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Home , Field of fractions

36 Rings of

Recall. If R is a PID then R is a UFD.

In particular

• Z is a UFD

• if F is a field then F[x] is a UFD.

Goal. If R is a UFD then so is R[x].

Idea of proof.

1) Find an R,→ F where F is a field.

2) If p(x) ∈ R[x] then p(x) ∈ F[x] and since F[x] is a UFD thus p(x) has a unique factorization into irreducibles in F[x].

3) Use the factorization in F[x] and the fact that R is a UFD to obtain a factorization of p(x) in R[x].

36.1 Definition. If R is a then a subset S ⊆ R is a multiplicative subset if

1) 1 ∈ S 2) if a, b ∈ S then ab ∈ S

36.2 Example.

Multiplicative subsets of Z:

1) S = Z

137 2) S0 = Z − {0}

3) Sp = {n ∈ Z | p - n} where p is a prime number.

Note: Sp = Z − hpi.

36.3 Proposition. If R is a ring, and I is a prime of R then S = R − I is a multiplicative subset of R.

Proof. Exercise.

36.4. Construction of a ring of fractions.

Goal. For a ring R and a multiplicative subset S ⊆ R construct a ring S−1R such that every element of S becomes a unit in S−1R.

Consider a relation on the set R × S:

0 0 0 0 (a, s) ∼ (a , s ) if s0(as − a s) = 0 for some s0 ∈ S

Check: ∼ is an equivalence relation.

Elements of S−1R = (equivalence classes of R × S under the relation ∼).

Notation. a/s := the represented by (a, s)

Note.

1) If 0 ∈ S then (a, s) ∼ (a0, s0) for all (a, s), (a0, s0) ∈ R × S, and so S−1R consists of only one element. 2) If R is an and 0 6∈ S then (a, s) ∼ (a0, s0) iff as0 = a0s.

138 Multiplication in S−1R: (a/s)(a0/s0) = (aa0)/ss0

Addition in S−1R: a/s + a0/s0 = (as0 + a0s)/ss0

Check: 1) The operations of and multiplication in S−1R are well defined. 2) These operations define a structure on S−1R. 3) The map −1 ϕS : R → S R, ϕS(r) = r/1 is a .

−1 Note. If s ∈ S then ϕS(s) = s/1 is a unit in S R:

(s/1)(1/s) = s/s = 1/1 = 1S−1R

36.5 Definition. If S is a multiplicative subset of a ring R then S−1R is called the ring of fractions of R with respect to S.

36.6 Theorem. If S is a multiplicative subset of a ring R and f : R → R0 is a ring homomorphism such for ever s ∈ S the element f(s) ∈ R0 is a unit, then there exists a unique homomorphism f¯: S−1R → R0 such that the following diagram commutes:

f R / R0 {= { { { ϕS { { f¯ { {  { S−1R

139 Proof. Define f¯(r/s) = f(r)f(s)−1. Check that

1) f¯ is a well defined ring homomorphism ¯ ¯ 2) f is the only homomorphism such that f = fϕS.

36.7 Examples.

−1 ∼ 1) If S ⊆ Z, S = Z then S Z = {0}.

−1 ∼ 2) If S0 ⊆ Z, S0 = Z − {0} then S0 Z = Q.

−1 3) If S0 ⊆ Z, S0 = Z − hpi then Sp Z is isomorphic to the of Q m consisting of all fractions n such that p - n.

−1 ∼ 4) If R = Z/6Z, S = {1, 3} ⊆ R then S R = Z/2Z (check!).

36.8 Proposition. If S is a multiplicative subset of a ring R then

Ker(ϕS) = {a ∈ R | sa = 0 for some s ∈ S}

Proof. We have ϕS(r) = 0/1 iff s(r · 1 − 0 · 1) = 0 for some s ∈ S.

