# Prove that $Y^n-13X^4$ is irreducible in $\mathbf{Q}[X,Y]$

Prove that $$Y^n-13X^4$$ is irreducible in $$\mathbf{Q}[X,Y]$$

Look at $$Y^n-13X^4$$ as a polyomial in $$(\mathbf{Q}[X])[Y]$$, then we can use Eisenstein with $$13\in\mathbf{Z}[X]$$ (which is prime) and $$169\not\mid 13X^4$$. Then the polynomial is irreducible over $$Y^n-13X^4$$ and by Gauss' Lemma also in $$\mathbf{Q}[X,Y]$$.

Is this the correct way to approach it? I am having trouble applying Eisenstein over polynomial rings in more variables.

• Your approach is entirely correct, except that the statement "irreducible over $Y^n-13X^4$" doesn't make any sense (to me). Nov 14 '18 at 14:39
• Is that really true? Shokran. I meant to say over $\mathbf{Z}[X,Y]$ and by Gauss' Lemma also in $\mathbf{Q}[X,Y]$. @Servaes Nov 14 '18 at 14:48
• It is really true, but if you are in doubt then you can verify the proof of Gauss' lemma for $\Bbb{Q}[X,Y]$ yourself; it is exactly the same as for $\Bbb{Q}[X]$. Nov 14 '18 at 14:52

If it were reducible in $$\mathbb Q[X,Y]$$ then you could substitute $$X$$ for $$Y$$ and you'd have a factorisation of $$X^n-13X^4$$, hence of $$X^{n-4}-13$$, which is impossible. (Some powers of $$X$$ will need borrowing if $$n<4$$, and if $$n=4$$ it's clear.)