What is the Zariski closure of the unitary group?

Question

Let $U(n,\mathbb{C})\subseteq GL(n,\mathbb{C})$ be the group of unitary $n\times n$ matrices. It is well-known that $U(n,\mathbb{C})$ is not Zariski closed (Why the unitary group is not a complex algebraic variety?). What is the Zariski closure of $U(n,\mathbb{C})$?

Answer 1

The Zariski closure of $U(n,\mathbb{C})$ is all of $GL(n,\mathbb{C})$. First, it is easy to see the Zariski closure of $U(n,\mathbb{C})$ contains all diagonal matrices. (This amounts to saying that if a polynomial in $n$ variables vanishes on $(S^1)^n$ then it must vanish on $\mathbb{C}^n$, which you can prove by induction as in this answer, for instance.) But every matrix can be written as a product of diagonal matrices and unitary matrices (for instance, via a singular value decomposition). Since $U(n,\mathbb{C})$ is closed under multiplication, so is its Zariski closure, and thus the Zariski closure is all of $GL(n,\mathbb{C})$.