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Let $R$ be an integral domain and $f = X^2 - Y^n \in (R[Y])[X]$ with $n \geq 1$. I'd like to prove the following: $f$ is irreducible in $(R[Y])[X]$ $\Leftrightarrow$ $n$ is odd.

If $n = 2k$ is even, then $f = X^2 - Y^{2k} = (X+Y^k)(X-Y^k)$ is reducible in $(R[Y])[X]$. But how can I show that $f$ is irreducible in $(R[Y])[X]$ if $n$ is odd? The tools I have for determining the irreducibility of $f$ (e.g. Eisenstein's criterion in conjunction with Gauss's lemma) only give me insights on the irreducibility of $f$ in the space of fractions above $(R[Y])[X]$, but not $(R[Y])[X]$ itself.

user26857
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Wilfred Montoya
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1 Answers1

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If it is reducible in $R[Y][X]$, then it is reducible in $K(Y)[X]$, where $K$ is the field of fractions of $R$. This means that there is $a\in K(Y)$ such that $a^2-Y^n=0$. Write $a=u(Y)/v(Y)$ with $u,v\in K[Y]$, and get $u(Y)^2=Y^nv(Y)^2$. But the degree on the left side is even, while on the right side is odd, a contradiction.

user26857
  • 53,190