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Let $k$ be a field. Let $\bar k$ be an algebraic closure of $k$. Let $G = \mbox{Aut}(\bar k/k)$. Let $n \ge 1$ be an integer. $G$ acts on $\bar k^n$ in the obvious way. Let $V$ be an irreducible algebraic set in $\bar k^n$. Let $\sigma \in G$. It is easy to see that $\sigma(V) = \{\sigma(x)\colon x \in V\}$ is an irreducible algebraic set. Hence $G$ acts on the set of irreducible algebraic sets in $\bar k^n$.

Is the following proposition true? If yes, how do we prove it?

Proposition. Let $\mathfrak p$ be a prime ideal of the polynomial ring $k[X_1, \dots, X_n]$. Let $V$ be the algebraic set in $\bar k^n$ defined by $\mathfrak p$. Let $V_1, \dots, V_r$ be the irreducible components of $V$. Then $G$ acts transitively on the set $\{V_1, \dots, V_r\}$. Moreover, $\dim V_i = \dim k[X_1, \dots, X_n]/\mathfrak p$ for all $i$.

Conversely let $W$ be an irreducible algebraic set in $\bar k^n$. Then the $G$-orbit $\{\sigma(W)\colon \sigma \in G\}$ is finite. Let $V = \bigcup_{\sigma\in G} \sigma(W)$. Then there exists a prime ideal $\mathfrak p$ of $k[X_1, \dots, X_n]$ such that $V$ is the algebraic set defined by $\mathfrak p$.

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It would probably be useful if, in your questions, you noted whether you know the answer or not. I have noticed that I, for one, do not read them and that is one significant factor. –  Mariano Suárez-Alvarez Oct 22 '12 at 23:17
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@MarianoSuárez-Alvarez Could you please explain why that is one significant motivating factor of your not even reading my questions? –  Makoto Kato Oct 24 '12 at 12:34
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I find no motivation at all in answering your questions when you know the answers; I see no point in it, and the idea to build a «database» of sorts in a medium such as this site attracts me even less and, really, bores me : both seem to me to be quite at odds with what a site like this might be good at. This is probably just me, of course. As I am being candid, let me add that the fact that interacting with you feels often very close to interacting with ELIZA does not help. –  Mariano Suárez-Alvarez Oct 24 '12 at 15:47
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@MarianoSuárez-Alvarez Please just ignore my questions if you are not interested in them. –  Makoto Kato Oct 24 '12 at 16:16
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@navigetor23 Thanks. That's much better than posting nonconstructive comments. –  Makoto Kato Oct 27 '12 at 12:21

1 Answer 1

up vote 1 down vote accepted

After posting some nonconstructive comments I decided to post a constructive answer.

Set $A=k[X_1, \dots, X_n]$ and $B=\bar k[X_1, \dots, X_n]$. The ring extension $A\subset B$ is flat (why?). Let $\mathfrak p$ be a prime ideal of $A$. The irreducible components of $V$, the algebraic set in $\bar k^n$ defined by $\mathfrak p$, are defined by the prime ideals $P_1,\dots,P_r$ of $B$ which are minimal over $\mathfrak pB$. Since the extension $A\subset B$ is flat it follows that $P_i\cap A=\mathfrak p$ for all $i$. Moreover, the ring extension $A_{\mathfrak p}\subset B_{P_i}$ is also flat and now we can apply the dimension formula and get $\dim B_{P_i}=\dim A_{\mathfrak p}+\dim B_{P_i}/\mathfrak pB_{P_i}$. But $\dim B_{P_i}/\mathfrak pB_{P_i}=0$ (why?) and thus we get $\dim B_{P_i}=\dim A_{\mathfrak p}$, that is, $\mbox{ht}(P_i)=\mbox{ht}(\mathfrak p)$ and this is enough to show that $\dim V_i = \dim k[X_1, \dots, X_n]/\mathfrak p$.

The extension $A\subset B$ is an integral extension of integrally closed domains. Furthermore, the field extension $K\subset L$ is normal, where $K$ and $L$ are the fields of fractions of $A$, respectively $B$, so we can apply Theorem 5(vi), page 33, from Matsumura, CA, that says the following: any two prime ideals $P_1$ and $P_2$ of $B$ lying over the same prime ideal $\mathfrak p$ of $A$ are conjugate, i.e. there is $\bar\sigma\in G(L/K)$ such that $\bar\sigma(P_2)=P_1$. Now take $\sigma=\bar\sigma_{\mid \bar k}\in G$. If I'm not wrong, $\sigma(V_2)=V_1$.

For the converse, let $P$ be a prime ideal of $B$ that defines $W$ and $\mathfrak p=P\cap A$. It remains to prove that $W$ is the algebraic set of $\bar k^n$ defined by $\mathfrak p$. This can be reformulated as follow: the prime ideals of $B$ lying over $\mathfrak p$ are the minimal elements of the set of the prime ideals of $B$ containing $\mathfrak p$, and this is easy to prove by using the well-known properties of integral extensions.

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You don't need to invoke the dimension formula for a flat ring extension. The equality $\mbox{ht}(P)=\mbox{ht}(\mathfrak p)$ for any prime ideal $P$ lying over $\mathfrak p$ follows immediately from Theorem 5(vi) of Matsumura you referred to. –  Makoto Kato Nov 28 '12 at 18:33

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