Showing that $x^2+y^2$ is irreducible in $k[x,y]$ provided that $\sqrt{-1}\notin k$

Question

Let $k$ be any field. I know how to prove "by hand" the fact that $x^2+y^2$ is irreducible in $k[x,y]$ assuming that the polynomial $x^2+1$ has no roots over $k$. For that, one has to impose the equation $$ x^2+y^2=(ax+by)\cdot(a'x+b'y), $$ expand the RHS and then, after solving the system of equations for $a,a',b,b'$, one finds that $a'a=1$ and $a^2+b^2=0$, so that $0=(a')^2(a^2+b^2)=1+(a'b)^2$, and arrives a contradiction.

But is there a better way of doing it? By "better" I mean a faster/more elegant/more technological way. For example, I don't see how Eisenstein criterion could be used here.

Answer 1

It's generally true that the homogenization of an irreducible polynomial is irreducible, see here for example. As $x^2+y^2$ is the homogenization of $x^2+1$, we can conclude.

Answer 2

In general, if the image of your polynomial $x^2+y^2$ after quotienting the base ring $k[y]$ by a proper ideal is irreducible, then $x^2+y^2$ was irreducible. So quotient by $(y^2-1)$; the image of your polynomial in this case is $x^2+1$ which you have supposed to be irreducible. So the original polynomial was irreducible.

Answer 3

If it were reducible, then so it is in $k(y)[x]$. Since $y$ is a unit, so would $(x/y)^2 + 1$. Replace $k(y)[x]$ by $k(y)[xy^{-1}]$, then equivalently $x^2 + 1$ is reducible in $k(y)[x]$. Then $\sqrt{-1}$ exists in $k(y)$. Let $p,q \in k[y]$ coprime and $q$ monic such that $p(y)^2/q(y)^2 = -1$, or $p(y)^2 + q(y)^2 = 0$. Looking at the leading coefficient yields a root of $X^2 + 1$ over $k$, contradction.