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
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