I was told that

$x^3-y^2$ is irreducible in $\Bbb C[x,y]$.

But I cannot really give a sounding argument. I supposed that it may be factored as $g_1, g_2$, and considered those monomials divisible by $y$. But possible cancellations makes it hard.

EDIT: Thoughts:

Suppose $(x^3-y^2)$ is reducible in $k[y][x] \cong k[x,y]$, then it would factorize as a polynomial in $x$ over the ring $k[y]$. As the degree is $3$, there must exists a monomial factorization. Hence there is a polynomial solution $x=P(y)$, which is impossible.

  • 2
    $\begingroup$ Your thoughts are correct. Another way is to factorize it in $k[x][y]$ and observe that $4x^3$ is not a square. $\endgroup$ – Saucy O'Path Oct 12 '18 at 23:04

A variant :

It is equivalent to show $y^2-x^3$ is irreducible. If it were not, it would factor as the product of two linear factors in $y$: $(y-p(x))(y-q(x))$. Expanding, we obtain by identification $$p(x)q(x)=-x^3, \qquad p(x)+q(x)=0,$$ whence $p(x)^2=x^3$, which is impossible for degree reasons.


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