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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.

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    $\begingroup$ Your approach is entirely correct, except that the statement "irreducible over $Y^n-13X^4$" doesn't make any sense (to me). $\endgroup$
    – Servaes
    Nov 14 '18 at 14:39
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    $\begingroup$ 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 $\endgroup$ Nov 14 '18 at 14:48
  • $\begingroup$ 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 $\Bbb{Q}[X]$. $\endgroup$
    – Servaes
    Nov 14 '18 at 14:52
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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.)

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