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I want to prove that: the integral domain $k[x,y]/ (x^3 - y^2)$ is not normal and find its integral closure in its field of fractions.

I know that I will use the idea that $UFD \implies$ normal and hence not normal means not $UFD$ so what I should show is that $k[x,y]/ (x^3 - y^2)$ is not a UFD and I have read a hint here How to show a ring is normal or not, and how to show the normalisation of the ring to consider $x = (y/x)^2.$but I can not see how that will show that it is not a UFD. could someone elaborate this to me please?

Also, what is the field of fraction of this integral domain?

I know that a unique factorization domain is an atomic integral domain in which factorization to irreducibles is unique (up to associates). And a commutative ring is atomic if each $r \in R$ is a finite product of irreducibles in R.

Also, a ring $R$ is a normal ring if $R$ is an integral domain that is integrally closed in $\operatorname{frac}(R).$

And $R$ is integrally closed in a commutative ring $S$ iff the integral closure of $R$ in $S$ (int($R$)) equals $R$. And int($R$) consists of all $x \in S$ that is integral over R. And note that an element is integral over $R$ if there exists a monic polynomial $f(x) \in R[x]$ such that $f(x) = 0.$

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    Wait, you have $\text{UFD}\Rightarrow\text{normal}$ but it doesn't mean $\text{not UFD}\Rightarrow\text{not normal}$. As for UFD, can't we use the identity $x\cdot x\cdot x=y\cdot y$ that holds in our ring? – richrow Feb 17 '21 at 08:59
  • @Richrow: I think you misread that first part. You are definitely right that $\overline{x}^3=\overline{y}^2$ yields very different factorizations of the same element. This can be used to directly show that $k[x,y]/(x^3-y^2)$ is not a UFD. – Mathematician 42 Feb 17 '21 at 09:00
  • @richrow I did not said that. –  Feb 17 '21 at 09:04
  • @Mathematician42 how can we do that ? –  Feb 17 '21 at 09:18
  • @brain-drain Fixing a commutative field $K$ and considering the unique morphism of unitary $K$-algebras $K[X, Y] \to K[T]$ given by $X \mapsto T^2$ and $Y \mapsto T^3$ it is not difficult to show that $K[X, Y]/\left(X^3-Y^2\right)$ is isomorphic to the subring $A$ of $K[T]$ constituted of all polynomials $f$ such that $f_1=0_K$, i.e. without a first degree term. The field of fractions of $A$ taken within $K(T)$ is precisely $K(T)$ and the integral closure of $A$ is easily shown to equal $K[T]$ since it contains the indeterminate $T$. – ΑΘΩ Feb 17 '21 at 09:33
  • @ΑΘΩ how can I show that isomorphism ? why it is not a UFD ? –  Feb 17 '21 at 10:43
  • See for example this duplicate why it is not a UFD. – Dietrich Burde Feb 17 '21 at 11:26
  • @brain-drain You can deduce the relation of isomorphism mentioned above by showing that the image of the $K$-algebra morphism also mentioned above is precisely $A$ and by showing that the kernel of the morphism in question is precisely the principal ideal $\left(X^3-Y^2\right)$. Having achieved this, an application of the fundamental theorem for quotient ring morphisms suffices to claim the desired isomorphism. $A$ clearly isn't a UFD as you have the decompositions $T^6=\left(T^2\right)^3=\left(T^3\right)^2$ with $T^2, T^3$ non-associated irreducibles. – ΑΘΩ Feb 17 '21 at 12:03
  • Counterexample $(y-x)(y+x)= y^2-x^2 = x^3-x^2 = x^2(x-1)$. – Wuestenfux Feb 18 '21 at 08:02

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Well, I think the first step is to find the standard basis of the quotient ring, which is

$\{x^iy^j\mid 0\leq i\leq 2, j\geq 0\}.$

Wuestenfux
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