sign in sign up
<<
Home , Polynomial ring, Principal ideal, Principal ideal ring, Rng (algebra)

Matematisk Institut Mat 3AL 2.1 Chapter II. Rings and

PREREQUISITES ABOUT RINGS AND IDEALS.

A R is a non-empty with two compositions + and · such that 1) (R, +) is an abelian ; 2) (R, ·) is associative; 3) The distributive laws hold: a · (b + c)=a · b + a · c (a + b) · c = a · c + b · c. The neutral element in R with respect to addition + is denoted by 0. A neutral element with respect to · is uniquely determined and is called the or element and is usually denoted by 1 (or sometimes by e). If (R, ·) is commutative, R is called commutative In the following we shall only consider commutative rings with identity element. We recall some basic definitions and theorems which should be known from Anders Thorup Mat 2ALG. For any element a of R the distributive law implies:

a · 0=a · (0 + 0) = a · 0+a · 0, hence a · 0=0.Thusa · b = 0 if either a =0orb =0. A for which the converse holds, i.e. a commutative ring for which the of two elements a and b can only be 0 if either a =0orb =0,is called an integral . A commutative ring R (with at least two elements) is called a field, if every non- a in R has an inverse, i.e. an element a−1 ∈ R such that a · a−1 =1. Such an inverse is necessarily uniquely determined. (Why?) A field is necessarily an . Indeed, assume a =0and a · b =0.Ifa has an inverse a−1 the equation a · b = 0 implies 1 · b = a · a−1 · b = 0; hence b =0. Exercise 2.1. Prove that an integral domain R (with at least two elements) is afieldifR has only finitely many elements. (Hint: if a is a non-zero element prove that the mapping x → a · x from R into itself is injective.) For any integral domain R there exists a fraction field K, i.e. a field K containing R as a such that every element x ∈ K is a quotient between two elements of R, i.e. there exist elements a and b of R, a =0forwhich a · x = b. Example 2.2. The rational numbers form a field Q which is the fraction field of the integral domain Z of all ordinary . Example 2.3.LetK be a field. The ring K[x]isanintegraldomain. K[x] is not a field; the field of rational functions K(x) is the fraction field of K[x]. Matematisk Institut Mat 3AL 2.2

The notion of “” is fundamental in . Definition 2.4. An additive I of (R, +), R being a commutative ring with an identity element, is called an ideal,ifRI ⊆ I, i.e. if r · a ∈ I for every r ∈ R and every a ∈ I. If I is an ideal of R the cosets (or residue classes) r + I = {r + a | a ∈ I} form a commutative ring, called the residue class ring (or )ofR with respect to I. This ring is denoted R/I. If r is an element of R the residue class containing r is often denoted byr ¯.If r¯ ands ¯ are residue classes containing the elements r and s from R the sum, resp. product ofr ¯ ands ¯ is defined byr ¯ +¯s = r + s,resp.¯r · s¯ = r · s. Amapφ from one ring R to another ring S is called a if φ sends sums into sums and products to products: φ(r1 + r2)=φ(r1)+φ(r2)and φ(r1 · r2)=φ(r1) · φ(r2) for all r1 and r2 in R. The subset I := {r ∈ R | φ(r)=0} of R is an ideal of R, called the of φ and denoted by Ker(φ). For elements in the residue class ring R/I we use the bar notation (as in (cf. Remark 1.51) by lettingr ¯ denote the residue class r + I The map R → R/I, defined by r → r¯ is obviously a homomorphism whose kernel is I. This map is called the canonical epimorphism from R to R/I. For ring there is an theorem quite analogous to Theorem 1.65 for groups. The proof is omitted since it is an obvious modification of the corresponding proof in the group theoretical case. Theorem 2.5. Let φ be a surjective homomorphism from the ring R onto the ring S.IfI is the kernel of φ, then there is an isomorphism ψ from R/I to S.Ifκ is the canonical homomorphism from R to R/I the isomorphism ψ can be chosen such that ψ = φ ◦ κ.

Exercise 2.6. In every ring R clearly R and the set {0} are ideals. Prove that a commutative ring R (with at least two elements) is a field if and only if R and {0} are the only ideals of R. Definition 2.7.AnidealI of R is called a if there exists an element a in I such that I = {ra | r ∈ R}.ThisisusuallywrittenI = Ra. Example 2.8. In the ring Z of integers, every ideal has the form Zn for some n in Z. Hence every ideal of Z is principal. Example 2.9.IfK is a field every ideal in the K[x] has the form K[x] · f(x) for some polynomial f(x)inK[x]. Hence every ideal of K[x]isprincipal. Definition 2.10. R is called a (”PIR”) if every ideal of R is a principal ideal. Matematisk Institut Mat 3AL 2.3

Definition 2.11. R is called a ,(”PID”)ifR is an integral domain and every ideal of R is a principal ideal. Example 2.12. Z and the polynomial ring K[x]overafieldK are principal ideal domains. The residue class ring Z/4Z is not an integral domain, but every ideal is principal. Hence Z/4Z is a principal ideal ring, but not a principal ideal domain. While the polynomial ring in one over a field is a principal deal domain, the polynomial ring in more than one variable is not a principal ideal domain. Actu- ally, in that case the ideal consisting of all polynomials with constant term 0 is not principal. Next we introduce some important notions for ideals reflecting properties of the corresponding residue class rings. Definition 2.13.AnidealI of the ring R, I = R, is called a prime ideal if the product of two elements a and b in R can only belong to I if either a ∈ I or b ∈ I. Theorem 2.14. An ideal I of the ring R is a prime ideal if and only if R/I is an integral domain. Proof. ”if”: Assume R/I is an integral domain. Let a and b be any two elements of R not lying in I. The corresponding residue classes in R/I represented by a and b are non-zero in R/I;sinceR/I is an integral domain the product of these two residue classes is non-zero. This product is represented by a · b, which therefore cannot lie in I. Hence I is a prime ideal. ”only if”: Assume I is a prime ideal. Let a and b in R represent two non-zero residue classesa ¯ and ¯b in R/I.SinceI is a prime ideal, a · b is not in I.Nowa · b represents the producta ¯ · ¯b of the two residue classes which cannot be zero because a · b/∈ I. Hence R/I is an integral domain. 

Example 2.15. In the ring Z of ordinary integers the prime ideals are the principal ideals generated by prime numbers and the zero ideal. Definition 2.16.AnidealI of the ring R, I R, is called maximal,ifI is maximal among the ideals of R that are = R. Theorem 2.17. An ideal I of the ring R is a if and only if R/I is a field. Proof. ”if”: Assume R/I is a field. Let M be an ideal of R containing I as a proper subset. Choose an element a ∈ M \ I. The residue classa ¯ (with respect to I)is=0, hence has an inverse, say ¯b,(b ∈ R) in the field R/I. Therefore a · b can be written as 1+(an element ∈ I). But this means that M contains 1 and therefore equals R. ”Only if”: Assume I is a maximal ideal. Leta ¯,(a ∈ R) be a non-zero residue class in R/I. Clearly a is not in I.ThesetM of elements of the form a · r + c, r running through the elements of R and c running through the elements of I,isanidealM of R.Sincea is not in I,theidealM contains I strictly and hence M = R.This Matematisk Institut Mat 3AL 2.4

implies that 1 = a · r + c for some r in R and some c in I. But this in turn means that 1=¯¯ a · r¯ in R/I which therefore is a field. 

Example 2.18. In the ring R = Z[x]theidealI = {f(x) ∈ Z[x] | f(0) = 0} is a non-maximal prime ideal. In this connection the following holds. We omit the proof. Theorem 2.19. Every non-zero prime ideal in a principal ideal domain is maximal.

Important example 2.20 In the ring Z every non-zero prime ideal is the principal ideal generated by a p. By the above result the residue class ring Z/Zp is a field. This can also be checked immediately since every finite integral domain (with more than one element) is necessarily a field. This field is sometimes denoted Zp and sometimes Fp. We shall also need the concept ”factorial ring” or ”ring with unique factorization”, denoted by UFD. We shall refrain from giving detailed proofs, but just explain the notions and the basic results in this subject. Let R be an integral domain. A non-zero element a in R which is not invertible is called an irreducible element if a has only trivial factorizations, i.e. a = r · s, r, s ∈ R implies that either r or s is invertible in R. Two non-zero elements a and b of an integral domain R are called associate if they generate the same principal ideal, i.e. if Ra = Rb. It is easily checked that a and b are associate if and only if a = b·(an invertible element) if and only if b = a·(an invertible element). R is called a UFD,ifR is an integral domain and every non-invertible non- zero element in a ”basically unique” way can be written as a product of irreducible elements. Here “basically unique” means the following: Assume an element a has two factorizations of irreducible elements:

a = π1 · π2 ···πn and a =˜π1 · π˜2 ···π˜m, then m = n and after a suitable permutation of the π’s every πi , i =1, 2,...,n, is associate toπ ˜i. A characteristic property of UFD’s is the following: If an irreducible element π divides a product a · b then π divides one of the factors a and b.Inotherwordsthe principal ideal generated by an irreducible element is a prime ideal. The classical examples of UFD’s are the ring Z of ordinary integers and the ring of polynomials in one variable over a field. Those two rings are PID’s. A general result is the following which we state without proof: Matematisk Institut Mat 3AL 2.5

Theorem 2.21. Every principal ideal domain is a unique factorization domain.

