this set, R is an integral domain. 1. Show that if F is a field, then R = F [x] is a PID. Hint: use the degree deg : F [x] → N and the division algorithm in F [x] to show that if f(x) 6= 0 is a polynomial of smallest degree in an ideal I 6= 0, then I = (f(x)). 2. Show that Z[x] is not a PID. (Hint: Show that I = (2, x) is not principal.) 3. Let R be a UFD with quotient field Q, and let f(x) ∈ R[x]. (a) Show that there is an element r ∈ R and a polynomial g(x) ∈ R[x], such that f(x) = rg(x) and the gcd of the coefficients of g(x) is 1. (r is called the content of f and we write c(f) = r). (b) Show that if f(x) = g(x)h(x) with g(x), h(x) ∈ Q[x], then there exists g1(x), h1(x) ∈ R[x] each having content 1, and elements 0 6= a, a1, b, b1 ∈ R such that a, a1 and b, b1 are coprime pairs, and ag(x) = a1g1(x) and bh(x) = b1h1(x). Thus, abf(x) = a1b1g1(x)h1(x). (c) Reduce ab and a1b1 in part (b) as necessary to obtain df(x) = eg1(x)h1(x) with the gcd(d, e) = 1. Show that if c(f) = 1, then d and e are units. (d) Conclude that f(x) ∈ R[x] is irreducible in R[x] if and only if f(x) is irreducible in Q[x]. 4. Let R be a UFD. Using Problem 1 and Problem 3 (d), show that R[x] is a UFD. Conclude that Z[x] is a UFD that is not a PID. 5. The Hilbert Theorem asserts that if R is noetherian, then so is R[x]. Assuming R is a noetherian domain, show that R[x] is noetherian. Hint: If J is an ideal in R[x], let In be the ideal of elements that occur as coefficients of xn in some polynomial of degree n in J. Observe that I0 ⊆ I1 ⊆ I2 ⊆ · · · . Use this to pick generators for J.

7. Primary Decompositions In this section R is a commutative (it need not be a domain, but the issue never arises). Since we are not restricting ourselves to domains, the symbol Q is in play and will be used here to denote a special of ideal. Recall that an ideal P 6= R is called a prime ideal, if ab ∈ P implies a ∈ P or b ∈ P . If one wishes, we can define the I divides J when I and J are ideals, to be the condition J ⊆ I. It is an easy exercise 14 RINGS AND FIELDS ‘07 to show that P 6= R is prime if and only if P divides I · J implies P divides I or P divides J, for ideals I and J. We say that an ideal Q 6= R is primary, if for any a, b ∈ R, ab ∈ Q and a∈ / Q, implies bn ∈ Q for some natural number n. While primary integers are just the powers of primes, this is not the case for ideals in general rings of our type. The radical of an ideal I, rad(I), is defined to be rad(I) = {r ∈ R | rn ∈ I for some n ≥ 1}. Proposition 13. (1) For any ideal I, there exists an n ≥ 1 such that rad(I)n ⊆ I. (2) If Q is primary, then rad(Q) is prime. Proof. (1) It is easy to see that rad(I) is an ideal, and since Ris noetherian, there exists a1, a2, . . . , am ∈ rad(I) such that rad(I) = ha1, a2, . . . , ami. For each j there exists an positive integer ki for which ki ai ∈ I. Take n = k1 + k2 + ··· + km. Then, the general element n r1a1 +r2a2 +···+rmam of rad(I) satisfies (r1a1 +r2a2 +···+rmam) = n! i1 i2 im Σ (r1a1) (r2a2) ··· (rmam) where the sum is over all non- i1!i2!···im! negative integers i1, i2, . . . , im that sum to n (Multinomial Theorem). i1 i2 im 0 In any monomial (r1a1) (r2a2) ··· (rmam) , one of the ijs must be n greater than kj and so that monomial lies in I. Therefore, rad(I) ⊆ I. (2) Let P = rad(Q) and suppose ab ∈ P with a∈ / P . By definition (ab)n ∈ Q for some n. Since a∈ / P , an ∈/ Q and since Q is primary, (bn)m ∈ Q for some m. Then bnm ∈ Q and b ∈ P . ¤ A proper ideal I is called irreducible, if

I = J1 ∩ J3 =⇒ I = J1 or I = J2, for ideals J1,J2. Proposition 14. Any irreducible ideal is primary. Proof. Let I be irreducible and suppose that I is not primary. There exists a, b ∈ R, such that ab ∈ I, a∈ / I and bn ∈/ I for every n ≥ 1. n n n+1 Let In = {r ∈ R | rb ∈ I}; rb ∈ I implies rb ∈ I and so In ⊆ In+1 for all n. Since R is noetherian, there is an index m such 0 m that In = Im for all n ≥ m. Let I = {rb + c | r ∈ R, c ∈ I}. 0 CLAIM: I ∩ Im = I. 0 m m 2m m m If s ∈ Im∩I , then s = rb +c and sb ∈ I. Then rb = cb −sb ∈ m 0 I and so r ∈ I2m = Im. Therefore, s = rb +c ∈ I. Clearly, I ⊆ I ∩Im, 0 and the claim is supported. But neither I nor Im equal I, resulting in a contradiction. ¤ RINGS AND FIELDS ‘07 15

Primary Decompositions of Ideals Exist . Any proper ideal I of R can be expressed as

I = I1 ∩ I2 ∩ · · · ∩ Im, with Ij primary ideals of R. Proof. If there are ideals that do not possess primary decompositions, let I be maximal with respect to not having a . By the previous proposition, I is not irreducible, and so I = J1 ∩ J2 with J1,J2 different (i.e., larger than) I. Hence J1,J2 have primary decompositions, and therefore so does I; a contradiction. ¤ There are serious problems with the uniqueness issue. However, a general context under which there is uniqueness, is for noetherian do- mains such that every nonzero prime ideal is maximal, once the primary decomposition has been adequately aligned. Proposition 15. If I and J are primary ideals with radical P , then I ∩ J is a primary ideal with radical P . Hence, when considering a primary decomposition

I1 ⊕ I2 ⊕ · · · ⊕ In, we can select the distinct primes among rad(Ij), j = 1, 2, . . . , n; call then P1,...,Pk, and set

Ji = ∩{Ij | rad(Ij) = Pi}. 0 We then have I1 ⊕I2 ⊕· · ·⊕In = J1 ⊕J2 ⊕· · ·⊕Jk, with the Jis primary with distinct radicals. We now only consider primary decompositions of the latter type, in seeking to show uniqueness.

Lemma 16. If I1,I2,...,Im are relatively prime (i.e., Ii + Ij = R for all i 6= j), then I1 ∩ I2 ∩ · · · ∩ Im = I1I2 ··· Im.

Proof. For two, I1 + I2 = R implies I1 ∩ I2 = (I1 ∩ I2)(I1 + I2) = (I1 ∩ I2)I1 + (I1 ∩ I2)I2 ⊆ I1I2. The other containment is obvious. The inductive step is left as Exercise 4.4. ¤ 16 RINGS AND FIELDS ‘07

Theorem 17. Let R be a noetherian domain such that every nonzero prime ideal is maximal. If I1,I2,...,Im and J1,J2,...,Jn are collec- tions of primary ideals with distinct radicals, such that

I = I1 ∩ I2 ∩ · · · ∩ Im = J1 ∩ J2 ∩ · · · ∩ Jn, then n = m and after reindexing, Ij = Jj for all j. Proof. Each primary ideal mentioned in the statement contains a power of its radical by Proposition 13. Let Pi be the radical of Ii, so that ki ki kj Pi ⊆ Ii for each i. Since Pi + Pj = R for i 6= j (Exercise 4.2), it follows that Ii +Ij = R for all i 6= j. Therefore, by the previous lemma,

I = I1I2 ··· Im = J1J2 ··· Jn. k k With k = max{k1, . . . , km}, P1 ··· Pm ⊆ I. Likewise, if Qj is the ` ` radical of Jj, then Q1 ··· Qn ⊆ I. For the sake of later discussions, let us assume that k ≥ ` (else, replace k by `). We now have, for any i, k k Q1 ··· Qn ⊆ I ⊆ Ii ⊆ Pi = rad(Ii).

