is $V$ Zariski dense in $K\otimes V$?

Question

Suppose $V$ is a vector space over infinite field $k$ and $K$ an extension of $k$. If we consider $(V)_K= K \otimes V$ as a $K$-vector space then in what sense $V$ is Zariski dense in $(V)_K$?

Answer 1

Yes, $V$ is Zariski dense in $K \otimes_k V$ if $k$ is infinite.
Indeed the closure of $V$ consists in the intersection $\cap_P V(P)$ of the zero sets $V(P)\subset K \otimes_k V$ for all polynomials $P\in K[T_1,\cdots,T_n]$ vanishing on $V$.
But $P$ is necessarily the zero polynomial so that $V(P)=K\otimes_k V$ and thus the Zariski closure of $V$ is indeed $K\otimes_k V$.

Remarks
1) The assertion on polynomials vanishing if they vanish on $V\cong k^n$ can be found here .
2) The result is false for every finite field $k$:
Indeed take $V=k^1=k$. Then for any strict field extension $k\subsetneq K$ the subset $V=k\subsetneq K\otimes _k V=K$ is finite, thus closed and thus not dense.