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Let $\mathbb{K}$ be an infinite field, $V, W$ be $\mathbb{K}$-vectorspaces and let $D: \operatorname{GL}(V) \to \operatorname{GL}(W)$ be a polynomial representation, that means $D$ is the restriction of a polynomial map $\tilde{D}: \operatorname{End}(V) \to \operatorname{End}(W)$ and $D$ satisfies $D(gh)=D(g)D(h)$ for all $g,h \in\operatorname{GL}(V)$.

Does it follow from this that $\tilde{D}(gh)=\tilde{D}(g)\tilde{D}(h)$ is also true for all $g,h \in \operatorname{End}(V)$? If so, why?

If $\mathbb{K} \in \{\mathbb{C}, \mathbb{R}\}$, then I guess we can say that $\tilde{D}$ is a continuous function, since it is polynomial, and since $\operatorname{GL}(V)$ lies dense in $\operatorname{End}(V)$ we can use an analytical argument to prove this.

But is there also a more algebraic way, for example by using the theory of polynomials, that works for other infinite fields as well? I know that if $p(x) = q(x)$ for infinitely many $x\in \mathbb{K}$, then the polynomials $p$ and $q$ are the same. But I guess this is not true for multivariate polynomials, so is there something similar that can be applied here?

Thanks in advance.

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    $\begingroup$ Yes, there is something that does essentially the same. This is called the Zariski topology, and it precisely allows you to make these sorts of arguments for polynomial maps. $\endgroup$ – Tobias Kildetoft Mar 6 '18 at 13:18
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First, note that if we can show this for an algebraically closed field, then it follows for any field by extension of scalars, so let $k$ be an algebraically closed field.

We will identify $\operatorname{End}(V)$ with $k^{n^2}$ where $n$ is the dimension of $V$. We give this the Zariski topology where the closed sets are the sets of common zeroes of some set of polynomials.

With this, we see that $\operatorname{GL}(V)$ is an open subset of $\operatorname{End}(V)$, since it is the preimage of the complement of the closed set $\{0\}$ in $k$ under the polynomial map $\operatorname{Det}$. The same argument shows that $\operatorname{GL}(V)\times\operatorname{GL}(V)$ is open in $\operatorname{End}(V)\times \operatorname{End}(V)$, by multiplying the determinants.

But in fact, the space $k^m$ (for any $m$) with the Zariski topology has some very important properties for us:
1) It is irreducible, which implies that all non-empty open subsets are dense.
2) It is separable, which means that the diagonal in $k^{m_1}\times k^{m_2}$ is closed when we identify $k^{m_1}\times k^{m_2}$ with $k^{m_1 + m_2}$ and give this the Zariski topology.

Now, returning to our specific case, we have a polynomial map $\tilde{D}: \operatorname{End}(V) \to \operatorname{End}(W)$ such that $\tilde{D}(gh) = \tilde{D}(g)\tilde{D}(h)$ for all $g,h\in \operatorname{GL}(V)$.

Let us note that by 1), $\operatorname{GL}(V)\times \operatorname{GL}(V)$ is dense in $\operatorname{End}(V) \times \operatorname{End}(V)$. So it suffices to show that the set of pairs $(g,h)\in \operatorname{End}(V)\times \operatorname{End}(V)$ such that $\tilde{D}(g)\tilde{D}(h) = \tilde{D}(gh)$ is a closed subset.

To see this, we consider the polynomial map $\operatorname{End}(V)\times \operatorname{End}(V) \to \operatorname{End}(W) \times \operatorname{End}(W)$ given by $(g,h)\mapsto (\tilde{D}(g)\tilde{D}(h),\tilde{D}(gh))$. The set of pairs $(g,h)$ with $\tilde{D}(g)\tilde{D}(h) = \tilde{D}(gh)$ is then the preimage of the diagonal in $\operatorname{End}(W)\times \operatorname{End}(W)$, which is closed by 2), and this shows that $\tilde{D}(g)\tilde{D}(h) = \tilde{D}(gh)$ for all $g,h\in \operatorname{End}(V)$.

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