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I would like to prove that $A:=\Bbb{C}[X,Y]/ (X^2-Y^3)$ is not a UFD. This is equivalent to find an irreducible element which is not prime.

I can prove that every element looks like $\overline{P_1X+P_0}$ for $P_1,P_0\in \Bbb{C}[Y]$. After that let $\overline{Y}$ be the "image" of $Y$ in the quotient. It's not prime because the ring $A/(\overline{Y})$ will be isomorphic to $\Bbb{C}[X]\ (X^2)$.

But I am stuck to prove that $Y$ is irreducible,

that is, I write: $$Y=P_1Q_1X^2+P_0Q_1X+P_1Q_0X+P_0Q_0 =P_1Q_1Y^3+P_0Q_0+(P_0Q_1+P_1Q_0)X$$

So by unicity on $\Bbb{C}(Y)[X]$, it's equivalent to $$Y=P_1Q_1Y^3+P_0Q_0$$ and $$(P_0Q_1+P_1Q_0)X=0.$$

Now taking the value at zero in the first equation, we get that $P_0(0)=0$ (by symmetry). It means that $Y$ divides $P_0$.

Question: How can we continue ?

user26857
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JeSuis
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  • You made a mistake when tried to show that $\overline{Y}$ is irreducible. The polynomial relation holds modulo the ideal $(X^2-Y^3)$, so things look even worse. – user26857 Dec 07 '15 at 15:11
  • @user26857 I don't think so because every element in the ring can be written uniquely as $P_1X+P_0.$ – JeSuis Dec 07 '15 at 15:16
  • $A$ non-UFD is not "equivalent to find an irreducible element which is not prime". – user26857 Dec 07 '15 at 16:17
  • @user26857 why ? The uniqueness property is equivalent at: $(p)$ is prime if and only if p is irreducible – JeSuis Dec 07 '15 at 16:33
  • In this case, since $A$ is noetherian every element of $A$ is a product of irreducibles, so the claim is ok, but in general no. There are integral domains with the property that every irreducible element is prime and they are not UFDs. (In fact, they are not even GCD domains.) – user26857 Dec 07 '15 at 16:35
  • @user26857 hum, so my teacher made simplifications in definitions.. – JeSuis Dec 07 '15 at 16:45
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    You can have a look at http://math.stackexchange.com/questions/222593/ring-where-irreducibles-are-primes-which-is-not-an-ufd/222700#222700 – user26857 Dec 07 '15 at 16:51

2 Answers2

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Let me denote by $x,y$ the residue classes of $X,Y$ modulo the ideal $(X^2-Y^3)$. Suppose $y=(ax+b)(cx+d)$. Then $Y=(aX+b)(cX+d)+(X^2-Y^3)e$ with $a,b,c,d\in\mathbb C[Y]$, $e\in\mathbb C[X,Y]$. Now send $X$ to $T^3$, $Y$ to $T^2$ and get $$T^2=[a(T^2)T^3+b(T^2)][c(T^2)T^3+d(T^2)].$$ This writes $$T^2=a(T^2)c(T^2)T^6+[a(T^2)d(T^2)+b(T^2)c(T^2)]T^3+b(T^2)d(T^2).$$ For $T\mapsto 0$ we get $b(0)d(0)=0$, so $b(0)=0$ or $d(0)=0$. Suppose $b(0)=0$ and then $b(Y)=Yb_1(Y)$. We now get $$1=a(T^2)c(T^2)T^4+[a(T^2)d(T^2)+b(T^2)c(T^2)]T+b_1(T^2)d(T^2).$$ For (odd) degree reasons $ad+bc=0$. It follows that $$1=a(T^2)c(T^2)T^4+b_1(T^2)d(T^2).$$ Then we must have $\deg a+\deg c+2=\deg b_1+\deg d$. But from $ad+bc=0$ we have $\deg a+\deg d=1+\deg b_1+\deg c$, so $\deg a=-\deg d+1+\deg b_1+\deg c$ and then $-\deg d+1+\deg b_1+\deg c+\deg c+2=\deg b_1+\deg d$, that is, $2\deg c+3=2\deg d$, a contradiction.

user26857
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  • As you can see, the "different method" is more or less involved in this answer, too. (If two rings are isomorphic, then it's hard to hide this.) – user26857 Dec 07 '15 at 16:39
  • Thanks, I don't follow the last part: then we must have... Can you explain? The degree of the sums is less than the max of $deg(a(T^2)c(T^2)T^4, b_1(T^2)d(T^2))$ but from here I don't see how can I get your 'claim'.. – JeSuis Dec 07 '15 at 16:41
  • @JeSuis In order to get $1$ you have to cancel $T$ on the right hand side (unless $a=0$ or $c=0$, and $b_1,d$ are constants, but this leads to a "trivial" decomposition of $y$) so the degrees of the two polynomials must be equal. – user26857 Dec 07 '15 at 16:42
  • Great, thank you for your time. +1 and accept> – JeSuis Dec 07 '15 at 16:47
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The ring is isomorphic to $\mathbb{C}[T^2,T^3] \subseteq \mathbb{C}[T]$ via $\overline{X} \mapsto T^3$ and $\overline{Y} \mapsto T^2$. Here, $T^2$ is irreducible, since the only non-trivial factors in $\mathbb{C}[T]$ are $\lambda \cdot T$ (with $\lambda \in \mathbb{C}^{\times}$), which do not lie in $\mathbb{C}[T^2,T^3]$.

Martin Brandenburg
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