Show $y^3+xy+x^2(x-1)^2\in\mathbb{R}[x,y]$ is irreducible

Question

I have tried applying Eisentein's Criterion to no avail, and I have also tried reducing modulo some proper ideal, with no luck either. Is there some standard way of showing polynomials in two variables are irreducible?

Answer 1

Suppose $f(x,y)=y^3+xy+x^2(x-1)^2$ and suppose we have the decomposition $$f(x,y)=g(x,y)h(x,y).$$ Since degree of $f(x,y)$ with respect to $y$ is $3$, then we must have $$g(x,y)=p(x)y+q(x)\quad \text{and}\quad h(x,y)=r(x)y^2+s(x)y+t(x)$$ where $p(x),q(x),r(x),s(x)\in\mathbb{R}[x]$ are polynomials with degree at most $4$.

Expanding $g(x,y)h(x,y)$ we reach $$f(x,y)=p(x)r(x)y^3+(p(x)s(x)+q(x)r(x))y^2+(p(x)t(x)+q(x)s(x))y+q(x)t(x)$$ which implies $p(x)=r(x)=1$ are constant polynomials because in $\mathbb{R}[x]$ only constant polynomials are invertible.

Now by comparing coefficient, we see $$s(x)=-q(x),\quad t(x)=-q(x)s(x)+x,\quad q(x)t(x)=x^2(x-1)^2$$ i.e. $$t(x)=q(x)^2+x$$ and therefore, $$q(x)(q(x)^2+x)=x^2(x-1)^2$$ which is impossible because LHS is a polynomial of degree multiple of $3$.

Answer 2

Perhaps consider mod 6, as then $y^3\equiv y$ and $y(x+1)+x^2(x-1)^2$ is irreducible