36.9 Corollary. If R is an integral domain and S ⊆ R is a multiplicative subset such that 0 6∈ S then the homomorphism

−1 ϕS : R → S R is 1-1. As a consequence in this case we can identify R with a subring of S−1R.

140 36.10 Note.

1) If R is an integral domain and S = R − {0} then the ring S−1R is a field. In this case S−1R is called the field of fractions of R.

2) If I is a of a ring R then the set S = R − I is a multiplicative subset of R. In this case the ring S−1R is called the localization of R at I and it is denoted RI .

36.11 Definition. A ring R is a local ring if R has exactly one maximal ideal.

36.12 Examples.

1) If F is a field then it is a local ring with the maximal ideal I = {0}.

2) Check: if F is a field then F[[x]] is a local ring with the maximal ideal hxi.

3) Check: if F is a field then F[x] is not a local ring since for every a ∈ F the ideal hx − ai is a maximal ideal in F[x].

4) Z is not a local ring.

36.13 Proposition. If R is a local ring then the maximal ideal J C R consists of all non-units of R.

Proof. Since J 6= R thus J does not contain any units.

Conversely, if a ∈ R is a non-unit then 1 6∈ hai so hai 6= R. Therefore by Theorem 29.1 hai is contained in some maximal ideal of R. Since J is the only maximal ideal we obtain hai ⊆ J, and so a ∈ J.

141 36.14 Proposition. If R is a ring and I C R is a prime ideal then the ring RI is a local ring, and the maximal ideal J C RI is given by J := {a/s | a ∈ I, s 6∈ I}

Proof. Exercise.

142 37 Factorization in rings of polynomials

Goal:

37.1 Theorem. If R is a UFD then so is R[x].

37.2 Lemma. If R is an integral domain then p(x) ∈ R[x] is a unit in R[x] iff deg p(x) = 0 and p(x) = a0 where a0 is a unit in R.

Proof. Exercise.

37.3 Lemma. If R is an integral domain and p(x) = a0 is a polynomial of degree 0 in R[x] then p(x) is irreducible in R[x] iff a0 is irreducible in R.

Proof. Exercise.

n 37.4 Definition. If R is a UFD and p(x) = a0 + a1x + . . . anx ∈ R[x], then p(x) is a primitive polynomial if gcd(a0, . . . , an) ∼ 1.

37.5 Lemma. If R is a UFD and p(x) ∈ R[x] is an such that deg p(x) > 0 then p(x) is a primitive polynomial.

n Proof. If p(x) = a0 + a1x + . . . anx is not primitive then

p(x) = gcd(a0, . . . an)q(x) for some q(x) ∈ R[x], deg q(x) > 0. Since both gcd(a0, . . . , an) and q(x) are non-units in R[x] the polynomial p(x) is not irreducible.

143 37.6 Lemma (Gauss). If R is a UFD and p(x), q(x) ∈ R[x] are primitive poly- nomials then p(x)q(x) is also primitive.

Proof. Assume that r(x) = p(x)q(x) is not primitive. Then we have r(x) = c · r˜(x) where c ∈ R is an irreducible element, and r˜(x) ∈ R[x]. Take the canonical π : R −→ R/hci This defines a homomorphism of rings of polynomials π¯ : R[x] −→ (R/hci)[x] given by

m n π¯(a0 + a1x + ... + anx ) := π(a0) + π1(a)x + ... + π(an)x Note: 1) Since c is irreducible element thus hci is a prime ideal and so R/hci is a domain. As a consequence (R/hci)[x] is a domain. 2) We have π¯(p(x))¯π(q(x)) =π ¯(c · r˜(x)) = 0 · π¯(˜r(x)) = 0 so either π¯(p(x)) = 0 or π¯(q(x)) = 0.

We can assume that π¯(p(x)) = 0. Then p(x) = c · p˜(x) for some p˜(x) ∈ R[x]. Since p(x) is primitive we get a contradiction.

