Show that $(x^2+y^2+a^2)^2-a^4-4a^2x^2$ is irreducible for $a\neq 0$ Suppose that $k$ is an algebraically closed field with characteristic not $2$. I must prove that in $k[x,y]$ this polynomial is irreducible ($a\in k$ non-zero): $$f(x,y):=(x^2+y^2+a^2)^2-a^4-4a^2x^2.$$ As I asked in my previous question (Question about irreducible polynomials in algebraically closed fields) I don't know how to approach in general this kind of problems; actually I shouldn't use theorems, since I'm not supposed to know any specific theorem for irreducible polynomials in my course, but I would like to know some strategies that I could use instead.
For this polynomial, the only thing that I noticed is that we can isolate easily the $y$, but I don't know if this can help. We could write $k[x,y]$ as $$k\left[x,\sqrt{\sqrt{f(x,y)+a^4+4a^2x^2}-a^2-x^2}\right].$$ Would it be intelligent to prove that the ring above is an integral domain? If not, can you give me an hint or some strategies to try in general?
I try using Eisenstein's criterion: I write $f$ as $$x^4+2(y+a)(y-a)x^2+y^2 (y+\sqrt {-2})( y-\sqrt {-2}). $$ Can I just say that $(y-a)$ divides the coefficient of $x^2$, doesn't divide $1$, and $(y-a)^2$ doesn't divide the last term (the one belonging to $k[y]$)? I'm using the generalized version of the criterion, that is on Wikipedia.
 A: I don't find an easy way to apply Eisenstein's criterion. If someone knows how to do it using Eisenstein, then I would be interested in seeing the answer.
I write the polynomial as $f(x) = x^4 + 2(y^2 - a^2)x^2 + (y^4 + 2a^2y^2)$ and treat it as a polynomial in $x$ with coefficients in $R = k[y]$. Suppose that it is not irreducible. As the coefficient before $x^4$ is $1$, we have only two cases:

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*$f(x) = (x - r)g(x)$ where $r \in R$ and $g \in R[x]$ has degree $3$. In this case, we have $$0 = f(r) = (r^2 + y^2 - a^2)^2 + 4a^2 y^2 - a^4.$$ An argument of degree shows that the degree of $r$ (as a polynomial in $y$) must be $1$, and we must have $r^2 = -(y + t)^2$ for some $t \in k$. We arrive that $$0 = (-2ty + t^2 - a^2)^2 + 4a^2y^2 - a^4.$$ Comparing $y^2$-terms, we get $t^2 = -a^2 (\neq 0)$. Then comparing $y$-terms leads to contradiction.


*$f(x) = (x^2 + ux + v)(x^2 + u'x + v')$ where $u, v, u', v' \in R$. Comparing $x^3$-terms, we get $u' = -u$. Comparing $x$-terms, we get $u(v - v') = 0$. Thus either $u = 0$ or $v' = v$.
If $u = 0$, then we have $v + v' = 2(y^2 - a^2)$ and $vv' = y^4 + 2a^2 y^2$. This leads to $$(v - v')^2 = (v + v')^2 - 4vv' = -16a^2y^2 + 4a^4.$$ This is not possible as the polynomial on the right hand side is not a square in $R$.
If $v' = v$, then comparing constant terms gives $v^2 = y^4 + 2a^2 y^2$, which is again impossible as the latter is not a square in $R$.
This completes the proof that $f$ is irreducible.