Since Pi is maximal (hence prime), each Pi contains and is therefore equal to some Qj. Conversely, each Qj coincides with some Pi, and since P1,P2,...,Pm are distinct and Q1,Q2,...,Qm are distinct, we conclude that n = m and after reindexing, Pi = Qi. We will now show that I1 = J1; the proof for the others follows from the indexing being arbitrary. By Exercise 4.3, there exists a t ∈ k (P2 ··· Pm) \ P1. It follows that t ∈ I2 ··· Im and that ta ∈ I for every j a ∈ I1, yet t∈ / I1 because t∈ / P1. Moreover, t ∈/ I1 for any j, because P1 is prime. Likewise, tr ∈ I for every r ∈ J1, yet t∈ / J1. Suppose r ∈ J1, so that tr ∈ I. Since tr ∈ I1 and I1 is primary, j j either r ∈ I1 or t ∈ I1 for some j. We’ve already observed that t ∈ P1 is impossible, so therefore, r ∈ I1. Similarly, I1 ⊆ J1, and the proof is complete. ¤ RINGS AND FIELDS ‘07 17

8. EXERCISE SET 4 1. Prove Proposition 15. 2. If I,J are comaximal ideals (i.e., I+J = R), show that Ik+J ` = R for any integers k, `.

3. Assume that I1,I2,...,Im are pairwise comaximal ideals of R. If k is a positive integer, show that k I1 + (I2 ··· Im) = R.

4. Finish the proof of Lemma 16: If I1,I2,...,Im are pairwise comaximal, then

I1 ∩ I2 ∩ · · · ∩ In = I1I2 ··· In.

5. Let R = Z[x]. Show that the ideal I = (x2, 2x) has the following primary decompositions with mutually distinct radicals: I = (x) ∩ (x2, 2) = (x) ∩ (x2, 2 + x) = (x) ∩ (x2, 2x, 4).

6. Let I 6= R be an ideal that contains a power of a maximal ideal. Then I is primary. Therefore, if R is a domain such that every nonzero prime ideal is maximal, the the primary ideals are precisely the ideals that contain a power of a maximal ideal.

7. Show that I = (x2, 2x) in R = Z[x] is not primary yet I contains (x)2. This shows the assumption in Exercise 4.6 that I contain a power of a maximal ideal cannot be weakened to a power of a prime. 18 RINGS AND FIELDS ‘07

9. Nakayama’s Lemma In this section R is still a commutative ring. Proposition 18. If I is an ideal in a commutative ring R, and I 6= R, then I is contained in a maximal ideal of R. As a consequence, and element 0 6= r ∈ R is contained in some maximal ideal if and only if r is not a unit. The (Jacobson) radical of R is defined to be J(R) = ∩{ M | M is a maximal ideal of R }. Nakayama’s Lemma . If K is a finitely generated module over a commutative ring R such that J(R)K = K, then K = 0. Proof. Suppose r ∈ J(R). If 1 − r is not a unit of R, then by the last proposition, 1 − r ∈ M for some maximal ideal M. But then 1 = 1 − r + r ∈ M, a contradiction. Thus, 1 − r is a unit. Let x1, x2, . . . , xn be a minimal generating set for K. By hypothesis, we can write x1 = Σjrjxj −1 n with rj ∈ J(R) for all j. Therefore, x1 = (1−r1) Σj=2rjxj, contradict- ing the minimality of the generating set, unless n = 1 and x1 = 0. ¤ A ring R is called quasi-local, if R has a unique maximal ideal M. In this case J(R) = M. Kaplansky . If R is a quasi-, then any projective R-module is free. Proof. We will only include the proof that finitely generated projective modules are free. Let W be a finitely generated projective module, and let x1, x2, . . . , xn be a minimal generating set for W . There is a free module F of rank n and a natural epimorphism f : F → W given by f(ei) = xi where e1, e2, . . . , en is the for F . Let K = Ker f. Since W is projective, there is a monomorphism g : W → F such that F = K ⊕ Im g. Observe that K ⊆ MF , because if f(r1, r2, . . . , rn) = Σjrjxj = 0, and r1 (say) is not in M, then r1 is a unit, allowing a reduction in the size of the generating set; a contradiction. Thus, K ⊆ MF and since K = π(F ) where π : F → K is the coordinate projection map, it follows that MK = K. By Nakayama’s Lemma, K = 0. ¤ RINGS AND FIELDS ‘07 19

10. Subrings of Quadratic Number Fields Let us look to the classical examples of Dedekind to illustrate some of the discussions thus far. Call an element α ∈ C integral over Z if there is a 0 6= f(x) ∈ Z[x] that has α as a root. If f(x) has degree n, then n−1 R = Z[α] = {b0 + b1α + ··· + bn−1α | bj ∈ Z}, is an integral domain that is generated by 1, α, α2, . . . , αn−1 over Z. It turns out, α is integral if and only if α belongs to a subring R of C that is finitely generated as a Z-module. We say that α is algebraic (over Q) if α is a root to some polynomial 0 6= f(x) ∈ Z[x]. Integral elements are algebraic but not conversely. The number √1 is not integral, being a root to the irreducible polyno- 2 mial 2x2 − 1, which is not monic. Given an α, there is an in Z[x] having alpha as a root. Denote this polynomial by irrZ(α). It is easy to check that a α is integral if and only if it is algebraic and irrZ(α) is monic.√ The elements of the form n m where m, n are integers and n > 0 are integral. For our examples here we will consider the case n = 2 and m is square-free. The field √ √ Q[ m] = {a + b m | a, b ∈ Q} is called a quadratic number field. √ A subring R of a quadratic number field Q[ m] is said to be full, if R contains at least one nonrational√ number. In this case, R is called integrally closed if each α ∈ Q[ m] that is integral actually belongs in R. √ Example 7. A full subring R of Q[ m] that is finitely generated over Z is integrally closed if and only if √ R = Z[ m] when m ≡ 2 or 3 (mod 4), and √ 1+ m R = Z[ 2 ] when m ≡ 1 (mod 4). √ It then turns out that a full subring S of Q[ m] is integrally closed if and only if S = ∩P ∈P RP where R is one of the rings of the previous example, and P is a set of maximal ideals of R. By the of localizations, the ideals of S are IS where I is an ideal of R. Hence S is noetherian if R is noetherian. Since R is finitely generated as a Z-module, any ideal is 20 RINGS AND FIELDS ‘07

finitely generated as a Z-module and therefore as an R-module. Hence R and S are noetherian. −1 If P is a nonzero prime√ ideal of R, and 0 6= β ∈ P , then nβ ∈ R for some n (since Q[ m] is a field), and therefore nβ−1β = n ∈ P . Consequently, since R is an of Z ⊕ Z, R/P is an image of Z ⊕ Z/n(Z⊕Z) ∼= Z/nZ⊕Z/nZ. Therefore R/P is a finite integral domain, implying that R/P is a field. I.e., P is maximal. √ To recap, an integrally closed full subring of Q[ m] is (i) integrally closed, (ii) noetherian, and (iii) every nonzero prime ideal is maximal. Such rings are called Dedekind domains, after the person to first for- mally study these rings, and the initiator of formal commutative . √ Theorem 19. A full subring R of Q[ m] is a Dedekind domain if and only if every ideal I is a product prime ideals. Furthermore, these conditions are√ equivalent to the property that for any ideal I 6= 0 of R, I−1 = {t ∈ Q[ m] | tI ⊆ R} satisfies I−1I = R. The proof of this is within the scope of our course, but as time is of the essence, will be omitted. Class√ and Theory study the field Q[ m] by considering the of isomorphism classes of ideals of R (as in Example 7) where [I] · [J] = [IJ]; here, [I]−1 = [I−1]. √ Corollary 20. Let R be a full subring of Q[ m]. Then every primary ideal is a power of a prime ideal if and only if R is Dedekind. Proof. Since R is noetherian such that nonzero prime ideals are maxi- mal, any proper ideal I of R can be expressed as

I = I1I2 ··· Im, where I1,I2,...,Im are primary ideals with distinct radicals. If for each I, every Ij is a power of a prime, then I is a product of prime ideals and R is Dedekind by the theorem. Conversely, if R is Dedekind, then a primary ideal J must also be a product of prime ideals. Since J k contains a power of its radical P = rad(J), P ⊆ J = P1P2 ··· P` ⊆ Pj, implies P = Pi for any i. I.e., J is a power of a prime ideal. ¤ √ Corollary 21. Let R be a full subring of Q[ m]. Then R is a UFD if and only if R is a PID. In this case, R is Dedekind.

