Is $x^4+x^2y^2-y^2$ irreducible over $\mathbb C[x,y]$

Question

I want to brutely show that polynomial is irreducible which is okay. However is there some other methods that you can give me hint about?

For example if we think $x^4+x^2y^2-y^2$ as

$$(x^2)^2+(x^2)y^2-y^2$$

so we can substitute $a=x^2$ then

$$x^4+x^2y^2-y^2$$$$=$$$$1/4(x^2-(-y^2-\sqrt{y^4+4y^2}))(x^2-(-y^2+\sqrt{y^4+4y^2}))$$

which is not true in the polynomial ring. But how can it show it is not reducible by 2deg x 2deg polynomials is it enough and what about 3degx1deg reducibility?

Answer 1

Consider $\mathbb{C}[X,Y]= \mathbb{C}[X][Y]$. We can transform the polynomial $Y^2(X^2-1) + X^4$ by reversing the order of its coefficients. There is a statement that the original polynomial is irreducible iff. the polynomial with coefficients in reversed order is (see: Wikipedia.) Reversing the order of the coefficients gives $Y^2X^4 + (X^2-1)$. Then Eisenstein's criterion is fulfilled for the prime element $X+1$. Hence also the original polynomial is irreducible.

Answer 2

If you want to use "brute force" to show that the polynomial is irreducible, then recall that a polynomial of degree two is irreducible over a field iff it does not have a root in that field. In this case the field is $\mathbb C(x)$. Suppose that the rational algebraic fraction $\frac{x^4}{1-x^2}$ has a square root in $\mathbb C(x)$. It follows that $1-x^2$ has a square root in $\mathbb C(x)$, that is, there exist $f,g\in\mathbb C[x]$ with $\gcd(f,g)=1$ such that $\sqrt{1-x^2}=\frac{f(x)}{g(x)}$. It follows that $g(x)^2(1-x^2)=f(x)^2$. Then $f(\pm1)=0$, and this implies that $1-x^2\mid f(x)$. It follows that $1-x^2\mid g(x)$, a contradiction with $\gcd(f,g)=1$.

Since the polynomial $(x^2-1)y^2+x^4$ is primitive, by Gauss lemma we can conclude that the polynomial is irreducible over $\mathbb C[x]$.