Tricki
• LaTeX processing error: math_failure (math_lexing_error):. By [ref Proposition #redp] the image is a finite field
• LaTeX processing error: math_failure (math_lexing_error):, is injective but not surjective. This is manifestly nonsense on cardinality grounds. </div> </div> <div class="tricki_section_example"> <h3 id="ex2">Example 2: Jordan-Chevalley decomposition in algebraic subgroups of
• LaTeX processing error: math_failure (math_lexing_error): is any field (actually, it needs to be a [[w:Perfect_field|perfect field]], but the fields we mention below are all perfect). The [[w:Jordan-Chevalley_decomposition|Jordan–Chevalley Decomposition]</h3> <div class="tricki_section_example_content"> of
• LaTeX processing error: math_failure (math_lexing_error): may be conjugated to an upper-triangular matrix with ones on the diagonal. It is a well-known theorem of linear algebra that the Jordan-Chevalley decomposition exists and is unique. <div class="tricki_env_theorem" id="theorem_jnf_env_box"><a class="tricki_env_theorem_title" id="theorem_jnf">Theorem 3</a> Let
• LaTeX processing error: math_failure (math_lexing_error):.</div> We will deduce this from the following finite field version of the result. <div class="tricki_env_theorem" id="theorem_jnff_env_box"><a class="tricki_env_theorem_title" id="theorem_jnff">Theorem 4</a> Let
• LaTeX processing error: math_failure (math_lexing_error):.</div> Let us give the deduction of [ref Theorem #jnf], as this is main point of this discussion. Suppose that
• LaTeX processing error: math_failure (math_lexing_error):. This is a finitely-generated ring, and [ref Theorem #jnf] takes place ''over
• LaTeX processing error: math_failure (math_lexing_error):. By the first part of [ref Proposition #redp] the image
• LaTeX processing error: math_failure (math_lexing_error):. By [ref Theorem #jnff] (applied to the subgroup of
• LaTeX processing error: math_failure (math_lexing_error):. To conclude, we use the second part of [ref Proposition #redp], which asserted that the intersection of all maximal ideals
• LaTeX processing error: math_failure (math_lexing_error):. It remains to prove [ref Theorem #jnff]. We leave the details as an exercise to the reader, using the following three observations: (1) If

## Revision of Use finite fields from Wed, 15/04/2009 - 12:54

### Quick description

If you have an algebraic problem about the complex numbers, it might be possible to solve it by solving a similar, but easier, problem about finite fields.

### Prerequisites

Basic notions of commutative and linear algebra (more sophisticated ideas are required to understand the proofs of the results quoted, but this is not needed to apply the trick).

This trick depends on the following proposition.

Proposition 1 (Reduction mod primes) Suppose that is a finitely-generated subring of . Then the quotient of by a maximal ideal is always a finite field, and .

In fact for the first example below we only require the fact that has some maximal ideal such that the quotient is a finite field. The proof of this (and a fortiori of the first statement in the proposition) is not particularly easy to find in the literature. Some links are given below, and perhaps the standard reference would be Bourbaki's Algebre Commutative, Ch. V, Sec 3, no. 4, Corollaire I, p.64.

### Example 1: Ax-Grothendieck theorem] The discussion here is cribbed from this article by Serre and a [http://terrytao.wordpress.com/2009/03/07/infinite-fields-finite-fields-and-the-ax-grothendieck-theorem/ blog by Tao

that followed from it, together with comments on that blog.

Theorem 2 Suppose that is an injective polynomial map. Then is also surjective.

The proof is by contradiction. If is injective but not surjective then

• vanishes whenever does and

• there is some such that vanishes whenever does (i.e. never).

The rather odd way of expressing these statements is so that we may apply Hilbert's Nullstellensatz. The two statements imply, in turn, that

• some power lies in the ideal generated by and

• some power lies in the ideal generated by .

More concretely,

• There is a polynomial such that \tilde Q(x) such that R\CP,Q\tilde Q\mathfrak{m}R\pi : R \rightarrow R/\mathfrak{m}. By Proposition 1 the image is a finite field \mathbb{F}\pi\mathbb{F}P\mathbb{F}^n\mathbb{F}^n, is injective but not surjective. This is manifestly nonsense on cardinality grounds.

</div> </d...\mbox{GL}_d(\C)x \in \mbox{GL}_d(k)k is any field (actually, it needs to be a perfect field, but the fields we menti...xx = x_u x_s = x_s x_ux_sx_u\overline{k}t(t-1)^d = 0t may be conjugated to an upper-triangular matrix with ones on the diagonal. It is a well-known theor...G \subseteq \mbox{GL}_d(\C)d^2Gx \in Gx_sx_uG.</div> We will deduce this from the following finite field version of the result. <div class="tri...G \subseteq \mbox{GL}_d(\mathbb{F})\mathbb{F}Gx \in Gx_sx_uG.</div> Let us give the deduction of Theorem 3, as this is main point of this discussion. ...Gx \in GGf_1,\dots,f_mgg^{-1}x_s gRx_sx_ugf_1,\dots,f_m. This is a finitely-generated ring, and Theorem 3 takes place ''over R\mathfrak{m}R\pi : R \rightarrow R/\mathfrak{m}. By the first part of Proposition 1 the image R/\mathfrak{m}\pi(x)\pi(x_s)\pi(x_u). By Theorem 4 (applied to the subgroup of \mbox{GL}_d(\mathbb{F})\pi(x)\pi(x_s)\pi(x_u)\pi(x)x_s \equiv x^a \mod{\mathfrak{m}}x_u \equiv x^b \mod{\mathfrak{m}}a,bf_1,\dots,f_mGx^ax^bf_i(x_s) \equiv f_i(x_u) \equiv 0 \mod{\mathfrak{m}}i = 1,\dots,m. To conclude, we use the second part of Proposition 1, which asserted that the intersecti...\mathfrak{m}f_i(x_s) = f_i(x_u) = 0i = 1,\dots,mx_sx_uG. It remains to prove Theorem 4. We leave the details as an exercise to the reader, usin...x \in \mbox{GL}_d(\mathbb{F})xy\mbox{char}(\mathbb{F})yz\mbox{char}(\mathbb{F})zx_sx_ux\$.

### General discussion

There are connections between ideas of this kind and logic/model theory, but this author is not qualified to discuss them. More details may be found in the "Princeton Companion" article by David Marker, where explicit mention of the Ax-Grothendick theorem is made on p642.