Proof. Assume that R is a UFD and let r be a prime√ in R. Then (r) is a prime ideal, and because R is a full subring of Q[ m], (r) is maximal. RINGS AND FIELDS ‘07 21

Conversely, if M is maximal then M = (r) for some prime r (that is, if p1p2 ··· pn ∈ M, then some pj ∈ M and therefore M = (pj)). Because the maximal ideals of R are of the form (p) for some prime p, if b ∈ R is not a multiple of p, then Rb + Rp = R. Therefore, sb + tp = 1 for some s, t ∈ R. If I is a primary ideal, then (pn) ⊆ I ⊆ (p) for some prime p. Let k be the smallest index such that (pk) ⊆ I, so that pk−1 ∈/ I. Suppose there exists a ∈ I \ (pk). Since R is a UFD we can write a = pjb with p, b relatively prime. Note j < k since a∈ / (pk). Since b, p are coprime, there exists s, t ∈ R such that sb + tp = 1. Thus, pk−1 = spk−1b + tpk = spk−1−jpjb + tpk ∈ I, contrary to the choice of k. Therefore, I = (pk) and every primary ideal is a power of a prime (maximal) ideal. ¤ There are only finitely many known PID’s of the form of Example 7. Here are some interesting heavy-duty algebraic number theoretic results: Let R be as in Example 7:

(a) There are infinitely many integral primes p such that pR is prime.

(b) There are infinitely many integral primes p such that pR = P1P2 for distinct maximal ideals P1,P2. (c) There are only finitely many p0s for which pR = P 2; these are the prime divisors of m with the possible exception of p = 2.

Some research problems suggested by the topics that came up in this regard: √ 1. Which full subrings S of Q[ m] have every ideal generated by an integer? This pertains to a problem of P. Hill which I answered in the 1990’s. 2. If R ⊆ S ⊆ Q with R a domain, when is it the case that ideals of S are extendable from R; i.e., every ideal J of S is of the form IS where I is an ideal of R (i.e., I = J ∩ R)? I answered this to a degree in the late 90’s. 22 RINGS AND FIELDS ‘07

3. If R is noetherian, when is every overring of R inside Q of the form ∩P ∈P RP ? These turn out to be the Dedekind domains. 4. If R is a noetherian domains such that every nonzero prime ideal is maximal, under what conditions does the Krull-Schmidt The- orem hold relative to finitely generated, torsion-free modules? I.e., when is it the case that whenever ∼ A1 ⊕ A2 ⊕ · · · ⊕ An = B1 ⊕ B2 ⊕ · · · ⊕ Bn,

with Ai,Bj indecomposable, torsion-free finitely generated R- modules, then ∼ m = n and after reindexing Ai = Bi ∀ i? RINGS AND FIELDS ‘07 23

11. Semi-Simple Rings Let R be a ring (not necessarily commutative). In this section, we will consider right R-modules and and right ideals. A module refers to a right R-module. The case for left modules can be made indepen- dently and analogously. A module is called simple if is had no proper submodules. A module M is called semi-simple if it is a sum of simple submodules; M = ΣiNi, where each Ni is simple. Schur’s Lemma . If I and J are simple modules and f ∈ Hom(I,J), then either f = 0 or f is an isomorphism. In particular, EndR(I) = {f | f : I → I is a } is a . Proof. Let 0 6= f : I → J. Since Ker f, Im f are submodules of I and J respectively, and since I, J are simple,Im f = I and Ker f = 0. In particular, any g : I → I is an . Thus, g−1 : I → I is a well-defined homomorphism, implying that EndR(I) is a division ring. ¤ The of R, J(R), is defined to be J(R) = ∩{M | M is a maximal right ideal}. The ring R is called (Jacobson) semi-simple if J(R) = 0. R is called right artinian if R has the descending chain condition on right ideals: i.e., if I1 ⊇ I2 ⊇ · · · , are right ideals, then for some index m, In = Im for all n ≥ m. 24 RINGS AND FIELDS ‘07

Chinese Remainder Theorem . If P1,P2,...,Pk be parwise comax- imal ideals of R, then ∼ R/ ∩j Pj = R/P1 × R/P2 × · · · × R/Pk as rings.

Proof. Define a map θ : R → R/P1 ⊕ R/P2 ⊕ · · · ⊕ R/Pk by θ(r) = (r + P1, r + P2, . . . , r + Pk). Clearly Ker θ = ∩jPj so it remains to show that θ is an epimorphism. We have P1 +P2 = R, implying that P1P3 +P2P3 = P3, from which it follows that P1 + P2P3 = P1 + P1P3 + P2P3 = P1 + P3 = R. Continuing ˆ in this way, P1 + Πj≥2Pj = R. Now let Pj = Πi6=jPi (with the Pi’s appearing with increasing index moving from left to right). We can ˆ ˆ ˆ induct on k to show that P1 + P2 + ··· + Pk = R. The case k = 2 is obvious. Inductively, for n ≥ 3, let n−1 Qj = Πi6=j Pi. ˆ By induction Q1 + Q2 + ··· + Qn−1 = R. Note Pj = QjPn if j < n. Thus Q1Pn + Q2Pn + ··· + Qn−1Pn = Pn, and therefore ˆ ˆ ˆ ˆ ˆ P1 + P2 + ··· + Pn−1 + Pn = Pn + Pn = R, as claimed. ˆ Write 1 = x1 + x2 + ··· + xk where xj ∈ Pj. Let r1, r2, . . . , rk ∈ R be given, and take r = r1x1 + r2x2 + ··· rkxk. Modulo Pj,

r ≡ rjxj ≡ (rjx1 + ··· + rjxn) ≡ rj.

Therefore, the map from R → R/P1 ⊕ R/P2 ⊕ · · · ⊕ R/Pk is an epi- morphism which is clearly an R-module map. 2 2 2 2 Note 1 = 1 = Σi,jxixj which maps to (x1+P1, x2+P2, . . . , xk+Pk) = (x1+P1, x2+P2, . . . , xk +Pk). Multiplying 1 = x1+x2+···+xk through by r and s respectively, we obtain

r = rx1 + rx2 + ··· + rxk and s = sx1 + sx2 + ··· + sxk. 2 2 2 So, rs maps to (rsx1 +P1, rsx2 +P2, . . . , rsxk +Pk) = (rsx1 +P1, rsx2 + 2 2 2 P2, . . . , rsxk + Pk). The former term (rsx1 + P1, rsx2 + P2, . . . , rsxk + Pk) is equal to (rx1 + P1, rx2 + P2, . . . , rxk + Pk) · (rsx1 + P1, rsx2 + P2, . . . , sxk + Pk) ¤ RINGS AND FIELDS ‘07 25

Artin-Wedderburn Theorem . The following are equivalent for a ring R: 1. R is semi-simple as a module. 2. R is semi-simple artinian as a ring. 3. There exists natural numbers n1, n2, . . . , nk and division rings D1,D2,...,Dk such that ∼ R = Matn1 (D1) × Matn2 (D2) × · · · × Matnk (Dk).