37.7 Lemma. Let R be a UFD and K be the field of fractions of R. 1) For any non-zero polynomial p(x) ∈ K[x] there is an element c(p) ∈ K and a primitive polynomial p∗(x) ∈ R[x] such that p(x) = c(p)p∗(x)

144 2) If p(x) = c(p)p∗(x) and p(x) =c ˜(p)˜p∗(x) for some c(p), c˜(p) ∈ K, and some primitive polynomials p∗(x), p˜∗(x) ∈ R[x] then there exists a unit u ∈ R such that

c˜(p) = uc(p), p˜∗(x) = u−1p∗(x)

3) If p(x), q(x) ∈ K[x] are non-zero polynomials then for some unit u ∈ R we have

c(pq) = uc(p)c(q)(p(x)q(x))∗ = u−1p∗(x)q∗(x)

37.8 Definition. If p(x) ∈ K[x] then the element c(p) ∈ K is called the content of p(x).

Proof of Lemma 37.7.

1) If p(x) ∈ K[x] then there is there exists 0 6= c ∈ R such that cp(x) ∈ R[x]. Let b ∈ R be a greatest common divisor of coefficients of cp(x). Take

p∗(x) = c/b · p(x), c(p) = b/c

Check that p∗(x) ∈ R[x] and p∗(x) is a primitive polynomial.

2) We have c(p)p∗(x) = p(x) =c ˜(p)˜p∗(x) where c(p), c˜(p) ∈ K are non-zero elements and p∗(x), p˜∗(x) ∈ R[x] are primitive ∗ n polynomials. Let p (x) = a0 + a1x + ... + anx .

We have p˜∗(x) = c(p)˜c(p)−1p∗(x) Since c(p)˜c(p)−1 ∈ K there b, d ∈ R such that c(p)˜c(p)−1 = b/d. We can assume that gcd(b, d) ∼ 1. We have

∗ ∗ n dp˜ (x) = bp (x) = ba0 + ba1x + ... + banx

145 This gives: d | bai for i = 1, . . . , n. By (35.6) this gives that d | ai for all i. Since p∗(x) is a primitive polynomial it implies that d is a unit in R. By an analogous argument we obtain that b is also a unit in R.

As a consequence u = db−1 is a unit in R and we have

c˜(p) = uc(p), p˜∗(x) = u−1p∗(x)

3) For p(x), q(x) ∈ K[x] we have

c(p)c(q)p∗(x)q∗(x) = p(x)q(x) = c(pq)(p(x)q(x))∗

We have c(p)c(q), c(pq) ∈ K, (p(x)q(x))∗ ∈ R[x] is primitive by construction and p∗(x)q∗(x) ∈ R[x] is primitive by Lemma 37.6.

Applying part 2) we obtain that there is a unit u ∈ R such that

c(pq) = uc(p)c(q)(p(x)q(x))∗ = u−1p∗(x)q∗(x)

37.9 Note. Check:

1) Let p(x) ∈ K[x]. Then p(x) ∈ R[x] iff c(p) ∈ R.

2) p(x) ∈ R[x] is primitive iff p(x) = up∗(x) for some unit u ∈ R.

37.10 Proposition. Let R be a UFD and K be the ring of fractions of R.

1) A polynomial p(x) ∈ K[x] of non-zero degree is irreducible in K[x] iff p∗(x) is irreducible in R[x].

2) A polynomial p(x) ∈ R[x] of non-zero degree is irreducible in R[x] iff p(x) is primitive and it is irreducible in K[x].

146 Proof.

1) (⇐) If p(x) is not irreducible in K[x] then p(x) = q(x)r(x) for some q(x), r(x) ∈ K[x], deg q(x), deg r(x) > 0. By Lemma 37.7 we have

p∗(x) = uq∗(x)r∗(x) for some unit u ∈ R. Therefore p∗(x) is not irreducible in R[x].

