By Ash R.B.

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**Extra resources for A course in commutative algebra**

**Sample text**

We will show that the set M of nonunits of V is an ideal. If a and b are nonzero nonunits, then either a/b or b/a belongs to V . If a/b ∈ V , then a + b = b(1 + a/b) ∈ M (because if b(1 + a/b) were a unit, then b would be a unit as well). Similarly, if b/a ∈ V , then a + b ∈ M. If r ∈ V and a ∈ M, then ra ∈ M, else a would be a unit. Thus M is an ideal. 4. V is integrally closed. Let α be a nonzero element of K, with α integral over V . Then there is an equation of the form αn + cn−1 αn−1 + · · · + c1 α + c0 = 0 with the ci in V .

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}. Then f (x) = ai xi , we take h(f ) = J = h(I) is an ideal of F [x], necessarily principal. Say J = (j(x)). If j is nonconstant, it must have a root β in the algebraically closed ﬁeld C. We can then extend h to h : R[α] → C via h(α) = β, as desired.

By the basic correspondence between prime ideals in a ring and prime ideals in its localization, it suﬃces to show that P1 ST = P2 ST . We claim that P RT ⊆ (P1 ST ) ∩ RT ⊂ RT . The ﬁrst inclusion holds because P ⊆ P1 and RT ⊆ ST . The second inclusion is proper, for otherwise RT ⊆ P1 ST and therefore 1 ∈ P1 ST , contradicting the fact that P1 ST is a prime ideal. But P RT is a maximal ideal of RT , so by the above claim, (P1 ST ) ∩ RT = P RT , and similarly (P2 ST ) ∩ RT = P RT . Thus P1 ST and P2 ST lie over P RT .

### A course in commutative algebra by Ash R.B.

by Mark

4.5