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.
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 $g,h\in \operatorname{End}(V)$.