Proof. (1) → (3): If R is semi-simple as a right module, write R = ΣjIj, and in particular, 1 = xj1 +xj2 +···+xjm with xji ∈ Iji . If r ∈ R, then m r = xj1 r+xj2 r+···+xjm r. Furthermore, r = rak1 +···+rakm ∈ Σi=1Iki for any r ∈ R, and after relabeling Iki as Ii, we obtain

R = I1 + I2 + ··· + Im, a finite sum. 0 Reorder the Ijs so that I1,I2,...,Ik are pairwise non-isomorphic and ∼ if j ≥ k, then Ij = Ii for some i ≤ k. We have that

∼ n1 n2 nk R = I1 ⊕ I2 ⊕ · · · ⊕ Ik , with HomR(Ii,Ij) = 0 for all i 6= j by Schur’s Lemma. We now have the

∼ ∼ n1 n2 nk n1 n2 nk R = EndR(R) = HomR(I1 ⊕ I2 ⊕ · · · ⊕ Ik ,I1 ⊕ I2 ⊕ · · · ⊕ Ik )

∼ n1 n1 n2 n2 nk nk = HomR(I1 ,I1 ) × HomR(I2 ,I1 ) × · · · × HomR(Ik ,Ik ), and in turn this last ring is isomorphic to

Matn1 (D1) × Matn2 (D2) × · · · × Matnk (Dk). The first isomorphism is called the left of R and sends r ∈ R to λr ∈ EndR(R) where λr(a) = ra (note, this is a right module map). If f ∈ EndR(R), then f(a) = f(1)a and so f = λf(1). The rest of the details concerning this isomorphism are easy to check. The second isomorphism is an application of the obvious fact that if ∼ ∼ A = B as modules, then EndR(A) = EndR(B) as rings. Specifically, if θ : A → B is a right module isomorphism, then the map δ 7→ θδθ−1 is an isomorphism from EndR(A) onto EndR(B). The third isomorphism employs the principal that if Hom(A, B) = 0, then   f11 f12 ··· f1n    f f ··· f  Hom(⊕n A , ⊕n B ) =  21 22 2n  , i=1 i i=1 i  ......  fn1 fn2 ··· fnn 26 RINGS AND FIELDS ‘07 where fij : Ai → Bj. Such a matrix sends (a1, a2, . . . , an) ∈ ⊕jAj to (Σjf1j(aj), Σjf2j(aj),..., Σjfnj(aj)) ∈ ⊕jBj (via left multiplication by the matrix on (a1, a2, . . . , an)). The final isomorphism follows from Schur’s Lemma and properties of Hom. Let I be a simple right ideal of R, and D = EndR(I). Then m ∼ EndR(I ) = Matm(D), as rings. To examine this we will distinguish between different copies th of I; let Xi be the copy of I generating the i component. Then α ∈ m th EndR(I ) can be identified with the m × m matrix whose i, j entry th is πjα|Xi where πj : ⊕`X` → Xj is projection onto the j coordinate. Conversely, a matrix A in Matm(D) represents an of ⊕iXi via left multiplication by A on the m-tuples inside X1 ⊕ X2 ⊕ m · · · ⊕ Xm. In this way we identify elements of EndR(I ) with the corresponding elements of Matm(D). (3) → (2): Follows from Exercise 5.5. (3) → (1): Follows from Exercise 5.4. (2) → (3): According to the Chinese Remainder Theorem, it is sufficient to find comaximal two-sided ideals P1,P2,...,Pn whose in- tersection is zero such that R/Pi is a matrix ring over a division ring for each i. Let P be the right annihilator of a simple right R-module A. Then A(rP ) ⊆ (Ar)P = 0, implying that P is a two-sided ideal. Certainly P 6= R. To the end of showing that R/P is a matrix ring over a division ring, we embed R/P into EndD(A) where D = EndR(A) (is a division ring), via right mul- tiplication (i.e., r ∈ R maps to right multiplication by r in EndD(A)). We will establish a lemma below which will aide us. Assuming this for now, suppose A is not finite dimensional over D; say x1, x2,... are independent over D. Set Ij equal to the right annihilator of {x1, x2, . . . , xj}. Then I1 ⊇ I2 ⊇ · · · , is a chain of right ideals of R. By the lemma below, for each j, there exists r ∈ Ij such that xj+1r 6= 0, implying Ij 6= Ij+1. This contradicts R being right artinian. Hence A is finite dimensional. By Exercise 5.1, EndD(A) is the matrix ring Matn(D) where n = dimD A. We have the of R/P into EndD(A). It remains to show that this map is onto. Let x1, x2, . . . , xn be a basis for A over D and let y1, y2, . . . , yn ∈ A. Set Bj equal to the D-subspace of A generated by {x1, x2, . . . , xn}\{xj} and let Ij = annR (Bj)R. By the lemma below, xjIj 6= 0 and since A is right simple over R, xjIj = A for all j. RINGS AND FIELDS ‘07 27

Write yj = xjtj for tj ∈ Ij and set

r = t1 + t2 = ··· + tn. Then xjr = xjt1 + xjt2 ··· + xjtn = xjtj = yj, since xjri = 0 when i 6= j. This shows that the map R/P → EndD(A) is onto. Now, as R/P is a matrix ring over a division ring, P is the intersec- tion of finitely many maximal right ideals of R. Therefore ∩{ P | P is the right annihilator of a simple right R-module } ⊆ J(R) = 0. But, R is artinian, so there are finitely many distinct ideals P1,P2,...,Pm that are annihilators of simple right R-modules, such that

P1 ∩ P2 ∩ · · · ∩ Pm = 0.

The rings R/Pi are simple by Exercise 5.2 and what we have shown above, so each Pi is a maximal two-sided ideal. By the Chinese Re- mainder Theorem, ∼ R = R/ ∩i Pi = R/P1 × R/P2 × · · · × R/Pm, and R is a finite product of matrix rings over division rings. ¤ Lemma 22. Let R be right artinian, and A a simple R-module with right annihilator P . If B is a finite dimensional subspace of A over the division ring D = EndR(A), and I = annR BR, then aI = 0 implies a ∈ B.

Proof. The proof is by induction on dimD B = n. If n = 0, then R = annR BR, so aR = 0 implies a = 0. Let B be an n-dimensional subspace of A, and let B1 be a subspace of B of n − 1. By the induction hypothesis, with I1 = annR (B1)R, aI1 = 0 implies a ∈ B1 for any a ∈ A. We must show that aI = 0 implies a ∈ B where I = annR BR. Equivalently, cI 6= 0 for any c ∈ A \ B. Let b ∈ B \ B1, and let a ∈ A \ B. We must show that aI 6= 0. If br1 = 0 yet ar1 6= 0 for some r1 ∈ I1, then r1 ∈ I, yet 0 6= ar1 ∈ aI; we have finished. Suppose, on the other hand, that br1 = 0 implies ar1 = 0 for all r1 ∈ I1. In this case, br1 = br2 if and only if b(r1 − r2) = 0 if and only if a(r1 − r2) = 0 if and only if ar1 = ar2. Therefore, the map θ : bI1 → aI1 given by θ(br1) = ar1 is a well-defined R-. By induction, bI1, aI1 6= 0, and because A is simple, bI1 = aI1 = A. Thus, the map θ ∈ D = EndR(A). Note, θ(br1)−ar1 = 0 = (θ(b)−a)r1 for all r1 ∈ I1. By induction again, θ(b) − a ∈ B1 and consequently, a = a − θ(b) + θ(b) ∈ B since B is a D-linear . This contradicts a∈ / B, and therefore there exists r1 ∈ I1 such that br1 = 0 while ar1 6= 0; that is, aI 6= 0 as desired. ¤ 28 RINGS AND FIELDS ‘07 ∼ Corollary 23. R is simple artinian if and only if R = Matn(D) for some n and some matrix ring D. Proof. Assume that R is simple and artinian. As in the proof of the theorem, if P is the right annihilator of a simple right R-module, then ∼ R/P = Matn(D) for some n and some matrix ring D. But R is simple, so P = 0. The converse is Exercise 5.2. ¤ Corollary 24. The following are equivalent: (1) R is semi-simple as a right module. (2) R is semi-simple as a left module. (3) Every right R-module is projective. (4) Every left R-module is projective. (5) Every right R-module is injective. (6) Every left R-module is injective. Proof. For pedagogical reasons, the proof of this theorem will be carried out for the finitely generated case; the general case holds with very little adaptation. (1) → (3): Every module H has a projective :

⊕nR → H, where n is the size of a generating set for H. Write R = I1 ⊕I2 ⊕· · ·⊕Im for simple (right or left) ideals I1,I2,...,Im of R (Artin-Wedderburn Theorem). Since the image of a is simple,

H = ΣjXj, where Xj is isomorphic to one of the Ii’s. We can easily find a sub-collection of the Xj’s indexed by X1,X2,...,Xk (after reindexing) so that H = ⊕jXj. Since each Xj is isomorphic to ∼ `1 `2 `m ` some Ii, we can write H = I1 ⊕ I2 ⊕ · · · ⊕ Im . With F = ⊕ R, ` = max{`1, `2, . . . , `m}, F ∼= H ⊕ K,

k1 k2 km where K = I1 ⊕ I2 ⊕ · · · ⊕ Im with kj = ` − `j, for all j. (3) ↔ (5): Recall that a module A (respectively C) is injective (respectively projective) if and only if every short 0 → A → B → C → 0, splits. Hence, every module is injective if and only if every module is projective. (5) → (1): Let soc(R) (read, the of R) be the submodule of R generated by all of the simple right ideals of R. We need to show that soc(R) = R. By (5), R = soc(R) ⊕ X since soc(R) is injective. RINGS AND FIELDS ‘07 29

