The Stacks project

Lemma 29.55.1. Let $A \to B$ be a ring map inducing a dominant morphism $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ of spectra. There exists an $A$-subalgebra $B' \subset B$ such that

  1. $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism,

  2. given a factorization $A \to C \to B$ such that $\mathop{\mathrm{Spec}}(C) \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism, the image of $C \to B$ is contained in $B'$.

Proof. We will use Lemma 29.45.6 without further mention. Consider the commutative diagram

\[ \xymatrix{ B \ar[r] & B_{red} \\ A \ar[u] \ar[r] & A_{red} \ar[u] } \]

For any factorization $A \to C \to B$ of $A \to B$ as in (2), we see that $A_{red} \to C_{red} \to B_{red}$ is a factorization of $A_{red} \to B_{red}$ as in (2). It follows that if the lemma holds for $A_{red} \to B_{red}$ and produces the $A_{red}$-subalgebra $B'_{red} \subset B_{red}$, then setting $B' \subset B$ equal to the inverse image of $B'_{red}$ solves the lemma for $A \to B$. This reduces us to the case discussed in the next paragraph.

Assume $A$ and $B$ are reduced. In this case $A \subset B$ by Algebra, Lemma 10.30.6. Let $A \to C \to B$ be a factorization as in (2). Then we may apply Proposition 29.46.8 to $A \subset C$ to see that every element of $C$ is contained in an extension $A[c_1, \ldots , c_ n] \subset C$ such that for $i = 1, \ldots , n$ we have

  1. $c_ i^2, c_ i^3 \in A[c_1, \ldots , c_{i - 1}]$, or

  2. there exists a prime number $p$ with $pc_ i, c_ i^ p \in A[c_1, \ldots , c_{i - 1}]$.

Thus property (2) holds if we define $B' \subset B$ to be the subset of elements $b \in B$ which are contained in an extension $ A[b_1, \ldots , b_ n] \subset B $ such that (*) holds: for $i = 1, \ldots , n$ we have

  1. $b_ i^2, b_ i^3 \in A[b_1, \ldots , b_{i - 1}]$, or

  2. there exists a prime number $p$ with $pb_ i, b_ i^ p \in A[b_1, \ldots , b_{i - 1}]$.

There are only two things to check: (a) $B'$ is an $A$-subalgebra, and (b) $\mathop{\mathrm{Spec}}(B') \to \mathop{\mathrm{Spec}}(A)$ is a universal homeomorphism. Part (a) follows because given $n \geq 0$ and $b_1, \ldots , b_ n \in B$ satisfying (*) and $m \geq 0$ and $b'_1, \ldots , b'_ m \in B$ satisfying (*), the integer $n + m$ and $b_1, \ldots , b_ n, b'_1, \ldots , b'_ m \in B$ also satisfies (*). Finally, part (b) holds by Proposition 29.46.8 and our construction of $B'$. $\square$

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0H3J. Beware of the difference between the letter 'O' and the digit '0'.