Let \(k\) be a field of characteristic different from two, put \(R=k[x_1,\ldots,x_n]\), and let \(f\in R\) be square-free and nonconstant. Thus no irreducible factor occurs more than once in the factorization of \(f\). Let \(A=R[z]/(z^2-f)\). Show that \(A\) is an integrally closed domain and is the integral closure of \(R\) in its fraction field \(K\).
(The polynomial in the hint)
For \(h\ne0\), the element \(g+hz\) generates \(K\) over \(F\), since \(z=((g+hz)-g)/h\). Its minimal polynomial is therefore the quadratic displayed in the hint. For \(h=0\), the minimal polynomial is instead \(T-g\), whereas that quadratic is \((T-g)^2\). The source calls the quadratic minimal without excluding \(h=0\) [@Har10a, Exercise II.6.4].
Hint
Put \(F=k(x_1,\ldots,x_n)\). First show that \(K=F[z]/(z^2-f)\) is a quadratic Galois extension of \(F\), with nontrivial automorphism \(\sigma(z)=-z\). For \(\alpha=g+hz\in K\), with \(g,h\in F\), the polynomial \[ T^2-2gT+(g^2-h^2f) \] annihilates \(\alpha\); it is the minimal polynomial when \(h\ne0\). Show that \(\alpha\) is integral over \(R\) if and only if \(g,h\in R\).
Solution
For an irreducible polynomial \(q\in R\) and \(u\in F^\times\), let \(v_q(u)\) be the exponent of \(q\) in the numerator of a reduced fraction for \(u\), minus its exponent in the denominator. The ring \(R\) is a unique factorization domain by Gauss's lemma, and is therefore integrally closed [@Har10a, proof of Proposition II.6.2].
1. The ring \(A\) is a domain with fraction field \(K=F(z)\), where \([K:F]=2\) and \(\sigma(z)=-z\) generates its Galois group.
Because \(f\) is nonconstant and square-free, it has an irreducible factor \(q\) with \(v_q(f)=1\). Every square in \(F^\times\) has even \(q\)-valuation, so \(f\) is not a square in \(F\). The quadratic \(T^2-f\) is consequently irreducible over \(F\).
Division by this monic polynomial gives \(A=R\oplus Rz\) as an \(R\)-module. The natural map \(A\to F[T]/(T^2-f)\) is injective, since \(1,z\) are linearly independent over \(F\) in the target field. It follows that \(A\) is a domain, and its fraction field is \(K=F(z)\). The roots \(z,-z\) are distinct because \(f\ne0\) and \(\operatorname{char}k\ne2\). They both lie in \(K\), so \(K/F\) is Galois of degree two with the asserted nontrivial automorphism.
2. If \(\alpha=g+hz\in K\) is integral over \(R\), then \(g,h\in R\).
The automorphism \(\sigma\) fixes \(R\), so \(\sigma(\alpha)=g-hz\) satisfies the same monic equation over \(R\) as \(\alpha\). Thus it is integral as well. The sum and product of these two integral elements are integral: the ring \(R[\alpha,\sigma(\alpha)]\) is a finite \(R\)-module, and the determinant trick applied to multiplication by their sum or product gives a monic equation for that element. Consequently \[ \alpha+\sigma(\alpha)=2g,\qquad \alpha\sigma(\alpha)=g^2-h^2f \] are integral over \(R\). Both belong to \(F\), so integral closedness of \(R\) implies \(2g,g^2-h^2f\in R\). Since \(2\) is a unit in \(R\), we obtain \(g\in R\) and \(h^2f\in R\).
If \(h=0\), the conclusion for \(h\) holds. If \(h\ne0\), then for every irreducible \(q\in R\), \[ 2v_q(h)+v_q(f)=v_q(h^2f)\ge0. \] Square-freeness gives \(v_q(f)\in\{0,1\}\), and \(v_q(h)\) is an integer, so \(v_q(h)\ge0\). A reduced fraction for \(h\) therefore has no irreducible factor in its denominator. Its denominator is a unit, and \(h\in R\).
3. The integral closure of \(R\) in \(K\) is \(A\), and \(A\) is integrally closed.
If \(g,h\in R\), then direct substitution shows that \(g+hz\) satisfies the monic equation \[ T^2-2gT+(g^2-h^2f)=0. \] It is therefore integral over \(R\). Together with step 2, this proves \[ \boxed{\{\alpha\in K:\alpha\text{ is integral over }R\}=R\oplus Rz=A}. \]
1.Hence \(A[\beta]\) is a finite \(R\)-module containing \(1\).The determinant trick for multiplication by \(\beta\) gives a monic equation over \(R\), so \(\beta\) is integral over \(R\) and belongs to \(A\) by the displayed equality.Thus \(A\) is integrally closed in \(K\).
4. Q.E.D.
Step 1 supplies the domain and its fraction field, and step 3 proves both requested integral-closure assertions.