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I solved all parts of Exercise 4.6 of the book Undergraduate Commutative Algebra of Miles Reid except the last one.

Let $A=k[X]$ and $f\in A$ has a square factor but it is not a square polynomial itself, then what will be the normalization of $B=k[X,Y]/(Y^2-f)$ in terms of $f$?

What should I do for this part as I can't use any parametrization.

user26857
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H.W.
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1 Answers1

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(In the following $x,y$ denote the residue classes of $X,Y$ modulo the ideal $(Y^2-f)$.)

The field of fractions of $B$ is $K=k(x,y)$ and every element $t\in K$ can be written as $t=a+by$ with $a,b\in k(x)$. Obviously $t$ is a root of the polynomial $p_t(T)=T^2-2aT+a^2-b^2f$ which belongs to $k(x)[T]$. Since $x$ is algebraically independent over $k$, by Gauss' Lemma we can conclude that $t$ is integral over $A$ if and only if $p_t$ has the coefficients in $k[x]$. Now we assume $\operatorname{char}k\ne2$. Then we get $p_t\in k[x][T]$ if and only if $a\in k[x]$ and $b^2f\in k[x]$. If write $f=g^2h$ with $h$ square free we can easily get the integral closure of $B$: $k[x]+k[x]\frac{y}{g}$.

user26857
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    I was wondering if you could you expand a little on this answer? Namely, could you say how do you use the Gauss' Lemma to conclude that t is integral iff $p_t$ has coefficients in k[x]? And the last line is completely obscure for me. How do you conclude? I have been looking at this for quite some time now, and I would love to at least understand this problem, if I obviously can't solve it right now. Thanks. – baltazar Dec 09 '14 at 20:54
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    @baltazar $A$ is a UFD and Gauss' lemma says that a monic polynomial in $A[x]$ is irreducible iff it is irreducible in $K[x]$, where $K$ is the field of fractions of $A$. For the last part notice that given $u\in k(x)$ we have $u^2f\in k[x]$ iff $ug\in k[x]$. – user26857 Dec 09 '14 at 21:06
  • @user26857: Is have a follow-up question: Let's change the situation from $y^2-f(x)$ to $y^5-f(x)$. Then we may write $t$ as $t=a_0+a_1y+\ldots+a_4y^4$, which makes calculations a lot more difficult.

    But one can find a suitable $t$ like $t=y^3/x$. Now how do I prove that the Ring $k[x]+t\cdot k[x]$ is integrally closed?

     – Dan Feb 06 '15 at 10:52