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I'd like to know why $X+(X^2-Y^3)$ and $Y+(X^2-Y^3)$ are irreducible, but not prime in $K[X,Y]/(X^2-Y^3)$.

I failed in both. For the first, I tried to use an isomorphism $$\phi:K[X,Y]/(X^2-Y^3)\to K[X], a_1=0$$ by $X\to X^3, Y\to X^2$ and for the second I tried to construct a case similar to the factorization $3^2=(2+\sqrt-5)(2-\sqrt-5)$.

user26857
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Greg P.
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  • By the way, $\phi$ is not an isomorphism, since it is not surjective (you never get $\phi(p)=X$). However, you can show that $K[X,Y]/(X^2-Y^3)\simeq K[X^2,X^3]\leq K[X]$. – Raúl Alegre Nov 19 '14 at 00:53
  • It's what I meant with $a_1=0$, excuse the messy notation. – Greg P. Nov 19 '14 at 00:53
  • Oh, perfect. I prefer to use another letter on the other side "$K[X,Y]/(X^2-Y^3)\simeq K[T^2,T^3]$". Now you can show that $T^3$ and $T^2$ are both irreducible but not prime. – Raúl Alegre Nov 19 '14 at 00:56

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Notation. Define $x := X + (X^2-Y^3)$ and $y:= Y + (X^2-Y^3)$, so that $x^2=y^3$

Hint. Clearly $x\mid y^3$ whereas $x\not\mid y$. In the same vein $y\mid x^2$ whereas $y\not\mid x$. This shows that the elements are not prime.

Raúl Alegre
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  • Why can you simplify $x^2=X^2+(X^2-Y^3)=Y^3$? – Greg P. Nov 19 '14 at 00:55
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    $a+I = b + I \ \Leftrightarrow\ a-b\in I$. Thus $X^2+I=Y^3+I$ because $X^2-Y^3 \in I=(X^2-Y^3)$. – Raúl Alegre Nov 19 '14 at 00:59
  • :D of course. Thanks for your patience! – Greg P. Nov 19 '14 at 01:00
  • Assume $x = pq$ If you had $p$ and $q$ not units, they would have degree $\geq 1$. And the product would have degree $\geq 2$. (Of course, one must be careful because $x^2=y^3$, but anyway, you can not get $pq=x$) – Raúl Alegre Nov 19 '14 at 15:03
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    It is easier to show that $T^2$ is irreducible in $K[T^2, T^3]$. If $T^2 = PQ$, then $\deg P + \deg Q = 2$. But in this ring we must have $\deg P \neq 1 \neq \deg Q$. Then, one of the factors has degree $0$ and is a unit. – Raúl Alegre Nov 19 '14 at 15:05