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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?

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  • $\begingroup$ 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
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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 $K[X_1,\dotsc,X_n]$.

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

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