Remark 2.22. The converse is not true. The ring of polynomials in more than one variable over a field is not a principal ideal domain, but it can be shown to be a unique factorization domain.

In a UFD the principal ideal generated by an irreducible element is a non-zero prime ideal. If R is a PID an ideal I of R is a non-zero prime ideal exactly when I = Rπ for some irreducible element π in R. Combining this remark with Theorem 2.19 we get: Theorem 2.23. Let R be a principal ideal domain. A non-zero ideal of R is a prime ideal if and only if it has the form Rπ for some irreducible element π. For any non-zero prime ideal Rπ the residue class ring R/Rπ is a field.

FACTORIZATIONS OF POLYNOMIALS.

We consider the polynomial ring R[x], where R is either a field or a PID, (in particular for instance R being the ). The following observation is often useful: Let ϕ be a homomorphism of R onto R∗,whereR and R∗ are commutative rings. The mapping

n n Φ(r0 + r1x + ···+ rnx )=ϕ(r0)+ϕ(r1)x + ···+ ϕ(rn)x defines a homomorphism Φ of R[x]ontoR∗[x].  n Definition 2.24.Apolynomialf(x)=a0 + ···+ anx ∈ Z[x] is called primitive,if n the (a0,...,an) is 1. In particular, if f(x)=a0+···+anx ∈ Z[x], is monic, i.e. an =1,thenf(x) is primitive. Theorem 2.25. (Gauss). The product of two primitive polynomials f(x) and g(x) is primitive. Proof. Indirectly. Assume f(x) · g(x) were not primitive; then there would exist a prime number p such that p divides all coefficients of f(x)g(x). Consider the homomorphism ϕ : Z → Zp = Fp defined by sending an into its residue class modulo p.ThenΦ(f) =0andΦ( g) = 0, but Φ(f · g)=0,whichis a contradiction since Φ is a homomorphism and Zp[x] is an integral domain.  Lemma 2.26. Let f(x) be a primitive polynomial in Z[x],andq ∈ Q.Thenq = ±1 if qf(x) is a primitive polynomial. r ··· Proof. Write q = s ,wherer and s are mutually prime integers. Let f(x)=a0 + + n r n ra0 ran n anx .Thens (a0 + ···+ anx )= s + ···+ s x ∈ Z[x] ⇒ s|rai ∀i ⇒ s|ai ∀i ⇒ Matematisk Institut Mat 3AL 2.6 r a0 an n s = ±1(sincef(x) is primitive). Hence s f(x)=r s + ···+ s x is primitive ⇒ r = ±1. 

The proofs of the following lemmas are quite straightforward and are left to the reader.

Lemma 2.27. For any non-zero polynomial f(x) in Q[x] there exists a q such that qf(x) is primitive.

Lemma 2.28. Let f(x) be a polynomial in Z[x].Ifq is a rational number such that qf(x) is primitive then 1/q is an integer.

Lemma 2.29. If f(x) is primitive and f(x)=g(x)h(x), g(x) and h(x) being poly- nomials in Z[x],theng(x) and h(x) are primitive.

Remark 2.30. The invertible elements in Z[x]are±1.

Theorem 2.31. Let f(x) be a polynomial in Z[x].Then: i) Degree f(x)=0: f(x) is irreducible in Z[x] ⇔ f(x)=±p, p for a prime number p. ii) Degree f(x) > 0: f(x) is irreducible in Z[x] ⇔ f(x) is primitive and f(x) is irreducible in Q[x].

Proof. i) is clear. ii) ⇐ assume f(x) had a non-trivial factorization f(x)=g(x)h(x), g(x)and h(x) ∈ Z[x], where g(x), h(x)arenotinvertibleiZ[x]. Since f(x) is primitive the degrees of g(x)andh(x)mustbe> 0, hence f(x) would have a non-trivial decomposition in Q[x]. Contradiction! ⇒ It is clear, that f(x) must be primitive. Assume f(x) had a non-trivial factor- ization in Q[x]: f(x)=g(x)h(x), g(x)andh(x) not being constants. By lemma 2.27 there exist rational numbers q1 and q2 such that q1g(x)andq2h(x) are primitive. q1q2f(x)=[q1g(x)][q2h(x)] .

By Gauss’ theorem q1q2f(x) is primitive. Since f(x) is primitive, lemma 2.26 implies that q1q2 = ±1, and hence

f(x)=(± q1g(x)) · (q2h(x))

Thus f(x) would have a non-trivial factorization in Z[x]. Consequently f(x)must be irreducible as an element in Q[x].  Matematisk Institut Mat 3AL 2.7

Theorem 2.32. Z[x] is a UFD. Proof. 1) Every f(x) ∈ Z[x] is a product of irreducible polynomials. Indeed, if f(x) is constant, i.e. is an ordinary integer, then the assertion follows since Z is a UFD. If a polynomial in Z[x] has positive degree, it can be written as (a non-zero integer)·(a primitive polynomial). Therefore it suffices to show that any primitive polynomial f(x) is a product of irreducible polynomials. We proceed by induction on the degree of the polynomial. If the degree is one the assertion is clear. Assume the assertion has been proved for all primitive polynomials of degree

the homomorphism ϕ : Z → Zp and

the homomorphism Φ : Z[x] → Zp[x] Then 0 = Φ(f(x)g(x)) = Φ(f(x)) · Φ(g(x)). Since Zp[x]isanintegraldomain,Φ(f(x)) or Φ(g(x)) must be 0, i.e.: p|f(x)or p|g(x). ad ii) Since p(x)|f(x)g(x)inZ[x] it follows that p(x)|f(x)g(x)inQ[x]. Because p(x) is irreducible in Q[x]andQ[x]isUFD, we conclude that p(x)|f(x) or p(x)|g(x)inQ[x]. Assume for instance that p(x)|f(x)inQ[x]. Then we have f(x)=p(x)h(x) ,h(x) ∈ Q[x] . By lemma 2.27 there exists q in Q such that qh(x) is primitive. Consequently we get qf(x)=p(x) · (qh(x)) . As before (Gauss’ Theorem) qf(x) must be primitive. By lemma 2.28 we can write q =1/s for some integer s. Hence: h(x)=s(qh(x)) ∈ Z[x], and thus p(x)|f(x)in Z[x]. Similarly if p(x)|g(x)inQ[x] we conclude that p(x)|g(x)inZ[x]. 

Remark 2.33. With minor obvious modifications the above proof carries over to the case where Z is replaced by an arbitrary UFD. In other words, generally we have: Matematisk Institut Mat 3AL 2.8

Theorem 2.34. R is a UFD ⇒ R[x] is a UFD.

Corollary to Theorem 2.34. For any field K the polynomial ring K[x1,...,xn] is a UFD.

Remark 2.35. For later explicit applications a special case of Theorem 2.31 is very important: A in Z[x] is irreducible as an element in Q[x]ifand only if it irreducible as an element in Z[x].

Theorem 2.36 (Sch¨onemann-Eisenstein’s irreducibility criterion). Apoly- n n−1 nomial f(x) ∈ Z[x] of the form f(x)=x + an−1x + ···+ a1x + a0 is irreducible 2 in Q[x], if there exists a prime number p, such that p|an−1,...,p|a1, p|a0, p a0. Proof. By Theorem 2.31 (cf. the above remark 2.35) it suffices to show, that f(x)is irreducible in Z[x]. Assume there were a factorization f(x)=g(x) · h(x), where g(x)andh(x) ∈ Z[x] and g(x)andh(x) both were of degree

Φ(f(x)) = xn =Φ(g(x)) · Φ(h(x)) ⇒ Φ(g(x)) = xm ;Φ(h(x)) = xn−m

for a suitable m,1≤ m ≤ n − 1. This implies that all coefficients of g(x)andh(x) except the highest one are divisible by p;inparticularp would divide g(0) and h(0) 2 and thereby p |g(0)h(0) = f(0) = a0. Contradiction ! 

xn−1 Example 2.37 . x−1 is irreducible in Q[x] ⇔ n = prime number.