Suppose 0 6= r ∈ X. Let M be a right submodule of X maximal with respect to r∈ / M. Then every submodule of X/M must contain r +M. It follows that the submodule C of X/M generated by r + M is simple. By (5), the simple submodule C of X splits out of X, so that X = C ⊕ Y for some module Y . Thus, soc(R) ⊕ C is contained in R and is a sum of simple modules, contrary to the maximality of soc(R) as the sum of all simple modules. Therefore X = 0 and R = soc(R). The equivalence of (2), (4) and (6) is handled analogously. The equiv- alence of (1) and (2) is a consequence of the Artin-Wedderburn Theo- rem. ¤ 30 RINGS AND FIELDS ‘07

12. EXERCISE SET 5 1. Let V be the of dimension n as a left module over the division ring D; where V consists of all n-tuples with entries in D. (a) Viewing V as a collection of row vectors, show that S = Matn(D) is equal to EndD(V ) (which operates via right multiplication). (b) Show that V is a simple S-module. 2. A ring R is called simple if R has no proper 2-sided ideals. Show that R = Matn(D) where D is a division ring is simple and artinian on either side (you can pick a side and do that case).

3. Let R = R1 × R2 × · · · × Rm as rings. Given a side, show that R is artinian if and only if each Rj is artinian.

4. Let R = R1 ×R2 ×· · ·×Rm as rings. Show that R is semi-simple as a right (left) R-module if and only if each Rj is semi-simple as a right (left) Rj-module. 5. Show that a matrix ring over a division ring is semi-simple artinian (on either side). Conclude that a product of matrix rings over division rings is semi-simple artinian (on either side).

6. Let A1,A2 be simple right R-modules with endomorphism rings ∼ Di = EndR(Ai). Show A1 = A2 as R-modules if and only if ∼ D1 = D2 as rings. This in turn is equivalent to one of Di being isomorphic to a subring of the other. RINGS AND FIELDS ‘07 31

13. Theory Throughout this subject matter, F is a field with subfield K. We will investigate the relationship between K and a subfield E of F containing K. We assume that the notation K ≤ E acknowledges the relationship that K is a subfield of E. For fields K ≤ E, E is a vector space over K; the dimension of E over K, is also called the degree of E over K, and is denoted by

[E : K] = dimK E. Theorem 25. Let K ≤ E ≤ F be fields. Then [F : K] = [F : E] · [E : K].

Proof. Let A = { αi | i ∈ I} be a basis for E as a vector space over K, and B = { βj | j ∈ J} be a basis for F as a vector space over E. We claim that the elements αiβj for i ∈ I and j ∈ J form a basis for F over K. Suppose Σi,jai,jαiβj = 0, for ai,j ∈ K (almost all zero) and i ∈ I, j ∈ J. Since the sum is finite, we obtain Σj(Σiai,jαi)βj = 0, which implies Σiai,jαi = 0 for each j since Σiai,jαi ∈ E and B is a basis for F over E. But then, ai,j = 0 for all i, j because A is a basis for E 0 0 over K. Therefore αiβj = αi0 βj0 if and only if i = i and j = j , and

{αiβj | i ∈ I, j ∈ J }, is a K-independent set of |I| · |J|. If δ ∈ F , then δ = Σjbjβj for some bj ∈ E (almost all zero) since B is a basis for F over E. For each j, write

bj = Σiai,jαi, with ai,j ∈ K (almost all zero). Then

δ = Σj(Σiai,jαi)βj = Σi,jai,jαiβj. Therefore AB = { αβ | α ∈ A, β ∈ B }, is a basis for F over K of cardinality |A| · |B|. ¤ 32 RINGS AND FIELDS ‘07 √ Example 8. The element 2 + ι ∈ C is algebraic of degree 4 over Q. √ √ √ Proof. Let α = 2+ι. Then (α− 2)2 = −1, and so α2+2+1 = 2 2α. Thus, α is a root to (x2 + 3)2 − 8x2, and so the degree of α is less than or equal to 4. However, the degree of α coincides with the dimension of Q[α] over√Q. −1 2 Observe, 2 = α (√α + 3)/2 ∈ Q[α], and consequently ι ∈ Q[α] as well. We have Q ≤ Q[ 2] ≤ Q[α], and so √ √ [Q[α]: Q] = [Q[α]: Q[ 2]][Q[ 2] : Q]. √ √ Since [Q[α]: Q[ 2]] is√ not 1 (ι∈ / Q[ 2]), but is less than√ or equal to 2 (α is a root to x2 − 2 2x + 3), we conclude [Q[α]: Q[ 2] = 2, and α has degree 4. ¤ √ 3 There are possible distinctions between√ √ subfields. For example, Q[ 2] contains a root to x3 − 2 while Q[ 3 2, 3ι] contains all roots. Kronecker’s Theorem . Let f(x) ∈ K[x] be irreducible. (1) There exists a field extension E of K that contains a root α to f(x). (2) If K0 is a field isomorphic to K under σ : K → K0, and E0 is a field extension of K0 containing a root α0 of σf (where σf is the polynomial formed by applying σ to the coefficients of f), then there exists an isomorphism of fields K[α] ∼= K0[α0] with k ∈ K 7→ σ(k) and α 7→ α0. Proof. (1): Let E = K[x]/(f), a field containing K canonically. The root to g(y) is α = x + (g). Hence E = K[α] contains a root to f(y). (2): The extension of σ to a map from K[x] → K0[x] which sends g(x) ∈ K[x] to σg, the polynomial in K0[x] obtained by applying σ to each of the coefficients of g(x) is an isomorphism of rings; K[x] ∼= K0[x]. Hence, K[α] ∼= K0[α0], by the construction in (1) (necessarily σf is irreducible). ¤ Corollary 26. Given any polynomial f(x) ∈ K[x], there is a field extension E of K such that f(x) splits into a product of linear factors. Moreover, E = K[α1, α2, . . . , αm] where αj are the roots of f(x) in E. RINGS AND FIELDS ‘07 33

The field E in this corollary is called a splitting field for f(x) over K. It is an easy proof by induction on the degree of f(x) that splitting fields are unique up to isomorphism. The of F over K is

G = AutK (F ); that is, G is the group (under composition of maps) of K-linear au- tomorphisms σ : F → F . If σ ∈ G, then σ(k) = kσ(1) = k for all k ∈ K (i.e., each element of K is fixed under σ). Conversely, if each element of K is fixed under a field automorphism σ : F → F , then σ(k · a) = σ(k)σ(a) = kσ(a), so σ is K-linear. Corollary 27. If σ ∈ G and α ∈ F is a root to f(x) ∈ K[x], then σ(α) is also a root to f(x). Proof. The map σ induces an automorphism of F [x]. On the one hand, σf = f since σ fixes K point-wise, and on the other, f(x) = (x−α)g(x) for some g(x) ∈ K[x] and so f = σf = (x − σ(α))σg. ¤ As a consequence, the more roots to a given f(x), F possesses, the more F has. Note that if θ : F → F is an field au- tomorphism and F contains Q, then nθ(m/n) = f(m) = m and so θ(m/n) = m/n. I.e., θ fixes Q automatically. √ 3 Example 9. The√ extension√ field Q[ 2] of Q contains√ a single root 3 3 2kπι/3 3 of x − 2 while Q[ 2, 3ι] contains√ all roots e 2, k = 0, 1, 2 to 3 3 x − 2. The Galois group of Q[√2]√over Q is√ the√ trivial group,√ while 3 3 3 there√ is an automorphism σ : Q[ 2, 3ι] → Q[ 2, 3ι] sending 2 to e2πι/3 3 2. To see this, by the last theorem, there is an isomorphism √ √ θ : Q[ 3 2] ∼= Q[e2πι/3 3 2], √ √ sending 3 2] 7→ e2πι/3 3 2. By the theorem again, this isomorphism extends to an automorphism of √ √ Q[ 3 2, 3ι] = √ √ √ √ (Q[ 3 2])[ 3ι] = (Q[e2πι/3 3 2])[ 3ι] √ √ extending θ and sending 3ι 7→ 3ι.

