
Lemma 10.122.2. Let $R$ be a ring. Let $\varphi : R[x] \to S$ be a ring map. Let $t \in S$. Assume that (a) $t$ is integral over $R[x]$, and (b) there exists a monic $p \in R[x]$ such that $t \varphi (p) \in \mathop{\mathrm{Im}}(\varphi )$. Then there exists a $q \in R[x]$ such that $t - \varphi (q)$ is integral over $R$.

Proof. Write $t \varphi (p) = \varphi (r)$ for some $r \in R[x]$. Using euclidean division, write $r = qp + r'$ with $q, r' \in R[x]$ and $\deg (r') < \deg (p)$. We may replace $t$ by $t - \varphi (q)$ which is still integral over $R[x]$, so that we obtain $t \varphi (p) = \varphi (r')$. In the ring $S_ t$ we may write this as $\varphi (p) - (1/t) \varphi (r') = 0$. This implies that $\varphi (x)$ gives an element of the localization $S_ t$ which is integral over $\varphi (R)[1/t] \subset S_ t$. On the other hand, $t$ is integral over the subring $\varphi (R)[\varphi (x)] \subset S$. Combined we conclude that $t$ is integral over the subring $\varphi (R)[1/t] \subset S_ t$, see Lemma 10.35.6. In other words there exists an equation of the form $t^ d + \sum _{i < d} (\varphi (r_ i)/t^{n_ i}) t^ i = 0$ in $S_ t$ with $r_ i \in R$. This means that $t^{d + N} + \sum _{i < d} \varphi (r_ i) t^{i + N - n_ i} = 0$ in $S$ for some $N$ large enough. In other words $t$ is integral over $R$. $\square$

Comment #3630 by Brian Conrad on

The end of the proof is not processing on my screeen (maybe a MathJax bug?), though it looks fine in the .pdf output.

Comment #3646 by on

Thanks for noticing this! It was a combination of Gerby and MathJax causing this. When Johan does the next update it should be fixed.

Comment #3959 by Manuel Hoff on

I think that an arbitrary element of $\varphi(R)[1/t]$ doesn't necessarily write in the form as claimed in the proof. Instead it should be of the form $\sum_j \frac{\varphi(r_j)}{t^j}$. Of course this doesn't change the argument.

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