Proof. ⇒: assume n is composite n = n1n2,1

xn − 1 xn1n2 − 1 xn1 − 1 = = (xn1 )n2−1 + ···+ xn1 +1 x − 1 x − 1 x − 1 is reducible. ⇐ xp−1 ⇔ assume n is a prime number p.Thenf(x)= x−1 is irreducible f(x +1)is irreducible. Now by explicit expansion we get (x +1)p − 1 p p p f(x +1)= = xp−1 + xp−2 + ···+ x + . x 1 p − 2 p − 1 p 2 p Since p is a prime number, p divides j for 1 ≤ j ≤ p − 1, and p p−1 = p. xp−1  Hence Theorem 2.36 shows the irreducibility of x−1 . Matematisk Institut Mat 3AL 2.9

Remark 2.38. Theorem 2.36 easily carries over to polynomials over a UFD: Let R be UFD and K its field of fractions. Let π be an irreducible element in R n n−1 and f(x)=x + rn−1x + ···+ r0 a polynomial in R[x]forwhichπ divides ri, 2 0 ≤ i ≤ n − 1, while r0 is not divisible by π ,thenf(x) is irreducible in K[X]. Example 2.39. f(x, y)=xn +yn −1 is irreducible in Q[x, y] for any n. Indeed, just consider f(x, y +1)asanelementof(Q[y])[x] and use the above remark with π = y. n Exercise 2.40. f(x)=Πi=1(x − ai) − 1, where a1,...,an are distinct numbers in n Z, is irreducible in Q[x]. (Is there a similar result for Πi=1(x − ai)+1?) Finally we give an example where we use that every polynomial over Q has roots in the complex field C: n n−1 Example 2.41.Letf(x)=x + a1x + ···+ an−1x ± p be a polynomial in Z[x] where p is a prime number. Then f(x) is irreducible, if p>1+ | a1 | + ···+ | an−1 |. Proof. Assume f(x) had a non-trivial factorization f(x)=g(x)h(x)inZ[x]. Then f(0) = g(0)h(0) = ±p would imply that one of the factors g(0) or h(0), say g(0), were ±1. The product of the roots of g(x)wouldthenbe±1. In particular, there wouldbeatleastonerootα of absolute value  1 .Since α isalsoarootoff(x)we n n−1 would get: α + a1α + ···+ an−1α = ±p, which - by taking absolute values - would imply 1+ | a1 | + ···+ | an−1 | p. Contradiction!

Two philosophical questions: Let n be a “random” natural number. What is most probable: That n is a prime number or that n is a composite number? Let f(x) be a “random” polynomial in Q[x]. What is most probable: That f(x) is an or that f(x) is a reducible polynomial?

SYMMETRIC POLYNOMIALS.

Let K be a field. A polynomial f(x1, ..., xn) ∈ K[x1, ..., xn] is called symmetric, if it is invariant under every permutation of the indeterminates x1, ...xn, i.e. if

f(x1, ..., xn)=f(xσ(1), ..., xσ(n)) for every permutation σ ∈ Sn. Example 2.42. The polynomials ··· n s1 = x1 + + xn 1 terms ··· n s2 = x1x2 + x1x3 + + xn−1xn 2 terms ··· n s3 = x1x2x3 + + xn−2xn−1xn 3 terms ... n sn = x1x2 ···xn n terms Matematisk Institut Mat 3AL 2.10

are symmetric. This may either been seen directly or by considering the polynomial n n−1 n−2 n (T − x1)(T − x2) ···(T − xn)=T − s1T + s2T + ···+(−1) sn

which is invariant under every permutation of x1..., xn; therefore all the coefficients are also invariant under every permutation of x1, ..., xn.

These polynomials s1, ..., sn are called the elementary symmetric polynomials in x1, ..., xn. It is clear, that every polynomial in the elementary is symmetric in x1, ..., xn. The converse also holds true, as the following important theorem shows.

Theorem 2.43. The main theorem about symmetric polynomials.

Every symmetric polynomial f(x1, ..., xn) ∈ K[x1, ..., xn],whereK is an arbitrary field, can in exactly one way be written as a polynomial (with coefficients in K)in the elementary symmetric polynomials s1, ..., sn. In other words: for every symmetric polynomial f(x1, .., xn) ∈ K[x1, .., xn] there exists a unique polynomial g(y1, ..., yn) ∈ K[y1, ..., yn], such that

f(x1, ..., xn)=g(x1 + ···+ xn,x1x2 + ··· , ..., x1 ···xn).

Before giving the proof here are some general remarks about polynomials in several indeterminates. Every polynomial in x1, ..., xn can be written i1 in f(x1, ..., xn)= ai1,...,in x1 ...xn

i1,...,in

where ai1,··· ,in are elements in K. i 1 in  ··· The degree of a term ai1,...,in x1 ...xn ,ai1,...,in = 0 is defined as i1 + + in . By the degree of a non-zero polynomial f we mean the highest degree of a term (=0)in f. For polynomials in more than one the degree of the terms is not enough to determine an ordering of these. Therefore we introduce the concept sig- i 1 ··· in nature. By the signature of a term ai1,...,in x1 ...xn we mean the n- i =(i1, ..., in). We the set of signatures by first ordering them according to the degree and then ordering terms of the same degree lexicographically. This means that for two different signatures (i1, ..., in)and(j1, ..., jn)wehave

(i1, ..., in) ≺ (j1, ..., jn) if either i1 + ···+ in

Lemma 2.44. The product of two non-zero polynomials in K[x1, ..., xn],whereK is a field, is non-zero, and the term of highest signature in product is the product of the terms of highest signature in the two polynomials.

Proof. Exercise.

We now return to the proof of the main theorem. First the existence: Let f ∈ K[x1, ..., xn] be symmetric. We may assume f =0. i1 i2 in Let A = cx1 x2 ···xn be the term in f of highest signature. We remark that since f is symmetric, we get i1 ≥ i2 ≥ ... ≥ in. Indeed, assume iμ >iν for some μ>ν.Sincef is left fixed when permuting xμ and xν the polynomial i i1 i2 μ iν in f must also contain the term cx1 x2 ···xν ···xμ ···xn . But the signature of that i i1 i2 iν μ in term is higher than that of A = cx1 x2 ···xν ···xμ ···xn .

The terms of highest signature in the elementary symmetric polynomials s1,s2, ..., sn are x1, x1x2,..., x1x2 ···xn. By virtue of the above lemma for any non-negative integers j1,j2, ..., jn the term of j1 j2 jn highest signature in the product s1 s2 ···sn , viewed as a polynomial in x1,x2, ..., xn, is j1 j2 jn x1 (x1x2) ···(x1x2 ···xn) The corresponding signature is

(j1 + j2 + ···+ jn,j2 + ···+ jn, ..., jn)

If we now choose j1 = i1 − i2,j2 = i2 − i3, ..., jn−1 = in−1 − in and jn = in, (which in view of the above remark are non-negative integers), this signature is just (i1,i2, ..., in). j1 j2 jn The difference f − cs1 s2 ...sn therefore is either 0 or a symmetric polynomial (= 0) for which the term of highest signature is less than (i1,i2, ..., in). If this difference j1 j2 jn is 0 the proof is done. Otherwise we apply the same procedure on f − cs1 s2 ...sn etc.. In this way we eventually reach the zero polynomial, since there are only finitely many signatures less than a given one. Now the uniqueness: It clearly suffices to show that for g(y1, ..., yn) ∈ K[y1, ..., yn],g(y1, ..., yn) =0the polynomial g(s1, ..., sn)=g(x1 + x2 + ···+ xn,x1x2 + ··· , ..., x1x2 ···xn)isnotthe zero polynomial. t1 t2 tn Let dy1 y2 ...yn ,d =0beatermin g.Ifwefory1,y2, ..., yn substitute the elementary symmetric polynomials s1,s2, ...sn the argument above shows that we get a polynomial in x1,x2, ..., xn in which the term of highest signature is

t1 t2 tn dx1 (x1x2) ···(x1x2 ···xn) . Matematisk Institut Mat 3AL 2.12

The signature of this term is

(t1 + t2 + ···+ tn,t2 + ···+ tn, ..., tn).

If we now first consider the terms in g for which t1 + t2 + ···+ tn is biggest, then among these terms those for which t2 + ···+ tn is biggest, etc. in g we get t1 t2 tn separated a certain term dy1 y2 ...yn with the property that this and only this term by substitution of the elementary symmetric polynomials gives rise to a term with signature (t1 + t2 + ···+ tn,t2 ···+ tn, ..., tn). This term can not be cancelled away against any other term. Consequently g(x1 + x2 + ···+ xn,x1x2 + ··· ..., ..., x1x2 ···xn) is not the zero polynomial.  Matematisk Institut Mat 3AL 2.13

ALGEBRAIC EXTENSIONS. Let K be a subfield of the field L. For an element α ∈ L we denote by K[α] the smallest subring of L containing K and α. The ring K[α] consists of all elements in n L that can be written in the form k0 + k1α + ···+ knα , ki ∈ K, n ∈ N. By K(α)wedenotethe smallest subfield of L containing K and α. The field K(α) consists of all elements in L that can be written in the form

n k0 + k1α + ···+ knα    n , k0 + k1α + ···+ knα

 ki, ki ∈ K, n ∈ N and the denominator =0. K(α) is clearly the field of fractions of K[α]. For a given α ∈ L, L  K we consider the homomorphism Φ : K[x] → K[α] defined by Φ(f(x)) = f(α). Clearly Φ is surjective. We distinguish between two cases: 1) Ker Φ = 0, i.e. K[α]  K[x] which is not a field. In this case K(x) is isomorphic to K(α) and we say that α is transcendent over K.

2) Ker Φ = 0. Since Ker Φ is an ideal in K[x]andK[x]isaPID,thekernel Ker Φ is the principal ideal K[x]p(x) generated by a polynomial p(x) =0.Bythe epimorphism theorem for rings (Theorem 2.5) we get

K[x]/ Ker Φ  K[α].

