Lemma 10.38.2. Let $\varphi : R \to S$ be a ring map. Let $I \subset R$ be an ideal. Let $A = \sum I^ nt^ n \subset R[t]$ be the subring of the polynomial ring generated by $R \oplus It \subset R[t]$. An element $s \in S$ is integral over $I$ if and only if the element $st \in S[t]$ is integral over $A$.

Proof. Suppose $st$ is integral over $A$. Let $P = x^ d + \sum _{j < d} a_ j x^ j$ be a monic polynomial with coefficients in $A$ such that $P^\varphi (st) = 0$. Let $a_ j' \in A$ be the degree $d-j$ part of $a_ j$, in other words $a_ j' = a_ j'' t^{d-j}$ with $a_ j'' \in I^{d-j}$. For degree reasons we still have $(st)^ d + \sum _{j < d} \varphi (a_ j'') t^{d-j} (st)^ j = 0$. Hence $s^ d + \sum _{j < d} \varphi (a_ j'') s^ j = 0$ and we see that $s$ is integral over $I$.

Suppose that $s$ is integral over $I$. Say $P = x^ d + \sum _{j < d} a_ j x^ j$ with $a_ j \in I^{d-j}$. Then we immediately find a polynomial $Q = x^ d + \sum _{j < d} (a_ j t^{d-j}) x^ j$ with coefficients in $A$ which proves that $st$ is integral over $A$. $\square$

## Comments (2)

Comment #8318 by Et on

I think the first paragraph has some typos. In the third sentence, instead of a_i it should be a_j. The end of the paragraph should also display a relation on s, not on st.

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