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I have some probelems with the following statement: Let $S, R$ be rings. Suppose $S$ is an integral extension of $R$ and $I$ is an ideal in $S$ then $S/I$ is an integral extension of $R/(R\cap I)$.

These are my thoughts so far. Let $i$ be embeddings and $\pi$ natural projections the following diagram commutes (right?) \begin{matrix} R & \overset{\pi _1}{\rightarrow } & R/(I\cap R)\\ i\downarrow & & i\downarrow\\ S& \overset{\pi _2}\rightarrow & S/I \end{matrix} For any $\overline{s}\in S/I$, there exists a monic polynomial $p(x)\in S[x]$ with $p(s)=0$, clearly $\pi _1 (p(s))=0$ in $R/(I\cap R)$ and $\pi _1(p(x))$ is still monic. Furthermore $\pi _2(i(p(x)))=i(\pi _1(p(x)))$ since $p(x)$ has coefficents in $R$.

I feel like one have to show that if $\overline{t}= \overline{s}$ then $p(t)\in R$.

If $p(t)\in R$ then $p(t)-p(s)\in (I\cap R)$ and hence $t$ is also a root of $\pi _1(p(x))\in R/(I\cap R)[x]$. Is this not needed? How can one explicit argue.

Michael Hardy
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harajm
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    Your proof is already complete. You don't have to prove $\overline{t}=\overline{s} \Rightarrow p(t) \in R$... – Martin Brandenburg May 27 '13 at 22:50
  • please explain why!! – harajm May 27 '13 at 22:51
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    I suggest using \cap instead of \bigcap when you intend it as a binary operation, e.g. $A\cap B$ instead of $A\bigcap B$. The \bigcap symbol is normally used for tthe intersection of a family of sets, e.g. $\bigcap_{k=1}^n A_k$. – Zev Chonoles May 27 '13 at 22:52
  • @ Chonoles -Thanks didn't know that! @ Brandenburg - for me it feels like there may be infinitely many representatives of $\overline{s}$ satisfying different monic polynomials in $R/(R\cap I)$... – harajm May 27 '13 at 22:57
  • @harajm: Please explain why you want to prove something which is not necessary (and besides, wrong). – Martin Brandenburg May 28 '13 at 21:36

1 Answers1

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Let $\bar{s} \in S/I$ with $s \in S$. By hypothesis, $s$ is integral over $R$, hence there are $a_i \in R$ such that $s^n+a_{n-1}s^{n-1}+\cdots+a_1 s + a_0=0$. Now this equation modulo $I$ becomes $\bar{s}^n+\bar{a}_{n-1}\bar{s}^{n-1}+\cdots+\bar{a}_1 \bar{s} + \bar{a}_0=0$ where we view the coefficients $\bar{a}_i$ as elements of the $R$-module $(R+I)/I$. Now by the third isomorphism theorem we have $(R+I)/I \cong R/(R\cap I)$ and we are done.

Alternatively and more consistently with your notation, let $\pi:S \rightarrow S/I$ be the natural projection. For $s \in S$ there exists monic $p(x) \in R[x]$ such that $p(s)=0$. Now $p(s)$ is an element of $S$ and its image under $\pi$ is $p^{\pi}(\pi(s))=0$, where $p^{\pi}$ denotes the monic polynomial that we obtain from $p(x)$ be applying $\pi$ to its coefficients. Now use the third isomorphism theorem to see that actually $p^{\pi}(x) \in R/(R\cap I)[x]$.

Manos
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  • Is it possible to argue without using the 2nd isomorphism theorems for rings explicitly, is this a correct argument: Let $\overline{p(x)}= \pi _2 (i(p(x)))$ be the image of $p(x)\in S/Q$ then $\overline{p(\overline{s})}=\pi _2 (i(p(s)))= i(\pi_1(p(s)))=0$ and hence $\overline{s}$ is a root of $\pi _1 (p(x))$. – harajm Jun 02 '13 at 20:14