Ker Φ is therefore a prime ideal =0.Since K[x] is a PID, Ker Φ is a maximal ideal (Theorem 2.19), therefore (Theorem 2.17) K[α] is a field and hence K(α)=K[α]. In this case α is called algebraic over K.Thatα is algebraic over K thus means, that α is root of a proper polynomial (i.e. = 0) with coefficients in K. Now Ker Φ = K[x]p(x), where p(x) is uniquely determined up to an invertible factor in K[x], i.e.: up to an constant in K. The uniquely determined monic (i.e.: highest coefficient = 1) polynomial in K[x] generates Ker Φ and is denoted Irr(α, K). This polynomial is irreducible in K[x], (cf. Theorem 2.23). Let f(x) ∈ K[x] be a polynomial for which f(α)=0.Thenf(x) = Irr(α, K) · g(x)forsomeg(x) ∈ K[x]. This implies, that Irr(α, K) can be characterized as the uniquely determined monic irreducible polynomial in K[x] having α as a root. Irr(α, K) can also be characterized as the uniquely determined monic polynomial in K[x] of lowest degree having α as a root. Therefore Irr(α, K) is sometimes called α’s minimal polynomial w.r.t. K. From the above we, in particular, conclude that a polynomial in K[x]hasα as a root if and only if it is divisible by Irr(α, K). Matematisk Institut Mat 3AL 2.14

Using the every polynomial f(x) ∈ K[x] can be written uniquely in the form

f(x) = Irr(α, K) · g(x)+r(x) , degree r(x)

If we in this equation insert α for x we get f(α)=r(α), hence every element in K[α] can be written

n−1 a0 + a1α + ···+ an−1α ,a0,...,an−1 ∈ K.

This representation is unique. Indeed, assume that an element β in K[α]hadthe representations n−1 β = a0 + a1α + ···+ an−1α

and    n−1 β = a0 + a1α + ···+ an−1α ,

  where a0,a1, ..., a0,a1, ... belong to K,thenα would be root of the polynomial    n−1 a0 − a0 +(a1 − a1)x + ···+(an−1 − an−1)x .Butα is not a root in any non-   zero polynomial i K[x]ofdegree

Definition 2.45.AnextensionL ⊇ K of fields is called algebraic, if every element in L is algebraic over K. We often just write: L/K is algebraic.

Before continuing we bring some general remarks, which we shall need a lot of times. If L is an extension field of K and α1, ··· ,αs are elements in L we denote by K(α1,... ,αs) the smallest subfield of L containing K and α1,... ,αs.Itisclear, that K(α1,... ,αs)=K(α1) ...(αs).

An extension field L of K can be considered as a vector space over K and thus has a dimension (i.e. the number of elements of a basis), which is denoted [L : K]. If this dimension is a finite number n any family of m elements ω1,... ,ωm, m>n, will be linearly dependent over K, i.e. there exist elements k1, ··· ,km ∈ K, not all 0 such that k1ω1 + ···+ kmωm =0. Matematisk Institut Mat 3AL 2.15

Theorem 2.46. Let L be a field extension of K for which [L : K] < ∞.ThenL/K is algebraic. Proof. Assume [L : K]=n<∞. For every element α in L the elements 1,α,α2,...,αn are linearly dependent over K; hence there exist elements k0,k1,...,kn in K, not all 0, such that n k0 + k1α + ···+ knα =0. Consequently α is a root of the non-zero polynomial

n k0 + k1x + ···+ knx ∈ K[x] .

Thus every element α in L is algebraic over K.  Theorem 2.47. (Transitivity Theorem). Let K  L  M be fields. Then [M : K]=[M : L][L : K],where[M : L] and [L : K] are assumed to be finite.

Proof. If α1,α2,...,αs form a K-basis for L and β1,β2,...,βt a L-basis for M,then αiβj,1 i  s,1 j  t,formaK-basis for M. For this we have to show two things: i) The elements αiβj ,1 i  s,1 j  t, are linearly independent over K. ii) Every element in M canbewrittenasaK- of the elements αiβj,1 i  s,1 j  t. Ad i) Assume kij αiβj =0, i,j where the kij ’s belong to K. This equation can be written ( kij αi)βj =0 j i

For each j the inner sum is an element in L. Since the elements βj, 1  j  n, are independent over L,weget kij αi =0 i for every j,1 j  n. Since the elements αi, 1  i  s, are linearly independent over K, we conclude that kij =0foralli, 1  i  s and all j, 1  j  t. Ad ii) Let ξ be an element in M.Sinceβj, 1  j  t form a L-basis for M,we can write ξ as ξ = jβj , j Matematisk Institut Mat 3AL 2.16

where each j lies in L.Sinceαi, 1  i  s,formaK-basis for L,each j can be written j = kij αi, i where every kij lies in K. This implies ξ = kij αiβj . i,j

Thus ii) has been proved. 

Remark 2.48. The transitivity theorem is extremely important in explicit compu- tations. A particular application is the following: Let M/K be a finite extension and L any field between K and M, K  L  M.Then[L : K] divides [M : K]. To take an explicit example√ let α be the real root of any irreducible polynomial in Q[x]ofdegree3.Then 2 is not contained in Q(α). (A direct verification without use of the above remark is not completely trivial.) Another example is the following. Let a and b be rational√ numbers√ such√ that neither a, b or ab is the of a rational number. Then a/∈ Q and b/∈ (Q( a)) (why?).√ √ This implies that√ √ √ √ [Q( a, b):Q]=[Q( a, b):Q( a)] · [Q( a):Q]=2· 2=4. Theorem 2.49. Let L beafieldextensionofK,andletα and β be elements in L which are algebraic over K.Thenα + β, α − β, α · β and α/β,(β =0 )arealso algebraic over K. Proof. K(α)=K[α] is a finite dimensional vector space over K.Sinceβ in particular is algebraic over K(α)=K[α], it follows that (K(α))(β)=K(α, β) is finite dimen- sional over K(α). By theorem 2.47 K(α, β) has finite dimension over K. Therefore by theorem 2.46 the field K(α, β) is algebraic over K. Since α + β, α − β, α · β and α/β,(β = 0) lie in K(α, β), these elements are algebraic over K. 

Definition 2.50.LetL be a field extension of K. The subset of L consisting of the elements that are algebraic over K (which by the above theorem is a subfield of L) is called the algebraic closure of K in L and is denoted K.

Example 2.51. There exist infinite dimensional algebraic field extensions.√ Let L = R and K = Q. For every natural number n theorem 2.36 implies that [Q( n 2) : Q]=n. Therefore the algebraic closure of Q in L = R has infinite dimension over Q. Theorem 2.52. Let K ⊆ L ⊆ M be fields. If L/K is algebraic, every element ξ ∈ M that is algebraic over L is also algebraic over K. Matematisk Institut Mat 3AL 2.17

Proof. ξ is root of a non-zero polynomial in L[x]:

n n−1 ξ + a0ξ + ···+ an−1 =0,a0,...,an−1 ∈ L.

Since a0 is algebraic over K, the dimension [K(a0):K] is finite. Since a1 is algebraic over K,inparticularoverK(a0), the dimension [K(a0,a1): K(a0)] is finite. Since a2 is algebraic over K,inparticularoverK(a0,a1), the dimension [K(a0,a1,a2): K(a0,a1)] is finite. etc.

Since an−1 is algebraic over K,inparticularoverK(a0,...,an−2), the dimension [K(a0,...,an−1):K(a0,...,an−2)] is finite. Since ξ is algebraic over K(a0,...,an−1), the dimension K(ξ,a0,...,an−1): K(a0,...,an−1)] is finite. By successive application of the transitivity theorem it follows that the dimension [K(ξ,a0,...,an−1):K] is finite; therefore Theorem 2.46 shows that ξ is algebraic over K.  Corollary 2.53. (Transitivity for algebraic extensions). Let K ⊆ L ⊆ M be fields. Then M/L algebraic ∧ L/K algebraic ⇒ M/K algebraic. = Corollary 2.54.. Let K ⊆ L be fields. Then K = K.

Exercise 2.55.Letα and β be complex numbers, that are algebraic over Q of degree p,resp. q. Show that α + β is algebraic over Q of degree pq,ifp and q are distinct prime numbers.

−−−−−−−−−−−−−−−

ADJUNCTION OF A ROOT. SPLITTING FIELDS. So far we have considered given existing field extensions and looked at polynomials vanishing on certain elements. Now we conversely consider a field K and a polynomial p(x) ∈ K[x] and are looking for extensions of K in which p(x)hasaroot. Theorem 2.56. Existence theorem concerning adjunction of a root of an irreducible polynomial. Let K be an arbitrary field and p(x)=a0 + a1x + ···+ n−1 n ∗ an−1x + x an irreducible monic polynomial in K[x]. Then there exists a field L containing a subfield K∗ with the following properties: 1) there exists an element α∗ in L∗ which is algebraic over K∗ such that L = K∗(α∗); 2) there exists an isomorphism ϕ of K onto K∗ such that Irr(α∗,K∗)=Φ(p(x)), where Φ is the isomorphism of K[x] onto K∗[x] induced by ϕ. Matematisk Institut Mat 3AL 2.18

Proof. Let L∗ be the residue class ring K[x]/K[x]p(x), which is a field, since p(x) is irreducible and K[x] is an PID. For a polynomial f(x) ∈ K[x] the corresponding residue class in L∗ is denoted f(x). Then:

∗ n−1 L = {k0 + k1x + ···+ kn−1x | k0,k1,...,kn−1 ∈ K} n−1 = {k0 + k1x + ···+ kn−1x | k0,k1,...,kn−1 ∈ K} .

