Is this polynomial irreducible? $h(x, y) = x^2\ − y^3 \in \mathbb{Q}[x,y]$

Question

$h(x, y) = x^2\ − y^3 \in \mathbb{Q}[x,y]$

I am trying to figure out if this polynomial is irreducible. I believe it is, but I am not sure how to show this. Any help would be great!

Answer 1

Hint: If $x^2-y^3=fg$, what can you say about the degrees of $f$ and $g$?

Answer 2

A sketch:

If $x^2-y^3$ were not irreducible, seen as a polynomial in $\mathbf Q[x][y]$, it would factor as $$x^2-y^3=\bigl(x-p(y)\bigr)\bigl(x-q(y)\bigr)=x^2-\bigl(p(y)+q(y)\bigr)x+p(y)q(y).$$ Deduce that $\;\deg y^3$ should be even.

Answer 3

Consider the ring homomorphism $\phi : \mathbb{Q}[x,y] \to \mathbb{Q}[t]$, $p(x,y) \mapsto p(t^3, t^2)$. We claim that the kernel of this homomorphism is the ideal generated by $h(x,y) = x^2 - y^3$.

Certainly, $h$ is in the kernel, so $\langle h \rangle \subseteq \ker(\phi)$. On the other hand, for any $p \in \mathbb{Q}[x, y]$, there exist some $r_0, r_1 \in \mathbb{Q}[y]$ such that $p \equiv r_0(y) + x r_1(y) \pmod{h}$. Then $\phi(p) = r_0(t^2) + t^3 r_1(t^2)$; in this expansion, $r_0(t^2)$ gives only even powers of $t$ whereas $t^3 r_1(t^2)$ gives only odd powers of $t$. Therefore, if $\phi(p) = 0$, then $r_0 = r_1 = 0$, so $p \equiv 0 \pmod{h}$, completing the proof that $\ker(\phi) \subseteq \langle h \rangle$.

It follows that $\mathbb{Q}[x,y] / \langle h(x,y) \rangle$ is isomorphic to the image of $\phi$, which is a subring of $\mathbb{Q}[t]$, and therefore an integral domain. (In fact, the image is easily seen to be $\mathbb{Q}[t^2, t^3]$ which is the polynomials with the coefficient of $t^1$ being 0.) That being the case, $\langle h(x,y) \rangle$ is a prime ideal, so $h(x,y)$ is irreducible.

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(I realize this answer is a bit roundabout compared to the others; I just thought it might give some insight to see why the quotient ring is necessarily an integral domain, by showing what the structure of that quotient ring actually is.)