# Showing product of normal varieties is normal

let $X, Y$ be two affine normal varieties, over an algebraically closed field $k$, I want to show that $X\times Y$ is still a normal affine variety.

There is an answer for "Does the fiber product of two normal varieties remain normal?", but I haven't learned scheme theory, so I am looking for a proof I am able to deal with.

So far, I can prove the result assuming $Y$ is affine space $Y=\mathbb A^n_k$. I list out sketch of my proof below. The strategy is that use the fact that the affine variety over a over an extended field $\mathbb A^n_{k(X)}$ is normal, (or equivalently affine space is geometrically normal). Using this fact, if we have an element in $k(X\times \mathbb A^n_k)$ integral over $k[X\times \mathbb A^n_k]$, we can make the denumerator very simple (only relies on variables in $X$). However, this method cannot be generalized, so I have no idea to prove the general case $X\times Y$ is normal.

Any suggestion or reference would be thankful.

proposition Let $k$ be an algebraically closed field, $X\subset \mathbb A^m_k$ a normal affine variety, then the product variety $X\times \mathbb A^n_k$ is affine normal.

sketch of proof: First $k[X\times \mathbb A^n_k]=k[X][y_1,y_2,...,y_n]$, $k(X\times \mathbb A^n_k)=k(X)(y_1,y_2,...,y_n)$. We want to prove that the coordinate ring $k[X][y_1,y_2,...,y_n]$ of $X\times \mathbb A^n_k$ is integrally closed in its fraction field $k(X)(y_1,y_2,...,y_n)$. Let $f\in k(X)(y_1,...,y_n)$ be an element which is integral over $k[X][y_1,y_2,...,y_n]$. We break up the proof into two steps:

Step 1: Show $f\in k(X)[y_1,y_2,...,y_n]$

This is easy because polynomial ring over a field is always a normal domain.

Step 2: Show $f\in k[X][y_1,y_2,...,y_n]$

Since step 1 is down, now $f=\sum_{i=1}^l u_i v_i$ for $u_i\in k(X)$, and $v_i\in k[y_1,...,y_n]$, also $v_i$'s are linearly independent over $k$. Note that if we restrict the integral relation $$f^k+a_{k-1}f^{k-1}+...+a_1f+a_0, \ \ \ \ a_i\in k[X][y_1,...,y_n]$$ to a fixed point $p$ in $\mathbb A^n_k$, then it gives a integral relation of $f(p)=\sum_{i=1}^l u_iv_i(p)\in k(X)$ with coefficients $a_i(p)\in k[X]$:$$f(p)^k+a_{k-1}(p)f(p)^{k-1}+...+a_1(p)f(p)+a_0(p), \ \ \ \ a_i(p)\in k[X]$$ By normality assumption of the variety $X$, $f(p)\in k[X]$.

Now, we claim that we can always pick $l$ points $p_1,...,p_l$ in $\mathbb A^n_k$ such that the vectors $(v_1(p_i),...,v_l(p_i)), 1\le i\le l$ are linearly independent (this is an easy argument since $k$ is infinite), therefore $u_i$'s are linear combination of $f(p_i)$'s, thus $u_i\in k[X], \forall i$. Thus $f=\sum_{i=1}^l u_i v_i\in k[X][y_1,y_2,...,y_n]$, as desired. $\tag*{$\blacksquare$}$