We set K∗ = {k | k ∈ K}, which becomes a subfield of L∗. The map defined by ϕ(k)=k is an isomorphism of K onto K∗.Nowx is a root of the polynomial

n−1 n a0 + a1x + ···+ an−1x + x

which is irreducible in K∗[x]; hence

∗ n−1 n Irr(x, K )=a0 + a1x + ···+ an−1x + x =Φ(p(x)).

This proves the theorem, since x can be used as α∗. 

Remark to theorem 2.56 By identifying K with K∗ we may consider L∗ as an extension field of K in which p(x) has a root. An extension (in the more immediate sense of the word) can be obtained by considering the (set theoretical) disjoint union of K and L∗ \ K∗ and by the above construction ”transplanting” the operations addition and multiplication in this union; in this way one obtains a field containing K as a subfield and containing a root of p(x). Theorem 2.57. Uniqueness concerning adjunction of a root of an ir- reducible polynomial. Assume we have fields L, K, L∗,K∗, such that L ⊇ K, L∗ ⊇ K∗, L = K(α), L∗ = K∗(α∗),whereα is algebraic over K and α∗ algebraic over K∗. Assume further that there is an isomorphism ϕ from K onto K∗ such that Φ Irr(α, K) = Irr(α∗,K∗),whereΦ denotes the isomorphism from K[x] onto K∗[x] induced by ϕ. Then there exists a uniquely determined isomorphism ϕ from L onto L∗ such that 1) ϕRes,K = ϕ; 2) ϕ(α)=α∗.

Proof. Let p(x) = Irr(α, K)andp∗(x) = Irr(α∗,K∗) and assume these polynomials have the degree n. We first prove the uniqueness of ϕ.Evidently

n−1 L = {k0 + k1α + ···+ kn−1α | k0,k1,...,kn−1 ∈ K} .

If there exists an isomorphism ϕ with the properties 1) and 2), then

n−1 ∗ ∗ n−1 ϕ(k0 + k1α + ···+ kn−1α )=ϕ(k0)+ϕ(k1)α + ···+ ϕ(kn−1)(α ) . Matematisk Institut Mat 3AL 2.19

Therefore there is at most one possibility for ϕ. Next we prove the existence of ϕ. There is an isomorphism

ϕ1 : K[x]/(p(x) → L defined by ϕ1(f(x)) = f(α), f(x) ∈ K[x] and an isomorphism

∗ ∗ ∗ ϕ2 : K [x]/(p (x)) → L

∗ ∗ defined by ϕ2(g(x)) = g(α ) ,g(x) ∈ K [x]. Further the isomorphism

K[x] Φ- K∗[x] induces an isomorphism

e K[x]/(p(x)) Φ- K∗[x]/(p∗(x)) by Φ( f(x) modulo p(x)) = Φf(x) modulo p∗(x) , thus in particular we have

Φ(( x) modulo p(x)) = x modulo p∗(x) .

−1 ∗ The composite mapping ϕ2 ◦ Φ ◦ ϕ1 from L to L can be used as ϕ since

−1 ∗ ϕ2 ◦ Φ ◦ ϕ1 (α)=α ,

−1 and ϕ =ResK(ϕ2 ◦ Φ ◦ ϕ1 ).  By successive applications of the existence theorem (Theorem 2.56) we get Theorem 2.58. Let K beafieldandf(x) an arbitrary polynomial in K[x] of positive degree. Then there exists an extension field M of K in which f(x) splits completely into linear factors (i.e. polynomials of degree one). Proof. We prove the assertion by induction on the degree n of the prescribed poly- nomium. If n = 1 the assertion is clear: We can use K itself. Now let n be > 1. If f(x) splits into linear factors in K[x] there is nothing to prove. Otherwise f(x) must be divisible by at least one polynomial p(x)inK[x] which is irreducible of degree > 1. By Theorem 2.56 there exists an extension field L of K in which p(x) and thereby also f(x)hasarootα.InL[x] we therefore can write f(x)=(x − α)g(x), where g(x) is a polynomial in L[x]ofdegreen − 1. By the inductive assumption there exists an extension field M of L in which g(x) splits into Matematisk Institut Mat 3AL 2.20

linear factors. But M is also an extension field of K, and in this extension field f(x) splits into linear factors. 

If f(x) is a polynomial in K[x] of positive degree n an extension field of K in which f(x) splits into linear factors must contain elements α1,...,αn, such that f(x)up to a constant factor in K equals a product (x − α1) ···(x − αn). The field generated over K by these elements is an extension field M = K(α1,...,αn)ofK with the following two properties: 1) f(x) splits into linear factors in M[x]. 2) f(x) does not split completely into linear factors in any proper subfield of M containing K. Definition.LetK be a field and f(x) a polynomial in K[x]. An extension field M of K is called a splitting field for f(x)overK,iff(x) splits into linear factors in M[x], while no proper subfield of M containing K has this property. Theorem 2.58 implies that there exists a splitting field for every polynomial (of positive degree) over any field.

Remark 2.59. In general there will be a proper subset, {α1,...,αt},oftheroots of f(x), such that a splitting field equals K(α1,...,αt). In other words, for the formation of a splitting field some of the roots may be superfluous in the sense, that they ”automatically follow suit” by adjunction of the other roots. If the degree n of n−1 f(x)is> 1, the sum of the roots α1 + α2 + ···+ αn = − (the coefficient of x )and hence is an element in the base field K. Therefore K(α1,...,αn−1)=K(α1,...,αn). 4 Often even more roots may be superfluous. If e.g. f√(x)=√x − 2 a splitting√ field (over Q) can be obtained just by adjoining the roots 4 2og 4 2 · i where i = −1. We shall now prove the uniqueness of splitting fields. Theorem 2.60. Let K beafieldandf(x) a polynomial in K[x] of positive degree, M a splitting field for f(x) over K.LetK∗ beafieldisomorphictoK and let ϕ : K → K∗ be an isomorphism. Let f ∗(x)=Φ(f(x)),whereΦ is the isomorphism of K[x] onto K∗[x] induced by ϕ.LetM ∗ be a splitting field for f ∗(x) over K∗.Then ϕ can be extended to an isomorphism of M onto M ∗. Proof. We use induction on the number of roots in the relative complement {the splitting field \ thebasefield}, i.e. the number of roots lying in the splitting field but not in the base field. Let us first assume that this number is = 0. In that case f(x) splits into linear factors in K and hence M = K and f ∗(x) splits into linear factors in K∗ so that M ∗ = K∗. Assume now the theorem has been proved when the number of roots in the above set { splitting field \ base field } is

If α ∈ M \ K, α is a root of f(x)thenp(x) = Irr(α, K) divides f(x) i.e.: f(x)= p(x) · h(x), where degree(p(x)) is > 1andh(x) is a polynomial in K[x]. In K∗[x]we get a analogous splitting f ∗(x)=p∗(x) · h∗(x) ,p∗(x) irreducible of the same degree as p(x) . Let α∗ ∈ M ∗ be a root of p∗(x). By the uniqueness theorem concerning adjunction of a root (Theorem 2.57) there exists a prolongation (actually uniquely determined)  - ∗ ∗  ∗ ϕ of ϕ : K(α) ϕ K (α ) such that ϕ (α)=α . In K(α)[x] the linear polynomial x − α divides f(x): f(x)=(x − α) ······ . Similarly in K∗(α∗)[x]wehave: f ∗(x)=(x − α∗) ······ . Now we notice M is a splitting field for f(x)overK(α) and M ∗ is a splitting field for f ∗(x)overK∗(α∗). The number of roots of f(x)in(M \ K(α)) < the number n of roots of f(x)in (M \ K). For the isomorphism Φ of K(α)[x]ontoK∗(α∗)[x] induced by ϕ we get Φ(f(x)) = f ∗(x). We now apply the inductive assumption on f(x)overK(α) and conclude that ϕ can be prolonged to an isomorphism from M onto M ∗. 

Comforting remark. The introduced abstract field extensions may at first seem a bit strange. However, in most cases (except the section about finite fields) one may assume that everything takes places inside the field C of complex numbers. In view of ”the fundamental theorem of algebra” (every non-constant polynomial over C splits into linear factors over C) the existence theorem concerning adjunction of roots and splitting fields become trivial. Hence, if we later consider splitting fields for polynomials over Q we may think of these fields as being subfields of the field of complex numbers.

GREATEST COMMON DIVISOR FOR POLYNOMIALS.

