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Tag 00R2

Chapter 10: Commutative Algebra > Section 10.6: Ring maps of finite type and of finite presentation

Lemma 10.6.3. Let $R \to S$ be a ring map of finite presentation. For any surjection $\alpha : R[x_1, \ldots, x_n] \to S$ the kernel of $\alpha$ is a finitely generated ideal in $R[x_1, \ldots, x_n]$.

Proof. Write $S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_k)$. Choose $g_i \in R[y_1, \ldots, y_m]$ which are lifts of $\alpha(x_i)$. Then we see that $S = R[x_i, y_j]/(f_j, x_i - g_i)$. Choose $h_j \in R[x_1, \ldots, x_n]$ such that $\alpha(h_j)$ corresponds to $y_j \bmod (f_1, \ldots, f_k)$. Consider the map $\psi : R[x_i, y_j] \to R[x_i]$, $x_i \mapsto x_i$, $y_j \mapsto h_j$. Then the kernel of $\alpha$ is the image of $(f_j, x_i - g_i)$ under $\psi$ and we win. $\square$

    The code snippet corresponding to this tag is a part of the file algebra.tex and is located in lines 556–561 (see updates for more information).

    \begin{lemma}
    \label{lemma-finite-presentation-independent}
    Let $R \to S$ be a ring map of finite presentation.
    For any surjection $\alpha : R[x_1, \ldots, x_n] \to S$ the
    kernel of $\alpha$ is a finitely generated ideal in $R[x_1, \ldots, x_n]$.
    \end{lemma}
    
    \begin{proof}
    Write $S = R[y_1, \ldots, y_m]/(f_1, \ldots, f_k)$.
    Choose $g_i \in R[y_1, \ldots, y_m]$ which are lifts
    of $\alpha(x_i)$. Then we see that $S = R[x_i, y_j]/(f_j, x_i - g_i)$.
    Choose $h_j \in R[x_1, \ldots, x_n]$ such that $\alpha(h_j)$
    corresponds to $y_j \bmod (f_1, \ldots, f_k)$. Consider
    the map $\psi : R[x_i, y_j] \to R[x_i]$, $x_i \mapsto x_i$,
    $y_j \mapsto h_j$. Then the kernel of $\alpha$
    is the image of $(f_j, x_i - g_i)$ under $\psi$ and we win.
    \end{proof}

    Comments (3)

    Comment #1099 by Filip Chindea on October 20, 2014 a 2:15 pm UTC

    It is obvious that $(f_j, x_i - g_i) \subset \ker(\alpha)$, but I don't know how to prove the reverse inclusion and I would be grateful for a hint. The proof in EGA is quite involved, and I would have liked to see an elementary alternative. Thank you.

    Comment #1132 by Johan (site) on November 4, 2014 a 6:32 pm UTC

    For any ring $A$ and element $a \in A$ we have $A[x]/(x - a) = A$. Does this help? Or am I misunderstanding the question?

    Comment #1158 by Filip Chindea on November 22, 2014 a 2:41 pm UTC

    Sorry for that, it was obvious: just consider the projection to $R[x_i, y_j]/(f_j, x_i - g_i)$ and see what happens after composition. Still the proof is amazingly short.

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