Conjugation is not expressible in terms of polynomials In order to convince myself that the set $U(n)$ of unitary matrices (matrices with columns that are orthonormal under the complex inner product) is not an affine variety in $\mathbb{C}^{n^2}$, I need that the conjugate of an entry $\overline{x}_{ij}$ cannot be expressed using polynomials. If we say that complex polynomials have to be finite sums of monomials of the form $cx_1^{e_1}\dots x_n^{e_n}$ where $c \in \mathbb{C}$, $e_i \geq 0$, then is this really enough? It seems a little slippery.
 A: Probably Perhaps the most conceptually straightforward explanation of why $U(n)$ is not an affine variety over $\mathbb{C}$ is the following: with respect to the complex topology (the usual topology -- not the Zariski topology!) $U(n)$ is compact, whereas it can be shown that a variety $V_{/\mathbb{C}}$ is such that $V(\mathbb{C})$ is compact (with respect to complex topology) iff $V$ is complete.  In particular, a complete affine variety is finite, which $U(n)$ evidently is not.
Of course all of these facts require (nontrivial) proof.  If I am remembering correctly, they are well treated in Shafarevich's text on algebraic geometry, for instance.
A: Here is a proof that $U(n)$ is not an affine variety in $\mathbb{C}^{n^2}$: If $X \subset \mathbb{C}^N$ is a smooth affine subvariety of $\mathbb{C}^N$, then, at every point of $X$, the tangent space to $X$ will be a complex sub-vector-space of $\mathbb{C}^N$. (Proof: The tangent space to a polynomial hypersurface is a complex sub-vector-space; the tangent space to an intersection is the intersection of tangent spaces. This argument also works for singular varieties if you use the Zariski tangent space.)
At the identity, the tangent plane to $U(n)$ is the Hermitian matrices. Mutiplyling a Hermitian matrix by $i$ does not give you a Hermitian matrix, so $U(n)$ is not an algebraic subvariety. 
A: That conjugation is not expressible in terms of polynomials comes from the Cauchy–Riemann equations, but that is not what you need to prove.
You need to prove that the set $U(n)$ cannot be defined by a system of polynomial equations in the entries of the matrix.
Here is a simple proof. Consider the set $X$ of matrices $(a_{ij})$ such that $a_{ij}=\delta_{ij}$, except for $i=1, j=1$. Then $X$ is an affine variety in $\mathbb{C}^{n\times n}$. If $U(n)$ is an affine variety, then so is $Y=X \cap U(n)$. But $Y$ corresponds to the unit circle in $\mathbb C$. However, the unit circle is not an affine variety in $\mathbb C$, because those are either $\emptyset$, $\mathbb C$, or a finite set of points.