Let K be a field and f(x)andg(x) polynomials in K[x]. Since K[x]isaPID, f(x)andg(x) has a greatest common divisor which we – to emphasize that f(x)and g(x) are viewed as polynomials in K[x] – denote (f(x),g(x))K. A priori the greatest common divisor is only determined up to a constant factor in K. To make it uniquely defined we require (f(x),g(x))K to be monic, i.e. such that the highest coefficient is 1. For an extension K ⊆ L ofthebasefieldsthefollowingholds: Matematisk Institut Mat 3AL 2.22

Theorem 2.61. If K is a subfield of the field L,then(f(x),g(x))K =(f(x),g(x))L for all polynomials f(x), g(x) in K[x]. Remark. Roughly speaking Theorem 2.61 says that greatest common divisor of two polynomials does not change by base field extension.

Proof of Theorem 2.61. Let dK =(f(x),g(x))K and dL =(f(x),g(x))L .Bythe above definition both of them are monic. In K[x]wehavedK |f(x)anddK|g(x)and this also holds in L[x]. Therefore dK|dL (in L[x]). Since dK generates the ideal in K[x] generated by f(x)andg(x), we conclude that dK = f(x)a(x)+g(x)b(x)for some polynomials a(x)andb(x)inK[x]. In L[x] the polynomials f(x)andg(x)are divisible by dL; consequently the above equation implies that dL|dK (in L[x]). Thus dK and dL are monic polynomials for which dK|dL and dL|dK . Therefore dK = dL. 

Corollary 2.62. Let K be a subfield of the field L and f(x) and g(x) polynomials in K[x].Iff(x) divides g(x) inside L[x],thenf(x) also divides g(x) inside K[x]. Proof. W.l.o.g. we may assume that f(x)andg(x) are monic. If f(x) divides g(x)in L[x]then(f(x),g(x))L = f(x). By the above theorem (f(x),g(x))K =(f(x),g(x))L and thus (f(x),g(x))K = f(x) implying that f(x) divides g(x)inK[x]. 

CHARACTERISTIC OF A .

We briefly recall the notion of and basic facts about the characteristic of a field K. For an integer n ∈ Z and an element k ∈ K we define ⎧ ⎨⎪ k + ···+ k (n terms) for n>0 nk = 0forn =0 ⎩⎪ (−n)(−k)forn<0 .

For these the following rules hold:

(n1 + n2)k = n1k + n2k for all n1,n2 ∈ Z,k∈ K

n(k1 + k2)=nk1 + nk2 for all n ∈ Z,k1,k2 ∈ K

(n1k1)(n2k2)=(n1n2)(k1k2) for all n1,n2 ∈ Z,k1,k2 ∈ K Now let e denote the identity element in the field K. The mapping φ from Z to K defined by φ(n)=ne is in view of the above rules a . Matematisk Institut Mat 3AL 2.23

Since the image φZ is a subring of the field K,itmustbeanintegraldomain.By the isomorphism theorem for rings (cf. Anders Thorup RNG 3.7 i 2AL) Z/Ker(φ)  φZ; therefore Ker(φ) is a prime ideal of Z. There are two possibilities: 1) Ker(φ)=0. 2) Ker(φ) is the principal ideal Zp generated by a a prime number p. ad 1). In this case φ is injective and:

nk =(ne)k =0⇔ n =0ork =0.

In this case we say that K has characteristic 0. Now φZ  Z and K must contain the fraction field of φZ. This fraction field is isomorphic to the field of rational numbers Q. ad 2). In this case φ is not injective and: nk =(ne)k =0⇔ n ∈ Zp or k =0⇔ p | n or k =0. In this case we say that K has characteristic p. Here φZ  Z/Zp and K thus contains a subfield called the prime field,with exactly p elements. An important rule for fields of characteristic p is the following: Theorem 2.63. (”Freshman’s dream”). Let x and y be two arbitrary elements in a field of characteristic p.Then

(x + y)p = xp + yp

Proof. p p (x + y)p = xp + xp−1y + ···+ xyp−1 + yp. 1 p − 1 p   − p p p  Since p divides i for 1 i p 1weget(x + y) = x + y .

Remark 2.64.For a field K of characteristic p the mapping σ of K into itself defined by σ(x)=xp is a homomorphism, both w.r.t + and · and thus a ring homomorphism of K into itself. Since the kernel Ker(σ) clearly is 0, the above mapping σ is an injective ring homomorphism of K into itself. If K is a finite field σ will also be surjective and hence an isomorphism (an automorphism) of K onto itself. σ is called the Frobenius automorphism of the finite field. If K is not a finite field σ does not have to be surjective. Matematisk Institut Mat 3AL 2.24

MULTIPLE ROOTS, FORMAL DIFFERENTIATION AND SEPARA- BILITY.

We briefly recall the notions ”multiple roots” and ”multiplicity”. If α is an element in the field K and is a root of the polynomial f(x) ∈ K[x] there exists a uniquely determined number t such that

f(x)=(x − α)t · g(x) where g(x) is a polynomial in K[x] which does not have α as a root. This number t is called the multiplicity of α as a root of f(x). If t =1wecallα a simple root of f(x); if t>1wecallα a multiple root of f(x). For further investigation concerning the occurence of multiple roots we introduce the concept formal differentiation. Let K be any field. For a polynomial f(x) ∈ K[x]

n f(x)=a0 + a1x + ···+ anx

we define the formal f (x) by setting

 n−1 f (x)=a1 +2a2x + ···+ nanx The following rules which are analogous to those known from classical calculus can immediately be verified:

(f + g) = f  + g , (kf) = kf for all k ∈ K (f · g) = f · g + f  · g.

Many of the results concerning differentiation in classical calculus also hold for the above formal . But in some cases, in particular for fields of prime characteristic, one has to be more careful. Theorem 2.65. Let K be a field and

n f(x)=a0 + a1x + ···+ anx a polynomial in K[x].  1) If K has characteristic 0 then f (x)=0 ⇔ ai =0 for all i>0.  2) If K has characteristic p then f (x)=0 ⇔ ai =0for all i, which are not divisible by p. Proof. ad 1)  i−1 f (x)= iaix =0⇔ ai =0foralli>0, i>0 Matematisk Institut Mat 3AL 2.25

wherewehaveusedtherulesforiai being 0 when K has characteristic 0. ad 2)  i−1 f (x)= iaix =0⇔ ai =0foralli which are not divisible by p, i>0 wherewehaveusedtherulesforiai being 0 when K has characteristic p.  The following theorem holds for fields of arbitrary characteristic. Theorem 2.66. Let K be any field. If the polynomial f(x) ∈ K[x] has a multiple root α in some extension field M of K then α isalsoarootoff (x). Proof. We can write

f(x)=(x − α)ν g(x), where ν>1 and get

f (x)=(x − α)ν g(x)+ν(x − α)ν−1g(x)=(x − α)ν−1{(x − α)g(x)+νg(x)} which shows that α is a root of f (x). 

Theorem 2.67. Let K be a field of characteristic 0. An irreducible polynomial f(x) ∈ K[x] has no multiple roots in any extension field of K,inparticularnotin the splitting field for f(x) over K. Proof. Assume α were a multiple root of f(x) in some extension field M of K. Theorem 2.66 implies that (x−α)isadivisoroff (x)inM[x]. The greatest common  divisor dM =(f(x),f (x))M therefore has degree at least 1. By theorem 2.61 the greatest common divisor does not change by extension of the base field. Therefore  dM = dK =(f(x),f (x))K. Since the degree of dK thus is ≥ 1andf(x) is irreducible in K[x], then dK (up to a constant factor in K \{0})mustbef(x). In particular, f(x) must divide f (x).ButthisisimpossiblesinceK has characteristic 0 and f (x) therefore (in view of theorem 2.65) a non-zero polynomial, whose degree is smaller than the degree of f(x). 

Theorem 2.68. Let K be field of characteristic p. An irreducible polynomium f(x) in K[x] has multiple roots in its splitting field over K ⇔ f (x)=0. Proof. “⇒” Matematisk Institut Mat 3AL 2.26

If f (x) were a proper polynomial we could in exactly the same way as in the proof of Theorem 2.67 conclude that f(x) has no multiple roots in any extension field of K. “⇐”Iff (x) = 0, Theorem 2.65 implies that f(x) has the form:

p 2p kp f(x)=a0 + apx + a2x + ···+ akpx .

It is enough to show that f(x) has multiple roots in some ”suitably big” field, in which f(x) splits into linear factors. p p p We now adjoin to K elements b0,b1, ..., bk such that b0 = a0, b1 = ap,...,bk = akp. In K(b0,b1,...,bk)wethenhave

p p p p kp k p f(x)=b0 + b1x + ···+ bkx =(b0 + b1x + ···+ bkx ) .

k Let L be the splitting field for g(x)=b0 + b1x + ···+ bkx over K(b0,b1,...,bk), so that we in L have the splitting g(x)=bk(x − β1) ···(x − βk) and thus f(x)= p p akp(x − β1) ···(x − βk) (in L[x]). The above decomposition for f(x) also holds in the splitting field for f(x)over K. We see, that actually all roots of f(x) are multiple roots. 

Definition 2.69. An irreducible polynomial f(x)inK[x] is called separable if f(x) has no multiple root in its splitting field over K (and thereby no multiple root in any extension field of K). An arbitrary polynomial f(x)inK[x] is called separable if each of its irreducible factors is separable in the above sense. Remark 2.70.Letf(x) be irreducible. Theorems 2.67 and 2.68 imply that f(x)is separable ⇔ f (x) = 0. In particular every polynomial in K[x] is separable if the characteristic of K is 0.