Example√ 10.√ In the last√ example we√ constructed an√ automorphism√ of F = Q[ 3 2, 3ι] sending 3 2 to e2πι/3 3 2 and sending 3ι to 3ι. Like- wise, there are 5 other automorphisms of F ; calling the one mentioned above, σ1, we have: 34 RINGS AND FIELDS ‘07 √ √ √ √ 3 3 σ0 : √ 2 7→ 2√ √3ι 7→ √3ι 3 2πι/3 3 σ1 : √2 7→ e √2 √ 3ι 7→ √3ι 3 2πι/3 3 σ2 : √2 7→ e √2 √3ι 7→ −√ 3ι 3 4πι/3 3 σ3 : √2 7→ e √2 √ 3ι 7→ √3ι 3 4πι/3 3 σ4 : 2√7→ e √ 2 √3ι 7→ −√3ι 3 3 σ5 : 2 7→ 2 3ι 7→ − 3ι Note that F is the splitting field of x3 − 2 and is of degree 6 over Q, and there are 6 distinct automorphisms of F . The splitting field of xp − 1 has degree p. The splitting field in this case is Q[ρ] where ρ = e2πι/p. Since xp −1 is irreducible (undergraduate exercise; (x + 1)p − 1 is irreducible by Eisenstein’s Criterion) with ρ as a root, Q[ρ] has degree p over Q). So the obvious conjecture that the splitting field of an irreducible polynomial of degree n has degree n! is incorrect. The Fundamental Theorem of Galois is a detailed description of the correspondence between of G and subfields of F containing K. Given a H of G, H0 = { t ∈ F | θ(t) = t ∀ θ ∈ H }. Given a subfield E of F containing K, 0 E = AutE(F ) = {θ ∈ G | θ(t) = t ∀ t ∈ E }. In Hungerford, the field F is called a of K, provided G0 = K. √ 3 Example 11. The extension√ field Q[ 2] of Q has Galois group { 1F } 0 3 and { 1F } = F 6= Q, so√Q[√ 2] is not a Galois extension of Q. We will show shortly that Q[ 3 2, 3ι] is Galois over Q. Fundamental Theorem of Galois . Let F be a finite dimensional Galois extension of K, and let G be the Galois group of F over K. (i) There is a bijective, order-reversing correspondence between the subfields of F containing K and the subgroups of G given by 0 E 7→ AutE(F ) and H ≤ G 7→ H ≤ F. (ii) Under this correspondence, an intermediate field K ≤ E ≤ F corresponds to a normal subgroup of G if and only if E is Galois over K; and [E : K] is the index of AutE(F ) in G. RINGS AND FIELDS ‘07 35

14. EXERCISE SET 6 1. Let α ∈ F . Show that K[α] is a field if it has finite dimension over K. 2. Let α, β ∈ F be algebraic over K having degrees n and m respectively. Show that [K[α, β]: K] ≤ m · n, and equality holds if n and m are relatively prime. 3. Let f(x), g(x) ∈ K[x] and let F be an extension field of K. Show that the gcd of f and g is 1 in F [x] if and onlt if the gcd of f and g is 1 in K[x]. 4. A irreducible polynomial f(x) ∈ K[x] is called separable if f(x) = a(x−α1)(x−α2) ··· (x−αk) with a ∈ K and α1, α2, . . . , αk distinct, in E for any splitting field E for f(x). Show that an irreducible polynomial f(x) ∈ K[x] is separable if and only the gcd of f and its derivative f 0 is 1 in K[x]. 5. Recall that a finite field is a field extension of Zp for some p and consequently must have order pn for some n. Show that F is a finite field if and only if F is the splitting field of xpn − x for some prime p and natural number n. Consequently, two finite fields F1,F2 are isomorphic if and only if |F1| = |F2|. (Hint: ⇒ If |F | = pn, note that F ∗ = F \{0} is a finite of order pn−1. Furthermore, f = xpn −x has distinct roots (consider the gcd of f with its derivative f 0). ⇐ In the splitting field of xpn − x, the roots form a field; hence the splitting field is the collection of the roots.) 6. Let E,E0 be subfields of F containing K and suppose σ : E ∼= E0 is an isomorphism fixing K. Assume that F is the splitting field of a f(x) ∈ K[x]. Using Kronecker’s Theorem and induction on [F : E], argue that there are [F : E] automorphismsσ ˜ ∈ G such thatσ ˜|E = σ. 7. Diminishing Returns: Let G be the Galois group of F over 0 0 K. Show that (AutH0 (F )) = H for every H ≤ G and that 0 AutE(F ) = Aut(AutE (F )) (F ) for every intermediate field K ≤ E ≤ F . 36 RINGS AND FIELDS ‘07

15. The Fundamental Theorem of Galois

Before embarking on a proof of the Main Theorem of Galois, we will examine the hypothesis of that result. Let us recall the Fundamental Theorem of : If M is an m × n matrix over a field, then rank M + nullity M = n. To prove this, row reduce M. The number of columns in the row- reduced form that do not contain a leading nonzero entry of a row is the nullity of M (i.e., the dimension of the nullspace of A), and the number of columns containing a leading nonzero entry of a row is the rank of M (rank M = dimension of the column space of M = dimension of the row space of M). Assume [F : K] < ∞ throughout, and let

G = Gal(F/K) = AutK (F ). Remember that an automorphism σ : F → F fixes K if and only if σ is K-linear.

A Lemma of Dedekind . If σ1, σ2, . . . , σm ∈ G are distinct, then they are linearly independent.

Proof. The proof will be by induction on m; clearly the case m = 1 is settled in the affirmative. Suppose, for the sake of induction, that

(1) a1σ1 + a2σ2 + ··· + amσm = 0, 0 0 with not all ais equal to zero. By induction, we assume that all ais are −1 nonzero. Multiplying through by am we may assume that am = 1. Since σ1 6= σn, there exists t ∈ F such that σ1(t) 6= σm(t). Evaluating −1 (1) at ts for s ∈ F arbitrary, and multiplying both sides by σm(t) , we obtain −1 σm(t) [a1σ1(ts) + a2σ2(ts) + ··· + σm(ts)] = −1 −1 a1σm(t) σ1(t)σ1(s) + ··· + a1σm(t) σm−1(t)σm−1(s) + σm(s) = 0. Evaluating (1) at s and subtracting this last , we have −1 −1 a1[1 − σm(t) σ1(t)]σ1(s) + a2[1 − σm(t) σ2(t)]σ2(s)+ −1 ··· + am−1[1 − σm(t) σ1(t)]σm−1(s) = 0, RINGS AND FIELDS ‘07 37 for every s ∈ F . That is, −1 −1 a1[1 − σm(t) σ1(t)]σ1 + ··· + am−1[1 − σm(t) σ1(t)]σm−1 = 0.

This contradicts the inductive hypothesis since, in particular, a1[1 − −1 σm(t) σ1(t)] 6= 0. ¤

Lemma 28. If H ≤ G, then [F : H0] ≥ |H|.

Proof. Let H consist of σ1, σ2, . . . , σm, and let u1, u2, . . . , uk be a basis for F over H0. Suppose to the contrary that m > k, and consider the system of linear

σ1(u1)x1 + σ2(u1)x2 + ··· + σm(u1)xm = 0

σ1(u2)x1 + σ2(u2)x2 + ··· + σm(u2)xm = 0 ···············

σ1(uk)x1 + σ2(uk)x2 + ··· + σm(uk)xm = 0 over F . We know there exists a nontrivial solution (x1, x2, . . . , xm). 0 th For any β ∈ F , write β = Σjbjuj for bj ∈ H . Multiplying the i equation by bi and adding we obtain

σ1(β)x1 + σ2(β)x2 + ··· + σm(β)xm = 0 contradicting Dedekind’s Lemma. ¤

Theorem 29. If H ≤ G, then [F : H0] = |H|.

Proof. It remains to show that [F : H0] ≤ |H|.