(⇒) If p∗(x) is not irreducible in R[x] then

p∗(x) = q(x)r(x) where q(x), r(x) are non-units in R[x]. Since p∗(x) is primitive we must have deg q(x), deg r(x) > 0. Then

p(x) = c(p)q(x)r(x) and so p(x) is not irreducible in K[x].

2) (⇒) If p(x) ∈ R[x] is irreducible then it must be a primitive polynomial. Therefore p(x) = up∗(x) for some unit u ∈ R. Since p(x) is irreducible in R[x], thus p∗(x) is also irreducible in R[x], and so by part 1) p(x) is irreducible in K[x].

(⇐) Since p(x) is irreducible in K[x] by part 1) we get that p∗(x) is irreducible in R[x]. Also, since p(x) is primitive we have p(x) = up∗(x) for some unit u ∈ R. It follows that p(x) is irreducible in R[x].

Proof of Theorem 37.1. By Theorem 31.7 we need to show that:

1) Every non-zero, non-unit p(x) ∈ R[x] is a product of irreducible polyno- mials.

2) Every irreducible polynomial p(x) ∈ R[x] is a prime element of R[x].

147 1) Recall that irreducible polynomials of degree 0 in R[x] correspond to irreducible elements of R. Since R is a UFD it follows that if p(x) ∈ R[x] has degree 0 then p(x) is a product of irreducibles in R[x].

If deg p(x) > 0 then p(x) is a non-zero non-unit element of K[x]. Since K[x] is a UFD we have p(x) = q1(x)q2(x). . .qk(x) where q1(x), . . . , qk(x) are irreducible polynomials in K[x]. We have

∗ ∗ ∗ p(x) = (c(q1)c(q2). . .c(qk))q1(x)q2(x). . .qk(x) ∗ By Proposition 37.10 qi (x) are irreducible polynomials in R[x]. Moreover, by Lemma 37.7 we have c(q1)c(q2). . .c(qk) = uc(p) for some unit u ∈ R, and also since p(x) ∈ R[x] we have c(p) ∈ R. Therefore c(q1)c(q2). . .c(qk) ∈ R, and since R is a UFD we have

c(q1)c(q2). . .c(qk) = a1a2. . .al where a1, . . . , al ∈ R are irreducible elements in R[x]. As a consequence we obtain a factorization of p(x):

∗ ∗ ∗ p(x) = a1a2. . .alq1(x)q2(x). . .qk(x) ∗ ∗ where a1, . . . , al, q1(x), . . . , qk(x) are irreducible elements in R[x].

2) Let p(x) ∈ R[x] be an irreducible polynomial. We need to show that if for some q(x), r(x) ∈ R[x] we have p(x) | q(x)r(x) then either p(x) | q(x) or p(x) | r(x).

Exercise: this holds if deg p(x) = 0.

If deg p(x) > 0, then since p(x) is irreducible in R[x] by Proposition 37.10 we obtain that p(x) is primitive and it is irreducible in K[x]. Since K[x] is a UFD we obtain that p(x) is a prime element of K[x], and so p(x) | q(x) or p(x) | r(x) in K[x].

We can assume that p(x) | q(x) in K[x]. Then there exists h(x) ∈ K[x] such that p(x)h(x) = q(x)

148 We have c(p)c(h) = uc(q) for some unit u ∈ R. Also, since q(x) ∈ R[x] we have that c(q) ∈ R and therefore c(p)c(h) ∈ R. Finally, since p(x) is primitive thus c(p) is a unit in R. We obtain:

c(h) = c(p)−1uc(q) ∈ R

Therefore h(x) ∈ R[x], and so p(x) | q(x) in R[x].

37.11 Corollary. Z[x] is a UFD

37.12 Corollary. If R is a UFD then the ring of polynomials of n variables R[x1, . . . , xn] is also a UFD.

Proof. It is enough to notice that ∼ R[x1, . . . , xn] = R[x1, . . . , xn−1][xn]

(check!).

149