One point in the post of Terence Tao on Ax-Grothendieck theorem - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T03:44:18Z http://mathoverflow.net/feeds/question/57515 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57515/one-point-in-the-post-of-terence-tao-on-ax-grothendieck-theorem One point in the post of Terence Tao on Ax-Grothendieck theorem aglearner 2011-03-05T23:55:49Z 2011-03-07T01:38:30Z <p>I was trying to understand completely the post of Terrence Tao on Ax-Grothendieck theorem.</p> <p><a href="http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/" rel="nofollow">http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/</a></p> <p>This is very cute. Using finite fields you prove that every injective polynomial map $\mathbb C^n\to \mathbb C^n$ is bijective. It seems to me that the only point in the proof presented in the post that is not explained completely is the following lemma:</p> <p>Lemma. Take any finitely generated ring over $\mathbb Z$ and quotient it by a maximal ideal. Then the quotient is a finite filed. </p> <p>Is there some comprehensible reference for the proof of this lemma?</p> <p>In slightly different wording, the question is the following: assuming Nullstelensatz, can one really give a complete proof of Ax-Grothendick theorem in two pages, so that it can be completely explained in one (2 hours) lecture of an undergraduate course on algebraic geometry? </p> http://mathoverflow.net/questions/57515/one-point-in-the-post-of-terence-tao-on-ax-grothendieck-theorem/57517#57517 Answer by Daniel Litt for One point in the post of Terence Tao on Ax-Grothendieck theorem Daniel Litt 2011-03-06T00:08:55Z 2011-03-06T00:08:55Z <p>Let $R$ be a finitely generated $\mathbb{Z}$-algebra, and $\mathfrak{m}\subset R$ are maximal ideal. We wish to show $R/\mathfrak{m}$ is a finite field.</p> <p>Let $i: \mathbb{Z}\to R$ be the unique ring map; then $i^{-1}(\mathfrak{m})$ is a maximal ideal in $\mathbb{Z}$ (as $R$ is finitely generated over $\mathbb{Z})$, and thus $\mathbb{Z}/i^{-1}(\mathfrak{m})$ is a finite field $\mathbb{F}_p$ for some prime $p$. As $R$ is finitely generated over $\mathbb{Z}$, $R/\mathfrak{m}$ is finitely generated over $\mathbb{F}_p$. But all finite field extensions of $\mathbb{F}_p$ are still finite, completing the proof.</p> http://mathoverflow.net/questions/57515/one-point-in-the-post-of-terence-tao-on-ax-grothendieck-theorem/57518#57518 Answer by Qing Liu for One point in the post of Terence Tao on Ax-Grothendieck theorem Qing Liu 2011-03-06T00:33:58Z 2011-03-06T00:56:14Z <p>To prove Nullstellensatz over $\mathbb{Z}$: as the morphism $f: \mathrm{Spec}(R)\to\mathrm{Spec}(\mathbb Z)$ is of finite type, a theorem of Chevalley says that the image of any constructible subset is constructible. So the image of any closed point by $f$ is a point which is a constructible subset. This can not be the generic point of $\mathrm{Spec}(\mathbb Z)$, so it must be a closed point.</p> <p>Note that this does not hold in general. For example, over the ring of $p$-adic integers, the ideal $(pX-1)\mathbb{Z}_p[X]$ is maximal, but its preimage in $\mathbb{Z}_p$ is $0$ and it not maximal. </p> <p>[<b>EDIT</b>] Another proof using Noether's normalization lemma: <a href="http://mathoverflow.net/questions/42276" rel="nofollow">http://mathoverflow.net/questions/42276</a>: if a maximal ideal $\mathfrak m$ of $R$ is such that $\mathfrak m\cap \mathbb Z=0$, then $R/\mathfrak m$ is finite type over (and contains) $\mathbb Z$. So there exits $f\in\mathbb Z$ non-zero and a finite injective homomorphism $\mathbb Z_f[X_1,\dots, X_d]\hookrightarrow R/\mathfrak m$. But then $\mathbb Z_f[X_1,\dots, X_d]$ must be a field. This is impossible because the units of this ring are $\pm f^k$, $k$ relative integers. </p> http://mathoverflow.net/questions/57515/one-point-in-the-post-of-terence-tao-on-ax-grothendieck-theorem/57525#57525 Answer by Takehiko Yasuda for One point in the post of Terence Tao on Ax-Grothendieck theorem Takehiko Yasuda 2011-03-06T02:21:35Z 2011-03-06T02:21:35Z <p>This is not an answer to your question, but let me point out that the Ax-Grothendieck theorem is now easy to prove using E-polynomials (Hodge-Deligne polynomials). If $f:X \to X$ is an injective endomorphism of a complex algebraic variety, then $E(X) = E(f(X))=E(X)-E(X\setminus f(X))$. So $E(X\setminus f(X))=0$ and $X\setminus f(X) = \emptyset$, because the degree of a constructible set is twice its dimension. Since one supposes the mixed Hodge theory, this proof is not trivial at all. But at least for me, this looks more natural. </p> http://mathoverflow.net/questions/57515/one-point-in-the-post-of-terence-tao-on-ax-grothendieck-theorem/57528#57528 Answer by Guillermo Mantilla for One point in the post of Terence Tao on Ax-Grothendieck theorem Guillermo Mantilla 2011-03-06T03:23:43Z 2011-03-07T01:38:30Z <p>One can give a more elementary proof of the fact that $\mathfrak{m} \cap \mathbb{Z} \neq 0$ - By more elementary I mean a proof that only uses the Nullstellensatz over $\mathbb{Q}$. </p> <p>Notice that it is enough to verify the claim for $R=\mathbb{Z}[x_1,..,x_n]$, and $\mathfrak{m} \in Max(R)$. </p> <p>Suppose there is $\mathfrak{m} \in Max(R)$ such that $\mathfrak{m} \cap \mathbb{Z} =0$. Then, we may assume that $\mathbb{Z} \subseteq F :=\mathbb{Z}[x_1,..,x_n]/\mathfrak{m}$. If we denote by $\alpha_{i}=x_i+\mathfrak{m}$ we have that $F=\mathbb{Z}[\alpha_1,..,\alpha_n]$. Since $F$ is a field we conclude that $\mathbb{Z}[\alpha_1,..,\alpha_n]=\mathbb{Q}(\alpha_1,..,\alpha_n)$. </p> <p>Claim: $F/\mathbb{Q}$ is an algebraic extension.</p> <p>proof: $F/\mathbb{Q}$ is a finitely generated field extension- generated as an algebra- in particular $F$ is of the form $\mathbb{Q}[y_1,..,y_m]/M$ for some $M$ maximal ideal of $\mathbb{Q}[y_1,..,y_m].$ By the Nullstellensatz $M$ has a zero $(\beta_1,...,\beta_m)$ where each $b_i$ is algebraic over $\mathbb{Q}$, so $F=\mathbb{Q}(\beta_1,...,\beta_m)$ is algebraic over $\mathbb{Q}$.</p> <p>Since each $\alpha_{i}$ is algebraic, there are integers $q_i$'s such that $q_{i}\alpha_{i}$ is integral over $\mathbb{Z}$ for all $i$. In particular $F=\mathbb{Z}[\alpha_1,..,\alpha_n]$ is an integral extension of $\displaystyle \mathbb{Z}[\frac{1}{q_1},..,\frac{1}{q_n}]$. Since $F$ is a field we have that $\displaystyle \mathbb{Z}[\frac{1}{q_1},..,\frac{1}{q_n}]$ is a field, which is a contradiction( $p$ is not invertible for any prime not dividing $q_{1}...q_{n}$). </p>