Let H consist of σ1, σ2, . . . , σm, and suppose that F contains m + 0 1 elements u1, u2, . . . , um+1 which are linearly independent over H . Consider the system of linear equations

σ1(u1)x1 + σ1(u2)x2 + ··· + σ1(um+1)xm+1 = 0 ···············

σm(u1)x1 + σm(u2)x2 + ··· + σm(um+1)xm+1 = 0 Let j be the minimal number of nonzero components in a nontriv- ial solution to the system. Choose a nontrivial solution with j terms 0 and reorder the ujs (if necessary) so that this solution is of the form 38 RINGS AND FIELDS ‘07

(a1, a2, . . . , aj, 0,..., 0) such that a1, . . . , aj are nonzero, and aj = 1. 0 0 Note that j > 1. Because any σi is H -linear, not all of the ais 0 0 0 can belong to H . Assume (by reordering the uis) that a1 ∈/ H ; so, σk(a1) 6= a1 for some k. th Applying σk to the ` -row in the system

(1) σ`(u1)a1 + σ`(u2)a2 + ··· + σ`(uj)aj = 0 we obtain

σkσ`(u1)σk(a1) + σkσ`(u2)σk(a2) + ··· + σkσ`(uj)σk(aj) = 0.

But σkH = H, and so for every i there exists an ` such that σkσ` = σi. Subtract the equation

σi(u1)σk(a1) + σi(u2)σk(a2) + ··· + σi(uj)σk(aj) = 0. from the equation pointed to by (1) but corresponding to i to obtain a new system whose ith row is

σi(u1)[a1 − σk(a1)] + ··· + σi(uj)[aj − σk(aj)] =

σi(u1)[a1 − σk(a1)] + ··· + σi(uj−1)[aj−1 − σk(aj−1)] = 0, since aj = 1. But a1 − σk(a1) 6= 0, so we have found a solution to the original system with fewer than j nonzero terms; a contradiction. ¤ The hypothesis of the Fundamental Theorem of Galois poses a re- striction on the field extension F (unlike the previous results of this section). F is called a Galois extension of K if F satisfies any (hence all) of the conditions of the next result.

Galois Extension Theorem . The following are equivalent: (1) [F : K] = |G|. (2) G0 = K. (3) Every monic irreducible polynomial in K[x] that has a root in F , is separable and splits in F [x]. (4) F is the splitting field of a separable polynomial in K[x].

Proof. (1) → (2) By the last theorem, [F : G0] = |G|. Since |G| = [F : K] = [F : G0][G0 : K] = |G|[G0 : K], [G0 : K] must be 1. RINGS AND FIELDS ‘07 39

(2) → (3) By the previous theorem, [F : G0] = |G| = [F : K]. Let p(x) ∈ K[x] be monic, irreducible, with a root α in F , and consider

g(x) = Πj(x − αi), where α1, . . . , αk are the distinct members of the set {σ(α) | σ ∈ G}. If we apply any σ ∈ G to the coefficients of g(x) we find that σ fixes g(x), and therefore g(x) ∈ G0[x] = K[x]. Since the gcd of g(x) and p(x) in F [x] is not 1, the gcd of g(x) and p(x) in K[x] cannot be 1. Therefore, p divides g implying that p is separable. (3) → (4) If α ∈ F , then α is algebraic and is the root of a separable (irreducible) polynomial p(x). If the splitting field E of p(x) inside F , differs from F , let β ∈ F \ E; β is a root to a separable polynomial p2(x) ∈ K[x] ≤ E[x], and the splitting field of p(x)p2(x) inside F properly contains E. Since [F : K] < ∞, we will obtain F as a splitting field of a separable polynomial in K[x] after a finite number of steps. (4) → (1) By Exercise 6.6, there are [F : K] automorphisms which extend the identity map 1K : K → K. I.e., [F : K] = |G|. ¤ Let Sub G denote the set of subgroups of G and Int F/K the ( of) intermediate subfields of F containing K. Both sets Sub G and Int G have containment as the partial order.

Fundamental Theorem of Galois . Let F be a Galois extension of K. Then: (i) The map Sub G → Int F/K given by H 7→ H0 is an order- reversing with inverse E 7→ AutE(F ). 0 (ii) AutH0 (F ) = H and E = AutE(F ) . 0 (iii) [E : K] = [G : AutE(F )] and [G : H] = [H : K]. (iv) E is Galois over K if and only if AutE(F ) is normal in G.

0 0 0 0 0 Proof. If H1 = H2, then it is easy to see that (H1H2) = H1 = H2. Theorem 29 reveals 0 0 [F :(H1H2) ] = |H1H2| = [F : H1] = |H1|.

Thus, H1 = H1H2. Similarly H2 = H1H2 = H1. Note; the hypothesis was not needed for this part. We next argue that AutE1 (F ) = AutE2 (F ) implies E1 = E2. Now, E1E2 (consists of finite sums of products a1a2, ai ∈ Ei) is finite dimensional over K, hence is a subfield of F containing both E1,E2. It readily checks that AutE1E2 (F ) = AutEi (F ), so we may assume that E1 ⊆ E2, in order to show that E1 = E2. 40 RINGS AND FIELDS ‘07

If there exists an α ∈ E2 \ E1, then α is a root to an irreducible and separable polynomial p(x) ∈ K[x] which splits in F (Galois Extension Theorem). Let g(x) be an irreducible factor of p(x) in E1[x] that has α as a root. Then g(x) is separable (it divides p(x)) and has degree at least 2, so there exists another root α0 to g(x) in F . By Exercise 0 6.6, there exists σ ∈ G such that σ fixes E1 and σ(α) = α . Then

σ ∈ AutE1 (F ) but is not in AutE2 (F ). Thus, E1 = E2. To finish (i) we need to establish (ii). But (ii) follows from Exercise 0 0 6.7; we know H = (AutH0 (F )) and so by what we have shown above, 0 H = AutH0 (F ). Similarly, E = (AutE(F )) . We next verify the claims about the indices: By Theorem 29, [F : H0] = |H| and by Lagrange’s Theorem and Theorem 25, [H0 : K] = [G : H]. Putting in E = H0 we obtain [E : K] = [G : AutE(F )] since H = AutE(F ) from (ii). If E is Galois over K, then E is the splitting field of a separable polynomial g(x) in K[x]. If σ ∈ G, then σ must permute the roots of g(x) in some fashion, and so σ(E) ⊆ E. Therefore σ−1δσ(a) = −1 −1 σ σ(a) = a for every a ∈ E and σ δσ ∈ AutE(F ), for δ ∈ AutE(F ). Conversely, suppose p(x) ∈ K[x] is irreducible with a root α ∈ E (we already know that p is separable since F is Galois), yet another root α0 of p is in F \ E. Then, there exists σ ∈ G such that σ(α) = α0 and −1 0 σ fixes E by Exercise 6.6. For H = AutE(F ), H1 = σHσ fixes α . Since 0 0 [F : H1] = |H1| = |H| = [F : H ] = [F : E], 0 it is impossible for H1 to contain E. Hence H1 6= H. I.e., if E is not Galois, then AutE(F ) is not normal in G. ¤ RINGS AND FIELDS ‘07 41

16. The Galois Group, Finite Fields and Simple Extensions Every finite dimensional extension F of K is

F = K[α1, α2, . . . , αk], for some elements α1, α2, . . . , αk, algebraic over K. It turns out, in the case of a Galois extension, that F = K[α] for a single algebraic element α. This is also the case in the context Q ⊆ K. Lemma 30. Any finite subgroup of F ∗ = F \ 0 is cyclic. Proof. Observe that an abelian group C of order n is cyclic if and only if for each divisor d of n, C has at most 1 cyclic subgroup of order d. To see this, if C = hci is cyclic (written multiplicatively), then hcn/di is the only cyclic subgroup of order d. Conversely, write C = C1×C2×· · ·×Cj 0 with Ci primary. Then, C is cyclic if and only if the Cis correspond to distinct primes. But, if Ci and C` are p-primary for a given prime p, then C has two cyclic subgroups of order p contrary to the assumption. Let C be a subgroup of order n of the F ∗ and let d divide n. Any cyclic subgroup B of F ∗ of order d consists of the roots of xd − 1 in F . Therefore, there is at most one such subgroup B and so C can have at most one cyclic subgroup of order d. Thus, C is cyclic. ¤ Corollary 31. If F is a finite field, then F ∗ is cyclic. Primitive Element Theorem . If F is a finite dimensional Galois extension of K, then F = K[α], for some algebraic element α. Proof. By the lemma, it suffices to assume that F is infinite. It suffices to argue that for α, β ∈ F , that K[α, β] = K[δ] for some δ ∈ F . By the Galois Correspondence Theorem, there are only finitely many intermediate subfields of F containing K. On the other hand, there are infinitely many distinct elements of the form α + aβ for 0 6= a ∈ F . Choose two such elements; α + aβ, α + bβ with a 6= b, such that K[α + aβ] = K[α + bβ]. Evidently, α, β ∈ K[α+aβ], and so δ = α+aβ satisfies our purpose. ¤ 42 RINGS AND FIELDS ‘07

If F is the splitting field of a separable polynomial f(x) ∈ K[x] of degree n, then clearly |G| = [F : K] ≤ n!, since any σ ∈ G must permute the roots of f(x) and is determined by how σ permutes the roots of f(x). That is,

G ≤ Sn, the on n letters.

