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By Ash R.B.

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Proof. Without loss of generality, we may assume that R is a local ring and F = h(R) is a subfield of C. To see this, let P be the kernel of h. Then P is a prime ideal, and we can extend h to g : RP → C via g(a/b) = h(a)/h(b), h(b) = 0. The kernel of g is P RP , so by the first isomorphism theorem, g(RP ) ∼ = RP /P RP , a field (because P RP is a maximal ideal). Thus we may replace (R, h) by (RP , g). Our first step is to extend h to a homomorphism of polynomial rings. If f ∈ R[x] with h(ai )xi ∈ F [x].

Without loss of generality, we may assume that R is a local ring and F = h(R) is a subfield of C. To see this, let P be the kernel of h. Then P is a prime ideal, and we can extend h to g : RP → C via g(a/b) = h(a)/h(b), h(b) = 0. The kernel of g is P RP , so by the first isomorphism theorem, g(RP ) ∼ = RP /P RP , a field (because P RP is a maximal ideal). Thus we may replace (R, h) by (RP , g). Our first step is to extend h to a homomorphism of polynomial rings. If f ∈ R[x] with h(ai )xi ∈ F [x]. Let I = {f ∈ R[x] : f (α) = 0}.

By definition of I we have rf (s) = 0, and by hypothesis, r is not a zero-divisor of S. Therefore f (s) = 0, so f ∈ I. Step 3 : f generates I. Let q ∈ I ⊆ IK[x]. 3. GOING DOWN 9 write q = q1 f /r1 with 0 = r1 ∈ R and q1 ∈ R[x]. Thus r1 q = q1 f , and if we pass to residue classes in the polynomial ring (R/Rr1 )[x], we have q1 f = 0. Since f is monic, the leading coefficient of q1 must be 0, which means that q1 itself must be 0. Consequently, r1 divides every coefficient of q1 , so q1 /r1 ∈ R[x]. Thus f divides q in R[x].

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A course in commutative algebra by Ash R.B.


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