Definition 2.71 .AfieldK is called perfect if all polynomials in K[x] are separable. The above theorems in particular imply that every field of characteristic 0 is perfect. Theorem 2.72. Let K be a field of prime characteristic p.Then:K is perfect ⇔ the mapping σ : α → αp which sends every element in K into its p-th power is surjective. Proof. ⇒:Letβ ∈ K and f(x)=xp − β. Let γ be a root of f(x) in its splitting field over K. Then: xp − β =(x − γ)p. Now Irr(γ,K) is irreducible and by assumption it has no multiple roots in the above splitting field. Since moreover Irr(γ,K) divides xp − β we conclude that Irr(γ,K)=x − γ; therefore γ must belong to K,inotherwordsβ = γp. Matematisk Institut Mat 3AL 2.27

⇐: It suffices to prove that every polynomial f(x) ∈ K[x] of positive degree is reducible if f (x) = 0. From theorem 18 it follows that f (x) = 0 implies that f(x) has the form p kp f(x)=a0 + a1x + ···+ akx . p Since the mapping σ : α → α is surjective there exist elements b0,...,bk ∈ K such p p p that a0 = b0, a1 = b1,...,ak = bk,andtherfore

k p f(x)=(b0 + b1 + ···+ bkx ) , showing that f(x) is reducible. 

Remark 2.73. Since the mapping σ : α → αp is injective for every field of char- acteristic p (cf. earlier remark) the above theorem implies that every finite field is perfect.

Example 2.74. A non-perfect field. Let K = Zp(t) (i.e. the field of all rational functions in one indeterminate over the field Zp). The element t is not a p-th power of an element in K since every p-th power must be of the form

p 2p a0 + a1t + a2t + ... p 2p b0 + b1t + b2t + ...

(ai,bi ∈ Zp).

Definition 2.75.LetL  K be fields. An element α ∈ L is called separable over K if α is algebraic over K and Irr(α, K) is a separable polynomial in K[x]. L ⊇ K is called a separable extension if all elements in L are separable over K. ABEL-STEINITZ’S THEOREM. Most of the field extensions that we are going to consider are ”simple”. The precise definition is the following. Definition. An algebraic extension L/K is called simple if there is an α ∈ L such that L = K(α). Such an α is called a primitive element for L/K. Theorem 2.76 (Abel, Steinitz). Let L/K be a finite (and thus in particular an algebraic) extension which is separable. Then L/K is simple, i.e. there exists a primitive element for the extension L/K. Proof. If K is finite so is L. As known from group theory L \{0} is a . If α is a generator for L \{0},theninparticularL = K(α). We may therefore assume that K has infinitely many elements. In this case Abel- Steinitz’s theorem can be obtained by succesive applications of If K ⊂ M, α, β ∈ Mandαand β are separable over K then there exists γ ∈ M such that K(α, β)=K(γ). Matematisk Institut Mat 3AL 2.28

Proof. Let f(x) = Irr(α, K), g(x) = Irr(β,K). We now work inside an extension field N of M in which f(x)andg(x) splits into linear factors. We may e.g. for N choose the splitting field for f(x) · g(x)overM.InN we may therefore write

f(x)= (x − α1) ...(x − αn) , wherewemayassumeα = α1

g(x)= (x − β1) ...(x − βm) , wherewemayassumeβ = β1 .

Since β is separable, the elements β1,...,βm are distinct. Since K has infinitely many elements there is an element c ∈ K such that α−α γ = α + cβ = αi + cβj 1  i  n  i c chosen such that c = βj −β 2  j  m 1 ≤ i ≤ n, 2 ≤ j ≤ m

We claim K(γ)=K(α, β). It is clear that K(γ)  K(α, β). To show the inverse inclusion we notice that f(γ − cx)hasβ as a root. Because ofthechoiceofc no βj ,2≤ j ≤ m, is a root of f(γ − cx). Therefore we may write:

f(γ − cx)=(x − β)t · h(x) , where t is an integer ≥ 1

and h(x) is a polynomial for which no βj,1≤ j ≤ m,isaroot. In N we thus have (f(γ − cx),g(x))N =(x − β) .

Now g(x) ∈ K[x] ⊆ K(γ)[x]andf(γ − cx) ∈ K(γ)[x]. By Theorem 2.61 we get (f(γ − cx),g(x))K(γ) =(f(γ − cx),g(x))N = x − β. This means that β ∈ K(γ). Since α = γ − cβ ∈ K(γ) we conclude that K(α, β) ⊆ K(γ); this taken together with the inverse (trivial) inclusion yields K(γ)=K(α, β). 

Remark 2.77. The assumption about separability in Abel-Steinitz’s theorem is important. Let L = Z2(x, y) (i.e. the field of rational functions in two indeterminates 2 2 x og y over the field Z2)andK = Z2(x ,y ) (i.e. the subfield of rational functions in x2 and y2). Here [L : K]=4since1,x, y, xy is a basis for L viewed as a vector space over K.ButL/K is not simple. Indeed α2 ∈ K for every element α ∈ L, hence [K(α):K]  2.

−−−−−−−−−−−−−−−

FINITE FIELDS. We first prove a theorem about the number of elements in a finite field. Matematisk Institut Mat 3AL 2.29

Theorem 2.78. The number of elements in a finite field K is the power of a prime number. Proof. Since K is finite the characteristic of K must be a prime number p. Therefore the field K must contain the field Zp as a subfield. Viewed as a vector space over Zp the field K has a finite dimension n where n is a natural number. Let ω1,... ,ωn be a basis for K over Zp. Every element in K has a unique representation of the form a1ω1 + ···+ anωn where a1,... ,an run through Zp.The field K thus has exactly pn elements.  Asakindofconversewenowprove: Theorem 2.79. For every power pn of a prime number p there exists one and up to isomorphism just one field with pn elements.

n Proof. 1) Existence.LetM be the splitting field for the polynomial f(x)=xp − x  n over Zp.Sincef (x)=−1 by Theorem 2.66 f(x) has exactly p distinct roots in M. n n Let α and β be two such roots of f(x). Then αp = α and βp = β. By repeated n application of ”Freshman‘s dream” (Theorem 2.63) we get (α + β)p = α + β, hence α + β is a root of f(x). Similarly we get that α − β is a root of f(x). Furthermore n n n from αp = α and βp = β we conclude that (α · β)p = α · β. Hence α · β is a root n n of f(x). If β =0then βp = β implies (1/β)p =1/β meaning that 1/β is a root of f(x). Consequently the pn roots of f(x) form a subfield K of M.ThusK is a field with exactly pn elements. This completes the existence proof. (Actually K = M;infact f(x) splits into linear factors in K and since f(x)haspn distinct roots and |K| = pn there is no proper subfield of K in which f(x) splits into linear factors. K must then be the splitting field M of f(x).) 2) Uniqueness.LetK be a field with pn elements. The characteristic of K must be a prime number q . The additive group (K, +) of K must contain the additive n group (Zq, +) of the prime field Zq. Thus the order q of (Zq, +) divides the order p of (K, +). Hence p = q and the prime field of K equals Zp. The elements of K∗ = K \{0} form a multiplicative group of order pn − 1. By n Lagrange’s theorem αp −1 =1forallα ∈ K , α = 0. This means that all elements n in K are roots of the polynomial xp − x. This polynomial has at most (actually n exactly) pn roots. Therefore K consists exactly of the roots of xp −x.Consequently pn K is a splitting field for x − x over Zp. By the uniqueness (up to isomorphism) of a splitting field (Theorem 2.60) K is uniquely determined up to isomorphism. 

Definition 2.80.Forapowerpn of a prime number p the above field with pn n elements is denoted GF(p )orFpn .(Forn = 1 there are thus three notations for the field with p elements: Zp,GF(p)ogFp!) Matematisk Institut Mat 3AL 2.30

Theorem 2.81. For a given prime p we have the inclusions: Fpm  Fpn ⇔ m|n. Here (“” should be read: “is isomorphic to a subfield of”).

Proof. “⇒”: Let d be the dimension of Fpn viewed as a vector space over Fpm .Then m d md the number of elements in Fpn is (p ) = p (cf. the proof of Theorem 2.78). Hence n = md, i.e. m | n. m n m n “⇐”: m|n ⇒ pm −1|pn −1 ⇒ (xp −1 −1) | (xp −1 −1) ⇒ (xp −x) | (xp −x) ⇒ pn pm the splitting field (over Zp)for(x − x)  the splitting field (over Zp)for(x − x). Hence (cf. the proof of theorem 2.79) Fpn  Fpm . 