Theorem 32. If F is the splitting field of an irreducible polynomial f(x) of degree p (a prime) over Q, and f(x) has exactly two non-real roots, then ∼ G = Sp. Proof. Since |G| = [F : Q] is divisible by p, G has an element σ of order p by Cauchy’s Theorem. We now argue that σ is a cycle of length p: Express σ as a product of disjoint cycles δ1 ··· δi; the order 0 of σ, p, is the least common multiple of the lengths of the δjs; hence one of the δ0s (hence σ) is a cycle of length p. Let E be the subfield of F generated by all of the real roots of f and let α andα ¯ be the two complex roots. Since F is the splitting field of f over E, the irreducible polynomial for α over E must have degree 2 (i.e., the irreducible factor of f in E[x] having α as a root must be of degree 2 since the only roots that do not split out of f over E are α andα ¯). There is an automorphism of F fixing E but sending α 7→ α¯. In summary, under the identification of G with Sp, G contains a transposition and a cycle of length p. Since Sp is generated by any such pair, G = Sp. ¤

Definition . Given a field extension K of Q, a polynomial f(x) ∈ K is said to be solvable by radicals if the roots of f(x) can be expressed using algebraic combinations and radical combinations of the coefficients of f(x).

Example 12. For f(x) = x3 +ax+b, the roots of f(x) are given below: A + B A − B √ A + B A − B √ x = A + B = − + −3 = − − −3, 2 2 2 2 r q r q 3 b b2 a3 3 b b2 a3 where A = − 2 + 4 + 27 , and B = − 2 − 4 + 27 . The general cubic x3 + ux2 + vx + w can be transformed into a cubic in the RINGS AND FIELDS ‘07 43 above form by setting 1 1 a = (3v − u2), and b = (2u3 − 9uv + 27w). 3 27

Example 13. Given the general quartic equation x4 +ax3 +bx2 +cx+d, the roots can be found as follows: Let y be any root to y3 − by2 + (ac − 4d)y − a2d + 4bd − c2; and s a2 R = − b + y. 4 If R 6= 0, take s 3a2 4ab − 8c − a3 D = − R2 − 2b + 4 4R and s 3a2 4ab − 8c − a3 E = − R2 − 2b − . 4 4R If R = 0, take s 3a2 q D = − R2 − 2b + 2 y2 − 4d 4 and s 3a2 q E = − R2 − 2b − +2 y2 − 4d. 4 The four roots of the original quartic are a R D a R E x = − + ± and − − ± . 4 2 2 4 2 2

Recall that a finite group G is solvable if there exists a chain

G0 = h1i < G1 < ··· < Gn = G such that Gi is normal in Gi+1, and Gi+1/Gi is cyclic for all i. 44 RINGS AND FIELDS ‘07

Big Theorem of Galois . Let K be a subfield of C, and let f(x) ∈ K[x]. Then, f(x) is solvable by radicals if and only if the Galois group of the splitting field of f(x) over K is a .

Example 14. The polynomial x5 − 4x2 + 2 ∈ Q[x] is irreducible by Eisenstein’s Criterion, and has exactly 2 non-real roots. Hence, the 5 2 Galois group of x − 4x + 2 over Q is S5; a group that is not solvable. Therefore, the polynomial is not solvable by radicals.

Example 15. The polynomial f(x) = xp − 1 splits into (x − 1)(xp−1 + ··· + 1). For any prime p, g(x) = xp−1 + ··· + x + 1 is irreducible. To see this observe that f(x + 1) = xg(x + 1) is equal to à ! à ! à ! p p p xp + xp−1 + xp−2 + ··· + x + 1 − 1. 1 2 p − 1 ³ ´ ³ ´ ³ ´ So g(x + 1) = xp−1 + p xp−2 + p xp−3 + ··· + p . 1 2 ³ ´ p−1 p The coefficient of xj, j < p − 1, is p−j and is divisible by p, while the constant term, p, is not divisible by p2. So g(x + 1), hence g, is irreducible. Therefore, the splitting field F = Q[ρ] of f(x) (and g) has degree p − 1 where ρ = e2πι/p. The Galois group of f (and g) is abelian ∗ and is isomorphic to the Z(p). The isomorphism sends ∗ j σ ∈ G to j ∈ Z(p) where σ(ρ) = ρ .

Theorem 33. Let K be a subfield of C. The splitting field of xn − 1 over K has an abelian Galois group of order at most ϕ(n) (where ϕ is the Euler phi-function). If K = Q, then G has order ϕ(n). Proof. The splitting field F is obtained as F = K[ρ] where ρ = e2πι/n. If σ ∈ G (the Galois group), then σ(ρ) = ρj. Evidently, ρj must be a primitive nth , and so j must be relatively prime to n. I.e., ∗ u ∗ the map from G into Z(n) has image in Z(n), the group of units in Z(n). This establishes the first part. Set gm(x) = Πδ(x − δ), where the product is indexed over all primitive mth roots of unity. Evidently n (x − 1) = Πd|ngd(x).

We will prove by induction that gd(x) ∈ Z[x]. RINGS AND FIELDS ‘07 45

For the sake of induction, set f(x) = Π{(x−δ) | δ is a primitive dth− root of unity, for d|n, d < n} = Πd|n,d

For a sketch of the proof of Galois’ Big Theorem, first assume that f is solvable by radicals. By the definition of f being solvable by radicals, the splitting field F of f is related to K as follows:

K = E0 ⊆ E1 = E0[α1] ⊆ E2 = E1[α2] ⊆ · · · ⊆ En = En−1[αn], such that En contains the splitting field F , and for each j there exists mj th mj with αj ∈ Ej−1. Note that if ρ is a primitive k root of unity and δ is a primitive `th root, then δρ is a primitive lcm{k, `}th root of unity. With regard to this chain, we will assume that α1 is a primitive m = lcm{m1, m2, . . . , mn} root of unity, and furthermore, that each mj is a prime. Because of this, note that each Ej is the splitting field mj over Ej−1 (it splits x − αj); and an easy proof by induction using the Galois Extension Theorem, shows that En is a splitting field over K. Because F is Galois over K, AutF (En) is a normal subgroup of the Galois group H of En over K, and the Galois group of f is isomorphic to H/AutF (En). Since quotient groups of solvable groups are solvable, it is enough to argue that H is solvable. Since each Ej is a splitting field over Ej−1 of prime degree, we obtain

AutEn (En) ≤ AutEn−1 (En) ≤ AutEn−2 (En) ≤ · · · ≤ AutK (En) = H, ∼ with AutEj−1 (En)/AutEj (En) = AutEj−1 (Ej) primary, cyclic. Hence H is solvable. 46 RINGS AND FIELDS ‘07

17. EXERCISE SET 7 √ √ 1. In the Galois extension F = Q[ 3 2, 3ι] over K = Q, find all intermediate subfields. Acknowledge the subfields that are Galois over K 2. Do the same as in 1. but for x4 − 5 over Q. 3. Determine the Galois group of (the splitting field for) (x3 − 2)(x2 − 3) over Q. 4. Assume that F is Galois over K. Show that [H1 : H2] = 0 0 [H2 : H1] for any subgroups H2 ≤ H1 of G, and [E1 : E2] =

[AutE2 (F ): AutE1 (F )] for any intermediate fields E2 ≤ E1 between F and K. 5. Let G be the Galois group of the polynomial x11 − 1 over the rational field Q. Determine this well-known group. Show all of the necessary work. 6. Determine the Galois group of the polynomial x10 − 1 over the rational field Q. Show all of the necessary work. 7. Determine the Galois group of x5 −6x+3 over the rational field Q. Show all of the necessary work.

Top View