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$\space$Suppose $R$ is a subring of the ring $S$ with $1\in R$ and assume $S$ is integral and finitely generated (as a ring) over R. If $P$ is a maximal ideal in $R$ then there is a nonzero and finite number of maximal ideals $Q$ of $S$ with $Q\cap R = P$.

$\space$I'm working on the verification of above corollary (Dummit and Foote, Abstract Algebra, p. 695).

$\space$In the proof described in the text, authors say "To prove that there are only finitely many possible $Q$ it suffices to prove that there are only finitely many homomorphism from $S$ to a field containing $R/P$ that extend the homomorphism from $R$ to $R/P$.", however, I can't understand the reason why this approach works.

$\space$Of course, for each maximal ideal $Q$, we can construct homomorphism from $S$ to $S/Q$ ,which is a field containing $R/P$, by natural projection. And $S/Q$ is isomorphic to one of the field extension of the form $(R/P)[\overline s_1,\overline s_2....,\overline s_n]$, and possible extension of such form is finitlely many.

$\space$ If $Q_1 \neq Q_2$ implies $S/Q_1 \ncong S/Q_2$ we have done. However, I can't prove this, and intuitively, it possibly be false... Could anyone help? Thanks.

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  • $\begingroup$ I read your post, as well as the solution by Moos. I just have one question. Why are there finitely many possible field extensions of the form you described? $\endgroup$ – Malkoun Mar 24 '17 at 8:33
  • $\begingroup$ @Malkoun Let me edit my answer. $\endgroup$ – MooS Mar 24 '17 at 8:34
  • $\begingroup$ @Moos: your edited answer now makes complete sense to me, thank you. $\endgroup$ – Malkoun Mar 24 '17 at 8:47
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    $\begingroup$ I know that what you want is to see how works a particular proof, but a more direct approach can be found here. $\endgroup$ – user26857 Mar 24 '17 at 10:06
  • $\begingroup$ Thank you very much for useful information! $\endgroup$ – Tetsu Hirosawa Mar 24 '17 at 12:11
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You have not really grasped which set exactly the authors claim to be finite. Let me spell it out.

Consider the set of pairs $(K,p)$ where $K$ is a field and $p$ an epimorphism $S \to K$, such that the composition $P \to R \to S \to K$ is zero.

First claim: This set is finite.

Second claim: This proves the desired statement. Let me proof this. Clearly any maximal ideal $Q$ with $Q \cap R = P$ yields such a pair $(S/Q,p)$ where $p$ is the projection $S \to S/Q$. And its also clear that different $Q$'s will yield different pairs, simply by the fact that $Q$ is the kernel of $p$, i.e. different $Q$'s yield different $p$'s.

Hence the set of such $Q$'s is smaller than the set of such pairs. So it suffices to prove the first claim, which you have already done.


Let me also proof the first claim, that there are finitely many pairs $(K,p)$: Let $s \in S$. Then we have a monic equation $$s^n+r_{n-1}s^{n-1}+ \dotsb + r_0=0.$$ Apply $p$ to get $$0=p(s)^n + p(r_{n-1})p(s)^{n-1} + \dotsb + p(r_0)$$ and note that for any $r \in R$, we have $p(r) = \overline r$, where $\overline r$ is the residue class in $R/P$, since $R \to S \to K$ factors over $R/P$.

Thus we have an equation

$$0=p(s)^n + \overline{r_{n-1}}p(s)^{n-1} + \dotsb + \overline{r_{0}},$$ which means that there are only finitely many choices for $p(s)$, namely at most the roots of this polynomial in $R/P[X]$.

Since $S$ is finitely generated over $R$, $p$ is uniquely determined by the images of the finitely many generators. We have just shown that each generator has only finitely many possibly images, so there only finitely many choices for $p$.

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  • $\begingroup$ Aah, very nice argument, thank you! $\endgroup$ – Malkoun Mar 24 '17 at 8:46
  • $\begingroup$ Thank you sooo much for informative suggestion and beautiful proof! Now, I understand! $\endgroup$ – Tetsu Hirosawa Mar 24 '17 at 12:15

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