In a proof of the weak Hilbert Nullstellensatz I read following statement (shortened):

Let $F$ be a field and let $F_i \in F[X_1,\ldots,X_n] \forall i=1,\ldots,m$ be fixed. Consider the map $$f:(A_1,\ldots,A_m) \mapsto A_1 F_1 + \ldots + A_m F_m$$

This is a $F$-linear map $V \to W$ between the finite dimensional subspaces $V \subset F[X_1,\ldots,X_n]^m , W \subset F[X_1,\ldots,X_n]$

It is obvious to me that $f$ is $K$-linear, but I don't see how $V,W$ are supposed to be finite dimensional, can anyone see that?

  • $\begingroup$ Without being too confident, is it possible that what is meant is $V=F^m$ and $W=\operatorname{Span}_F(F_1,\ldots,F_m)$? $\endgroup$ – Arthur Jul 29 '16 at 10:01
  • $\begingroup$ What do you mean by $V$ and $W$? It might help if you posted the source of the proof you are working through. $\endgroup$ – Claudius Jul 29 '16 at 10:10
  • $\begingroup$ As far as I can see, $V,W$ are not specified further. $\endgroup$ – flawr Jul 29 '16 at 10:30
  • $\begingroup$ The source is available here (page 35, below formula 1.6) $\endgroup$ – flawr Jul 29 '16 at 10:31

Instead of answering your question, I will try to explain how the argument in your source works.

So let $K$ be a field, $\overline K$ an algebraic closure of $K$. Let $I\subseteq K[X_1,\dotsc,X_n]$ be an ideal and suppose we have $F_1,\dotsc,F_m\in I$, $A_1,\dotsc,A_m\in \overline K[X_1,\dotsc,X_n]$ such that $$ A_1F_1+\dotsb+A_mF_m = 1.\tag{$*$} $$ The goal is to prove that $A_1,\dotsc,A_m$ can actually be chosen in $K[X_1,\dotsc,X_n]$.
For $P(X)= \sum_{i_1,\dotsc,i_n\ge0}a_{i_1,\dotsc,i_n}X_1^{i_1}\dotsm X_n^{i_n}$ denote by $\deg P := \max\{i_1+\dotsb+i_n\,|\, a_{i_1,\dotsc,i_n}\neq 0\}$ the degree of $P(X)$.

For $1\le i\le m$ let $V_i \subseteq K[X_1,\dotsc,X_n]$ denote the subspace of polynomials of degree $\le \deg A_i$. In particular, we have $A_i\in \overline K\cdot V_i \subseteq \overline K[X_1,\dotsc,X_n]$. Now, define $V:= V_1\times\dotsb\times V_m$.
Let $W\subseteq K[X_1,\dotsc,X_n]$ denote the subspace of all polynomials of degree $\le \max\{\deg A_iF_i\,|\, 1\le i\le m\}$. In particular, we have $1\in W$.

Now, $V$ and $W$ are finite dimensional subspaces and $f\colon V\rightarrow W$, $(P_1,\dotsc,P_m)\mapsto P_1F_1+\dotsb+P_mF_m$ is a well-defined $K$-linear map. Applying the Gauss algorithm to the corresponding matrix equation of ($*$) (with respect to any basis), and using that a solution exists over $\overline K$, it follows that there is already a solution over $K$, i. e. $A_1,\dotsc,A_m$ in ($*$) can be chosen in $K[X_1,\dotsc,X_n]$.

| cite | improve this answer | |
  • $\begingroup$ Oh now I see, the restriction of the degree is the key! Thank you very much! $\endgroup$ – flawr Jul 29 '16 at 12:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.