Hartshorne Exercise II.6.4.
Let $k$ be a field of characteristic $\ne 2$. Let $f\in k[x_1,…,x_n]$ be a squarefree nonconstant polynomial, i.e., in the unique factorization of $f$ into irreducible polynomials, there are no repeated factors. Let $A = k[x_1,…,x_n,z]/(z^2 – f )$. Show that $A$ is an integrally closed ring. [Hint: The quotient field $K$ of $A$ is just $k(x_1, … ,x_n)[z]/(z^2 – f)$. It is a Galois extension of $k(x_1, . . . ,x_n)$ with Galois group $\Z/2\Z$ generated by $z\mapsto -z$. If $\alpha=g+hz\in K$ ,where $g,h\in k(x_1,…,x_n)$, then the minimal polynomial of $\alpha$ is $X^2 – 2gX + (g^2 – h^2f)$. Now show that $\alpha$ is integral over $k[x_1,…,x_n]$ if and only if $g,h\in k[x_1,…,x_n]$. Conclude that $A$ is the integral closure of $k[x_1,…,x_n]$ in $K$.]
It is clear that if $g,h\in k[x_1,…,x_n]$ then $\alpha$ is integral over $k[x_1,…,x_n]$. Conversely, suppose $\alpha$ is integral over $k[x_1,…,x_n]$. Since $k[x_1,…,x_n]$ is a UFD, it is normal, so if $h=0$ then in fact $g\in k[x_1,…,x_n]$ too. Let $\alpha$ satisfy a monic polynomial $P(T)\in k[x_1,…,x_n](X)$, and denote the minimal polynomial of $\alpha$ by $p(X)\in k(x_1,…,x_n)(X)$. Then $p|P$ in $k(x_1,…,x_n)$. As $k[x_1,…,x_n]$ is a UFD, if a polynomial is irreducible in $k[x_1,…,x_n][X]$ then it is irreducible in $k(x_1,…,x_n)[X]$. It therefore follows that $\deg P =2$ and in fact $P=up$ for some unit $u\in k[x_1,…,x_n]^\times=k^\times$. We conclude that $p\in k[x_1,…,x_n]$, so $g\in k[x_1,…,x_n]$ so $h^2f\in k[x_1,…,x_n]$, and because $f\in k[x_1,…,x_n]$ is squarefree, in looking at a prime factorization of $h^2f$ it follows that $h\in k[x_1,…,x_n]$. Thus, if an element of $K$ satisfies a monic polynomial over $k[x_1,…,x_n]$ then it is an $A$ and if $\alpha=a+bz \in A$ then $\alpha$ satisfies a monic polynomial in $k[x_1,…,x_n]$, and we conclude that $A$ is the integral closure of $k[x_1,…,x_n]$ in $K$.
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