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The exercise that I'm having trouble with is the following.

Hartshorne II.6.4: Let $k$ be a field of characteristic $\neq 2$. Let $f \in k[x_1, \dots x_n]$ be a square free nonconstant polynomial, i.e. in the unique factorization of $f$ into irreducible polynomials, there are no repeated factors. Let $A=k[x_1 \dots x_n,z]/(z^2-f)$. Show that $A$ is an integrally closed ring. [Hint: The quotient field $K$ of $A$ is just $k(x_1, \dots x_n)[z]/(z^2-f)$. It is a Galois extension of $k(x_1, \dots x_n)$ with Galois group $\mathbb{Z}/2\mathbb{Z}$ generated by $z \mapsto -z$. If $\alpha=g+hz \in K$, where $g,h \in k(x_1, \dots x_n)$, then the minimal polynomial of $\alpha$ is $X^2-2gX+(g^2-h^2f)$. Now show that $\alpha$ is integral over $k[x_1, \dots x_n]$ if and only if $g,h \in k[x_1, \dots x_n]$. Conclude that $A$ is the integral closure of $k[x_1, \dots x_n]$ in $K$.]

Even with the hint I have two questions.

1) Why do I require $f$ to be square-free? I know that I do not want $f$ to be a square, because then $(z^2-f)$ would not be a prime ideal in $k[x_1, \dots x_n,z]$. I cannot see where square-free is used though.

2) For $\alpha=g+hz$ with $g,h \in k(x_1, \dots x_n)$, the minimal polynomial over $k(x_1, \dots x_n)[z]$ is exactly the one given above. However, why couldn't there be some OTHER monic polynomial $s \in k[x_1, \dots x_n][z]$ such that $s(\alpha)=0$? I don't understand why $\alpha$ integral over $k[x_1, \dots x_n]$ forces $g,h \in k[x_1, \dots x_n]$.

red_trumpet
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Marc Besson
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  • @Servaes In $k[x,y,z]$, $z^2-x^2y$ is a prime, though $x^2y$ is not square free. The correct argument is, if $f=p^2q$ for some prime $p$, then $z/p$ is integral, but not in the ring and thus the ring is not integrally closed. – Mohan Mar 06 '19 at 19:54
  • @Mohan You are right. – Servaes Mar 06 '19 at 20:17
  • See also https://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings-reid-exercise-4-6?noredirect=1&lq=1 – user26857 Mar 06 '19 at 21:28
  • Actually I'm right now confused why we need the hypthesis that $\text{char}(k) \neq 2$. I think this is needed to show that $Q(A)$ is a Galois extension of $k(x_1,\dots,x_n)$ (for normality, or $z \mapsto -z$ is identity), but I don't see why we even need this. Elements in $K$ are still of the form $g + hz$, $Q(A)$ is still an algebraic extension of $k(x_1,\dots,x_n)$, and the minimal polynimal still works out. – red_trumpet Mar 06 '19 at 22:07
  • I'm not very comfortable with inseparable extensions so I won't think about this very much, but if you find an answer I'd love to hear it. – Marc Besson Mar 06 '19 at 22:51
  • Or perhaps it is more simple: if we are in char$(k)=2$, then the minimal polynomial of $\alpha$ is $X^2+(g^2-h^2f)$; now perhaps it is possible for the last coefficient to be in $k[x_1, \dots x_n]$ without $g$ or $h$ being there. I'm not sure if this is right but the idea came from the stackexchange link above. – Marc Besson Mar 06 '19 at 22:58
  • Yeah right, that is where my argument breaks down. We cannot conclude $g \in k[x_1,\dots,x_n]$, if $\text{char }k = 2$. I*m not sure about a counterexample though – red_trumpet Mar 07 '19 at 09:02
  • You may find this answer helpful. – Viktor Vaughn Mar 07 '19 at 15:17
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    @red_trumpet Try $f=X^4+X^3+1$ over $\mathbb F_2$, and consider the element $\frac{X+1}{X}(1+\sqrt f)$ which is integral (why?) over $k[X][\sqrt f]$. – user26857 Mar 07 '19 at 23:50
  • @red_trumpet Please delete your last two comments in order to give some coherence to this chat. – user26857 Mar 07 '19 at 23:52

1 Answers1

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Regarding your first question, @Mohan posted a counterexample in the comments: $R = k[x,y,z]/(z^2 - x^2y)$ has an integral element $\frac{z}{x}$, which is not contained in $R$: $$\left(\frac{z}{x}\right)^2 = \frac{z^2}{x^2} = \frac{x^2y}{x^2} = y,$$ so an integral relation would be $t^2 - y$.

To see that square-free is sufficient, and to answer your second question we need a Theorem about how integral extensions of integral domains and algebraic extensions of their quotient fields relate. This can be found in Matsumura's Commutative ring theory, p.65:

Theorem 9.2. Let $A$ be an integrally closed domain, $K$ the field of fractions of $A$, and $L$ an algebraic extension of $K$. Then an element $\alpha \in L$ is integral over $A$ if and only if its minimal polynomial over $K$ has all its coefficients in $A$.

Here $A = k[x_1,\dots,x_n]$ which is a UDF, hence integrally closed in $K = k(x_1,\dots,x_n)$, and $L = k(x_1,\dots,x_n)[z]/(z^2 - f)$.

Now we can see the formal reason why $f$ has to be square-free: We have to show that if $\alpha = g + hz$ is integral, both $g$ and $h$ belong to $A = k[x_1,\dots,x_n]$. But the Theorem only tells aus that the coefficients of the minimal polynomial $m_\alpha = X^2 - 2gX + (g^2 - h^2f)$ belong to $A$: $g \in A$, because $2 \neq 0$ in $k$ (good) and $g^2 - h^2f \in A$, so at least $h^2 f \in A$. Now if $f$ is square-free, it cannot compensate any denominators in $h$, because those appear twice. Hence $h \in A$ follows.

red_trumpet
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