I have encountered this while going over the wikipedia page "irreducible polynomial".
A polynomial that is irreducible over any field containing the coefficients is absolutely irreducible. By the fundamental theorem of algebra, a univariate polynomial is absolutely irreducible if and only if its degree is one. On the other hand, with several indeterminates, there are absolutely irreducible polynomials of any degree, such as $x^{2}+y^{n}-1$, for any positive integer $n$.
How can I show that $x^2+y^n-1$ is absolutely irreducible for any positive integer $n$? I'm new so I don't have enough tools to tackle this. Any help is appreciated.
a) Over a field $k$ of characteristic $2$ and for $n=2s$ even, it is not true that the polynomial is irreducible.
Indeed, since $-1=1$ in characteristic $2$:$$x^2+y^{2s}-1=(x+y^s-1)^2$$ b) In all the other cases the polynomial $x^2+y^n-1$ is irreducible because $1-y^n$ is not a square in $k[y]$.
The reason is that in the decomposition in linear factors of $1-y^n$ over an algebraic closure $\bar k$ of $k$ some factor will appear with odd exponent:
(i) This is clear in characteristic $0$ since $1-y^n$ is separable, i.e. all its linear factors appear with exponent $1$.
(ii) And in characteristic $p\gt2$ you can write $1-y^n=(1-y^r)^{p^e}$ with $1-y^r$ separable (equivalently $p$ does not divide $r$), so that all linear factors appear with exponent $p^e$, an odd number, so that $1-y^n$ is not a square either