−−−−−−−−−−−−−−−

We give a quite concrete number theoretic application of the theorem above. We derive an explicit formula for the number π(n) of monic irreducible polynomials in Zp of degree n. n Let xp − x =Πp(x)wherep(x) runs through the monic irreducible factors (in pn Zp[x]) of x − x. pn Fpn is the splitting field for x −x over Zp.Ifα ∈ Fpn and p(α)=0then[Zp(α): d Zp]=thedegreed of p(x), hence: [Zp(α):Zp]=d and thus |Zp(α)| = p ⇒ d|n. Conversely every irreducible monic polynomial q(x) ∈ Zp[x] whose degree d di- pn vides n will be a divisor of x − x. Indeed: Let L = Zp(α)whereα is a root of q(x) d (or equivalently q(x) = Irr(α, Zp)). Then [L : Zp]=d ⇒|L| = p .Butthenα is a pd pd pn root of x − x and thus q(x) = Irr(α, Zp)|x − x|x − x. n Since xp − x has no multiple factors, the irreducible factors p(x) in the product pn x − x =Πp(x) are just the monic irreducible polynomials whose degrees divide n. n In this way we obtain the formula p = d|n d · π(d). To find an explicit formula for π(n) we shall need the number theoretical function, the M¨obius-function μ(n), defined on N in the following way: ⎧ ⎨⎪ 1forn =1, μ(n)=⎪ 0ifn is divisible by a square > 1, ⎩ r (−1) if n = p1 ···pr, where p1,... ,pr are distinct prime numbers.

Exampl 2.82 μ(1) = 1,μ(2) = μ(3) = −1,μ(4) = 0,μ(5) = −1,μ(6) = 1,μ(7) = −1,μ(8) = μ(9) = 0,μ(10) = 1. Theorem 2.83.  1 for n =1 μ(d)= d|n 0 for n>1. Matematisk Institut Mat 3AL 2.31

where the summation is extended over all positive divisors of n.

a1 ar Proof. For n = 1 the assertion is clear. For n>1, let n = p1 ...pr pi = pj for i = j; we only have to consider the square-free divisors of n and find:

r r μ(d)= (−1)ν =(1− 1)r =0. d|n ν=0 ν



From Theorem 2.83 we obtain ⎧ ⎫ ⎨ ⎬ n d · p μ(d)= μ(d) ⎩ δπ(δ)⎭ = d|n d|n δ| n d⎧ ⎫ ⎨ ⎬ · δπ(δ)μ(d)= δπ(δ) ⎩ μ(d)⎭ = d,δ n δ|n d| δ d·δ|n n · π(n) .

Consequently we find the following explicit for π(n): 1 n π(n)= · p d μ(d) . n d|n

This in particular shows that π(n) is positive for every natural number n.

DISCRIMIMANT OF A POLYNOMIAL.

We now introduce a classical invariant which is of great importance for explicit questions concerning the roots of a polynomial. For this purpose we first consider the following polynomial in n indeterminates x1, ···,xn: 2 n−1 1 x1 x1 ··· x1 2 n−1 1 x2 x2 ··· x2 Δ(x1,x2,...,xn)= (xi − xj)= ... n≥i>j≥1 2 n−1 1 xn xn xn which plays an important role by the introduction of discriminants. The explicit computation of the above determinant (called Vandermondes determinant)canbe found in an appendix of these notes. Matematisk Institut Mat 3AL 2.32

Clearly for every permutation σ ∈ Sn we get  Δ(x1,x2,...,xn)whenσ is even Δ(xσ(1),xσ(2),...,xσ(n))= −Δ(x1,x2,...,xn)whenσ is odd

2 In particular d(x1,x2,...,xn)=[Δ(x1,x2,...,xn)] is a symmetric polynomial. n n−1 Now let K be an arbitrary field and f(T )=T + b1T + ···+ bn a monic polynomial in K[T ]. If β1,β2,... ,βn are the roots of f(T ) in its splitting field K we define the discriminant, discrim(f), for f as 2 d(β1,β2,...,βn)= (βi − βj ) . n≥i>j≥1 d(x1,x2,...,xn) is a symmetric polynomial; the main theorem about symmetric poly- nomials implies that it can be written as a polynomial (with coefficients in K)ofthe elementary symmetric polynomials of x1,x2,...,xn. By substituting β1 for x1, β2 for x2,etc.weseethatd(β1,β2,...,βn) becomes a polynomial (with coefficients in K)ofb1,b2,...,bn since the elementary symmetric polynomials of x1,x2,...,xn by substitution of β1 for x1, β2 for x2,etc.uptofactors±1justgiveb1,b2 etc. In particular it follows that the discriminant discrim(f)ofapolynomialf with coefficients in the field K is an element of K. From the definition it is clear that all roots of f are simple if and only if discrim(f) =0. The following simple observation is often quite useful Theorem 2.84. Let K beafieldandMthe splitting field over K for a monic polynomial f(x) ∈ K[x] of degree n.Then discrim(f) is an element of M. Proof. Let β1,... ,βn be the roots of f(x). Clearly Δ(β1,... ,βn)isanelementof 2 M. Since discrim(f)=[Δ(β1,... ,βn)] , the square root discrim(f) lies in M. 

Let us compute the discriminants of quadratic and cubic polynomials. 2 2 It is immediate to check that discrim(T + a1T + a2)=a1 − 4a2. Among the cubic polynomials we just consider those of the form T 3 + pT + q.If β1,β2 and β3 are the roots we find

β1 + β2 + β3 =0 β1β2 + β1β3 + β2β3 = p β1β2β3 = −q.

For the power sums t t t Pt = β1 + β2 + β3 we find P1 =0 P2 = −2p Matematisk Institut Mat 3AL 2.33

and since 3 3 βi + pβi + q =0 i=1 and 3 4 2 (βi + pβi + qβi)=0 i=1 2 we conclude P3 = −3q og P4 =2p . Furthermore 3 2 discrim(T + pT + q)=[(β2 − β1)(β3 − β2)(β3 − β1)] = 2 2 1 β1 β1 3 P1 P2 30−2p 2 1 β2 β2 = P1 P2 P3 = 0 −2p −3q 2 2 1 β3 β3 P2 P3 P4 −2p −3q 2p = −27q2 − 4p3 .

For a cubic polynomial with real coefficients the number of real roots is determined by the discriminant. Indeed: Theorem 2.85. Let f(x) be a monic cubic polynomial with real coefficients. Then: discrim(f) > 0 ⇔ f has three distinct real roots; discrim(f)=0⇔ f has a multiple root; discrim(f) < 0 ⇔ f has exactly one real root. Proof. Exercise.

More generally the following holds: Theorem 2.86. Let f(x) be a monic polynomial in R[x] of degree n. i) discrim(f)=0exactly when f(x) has a multiple root. Assume all roots of f(x) are simple. Let r1 be the number of real roots and r2 the number of pairs of complex conjugate roots (i.e. r1 +2r2 = n). ii) discrim(f) is positive exactly when r2 is even. iii) discrim(f) is negative exactly when r2 is odd. Proof. The assertion i) is clear. Assume now f(x)hasn simple roots β1,... ,βn. Then the discriminant discrim(f)= 2 [Δ(β1,... ,βn)] . Now we have 2 n−1 1 β1 β1 ··· β1 2 n−1 1 β2 β2 ··· β2 Δ(β1,... ,βn)= ... 2 n−1 1 βn βn βn Matematisk Institut Mat 3AL 2.34

Application of complex conjugation on Δ(β1,... ,βn) gives rise to transpositions of r2 rows in the above determinant. If r2 is even Δ(β1,... ,βn) will be invariant under complex conjugation and there- 2 fore a . The square discrim(f)=[Δ(β1,... ,βn)] is then a positive number. If r2 is odd√ Δ(β1,... ,βn) will change sign by complex conjugation and therefore 2 of the form −1· (a real number). The square discrim(f)=[Δ(β1,... ,βn)] must then be a negative number. 

For explicit computation of discriminants the following is often quite useful n n−1 Theorem 2.87. Let f(T )=T + b1T + ···+ bn be a polynomial with the roots n n(n−1)  β1,...,βn. Then we have discrim(f)=(−1) 2 f (βi). i=1 n Proof. From the rules for formal differentiation we get for f(T )= (T − βi) i=1

 f (T )=(T − β2) ···(T − βn)+(T − β1)(T − β3) ···(T − βn)

+ ···+(T − β1)(T − β2) ···(T − βn−1) . n  Hence for the product f (βi) we find i=1

(β1 − β2) · (β1 − β3) ···(β1 − βn) (β2 − β1) · (β2 − β3) ···(β2 − βn) ... (βn − β1) · (βn − β2) ···(βn − βn−1)

n(n−1) n(n−1) =(−1) 2 d(β1,...,βn)=(−1) 2 discrim(f) . 

n n (n−1)(n−2) Theorem 2.88. The discriminant of the polynomial f(T )=T −1 is n (−1) 2 .  n−1 Proof. Clearly f (T )=nT . For the roots β1,...,βn the product β1 ···βn is (−1)n−1, so theorem 31 yields n n n(n−1) n(n−1) n−1 2  2 discrim(f)=(−1) f (βi)=(−1) (nβi ) i=1 i=1 n n−1 n(n−1) n n n(n−1) (n−1)2 =(−1) 2 n βi = n (−1) 2 (−1) i=1 n (n−1)(n−2) = n (−1) 2 . Matematisk Institut Mat 3AL 2.35



Exercise 2.89. Show that the discriminant of the polynomial T 4 + aT 2 + b is 16(a2 − 4b)2b.