# Showing a polynomial irreducible

How to show that the polynomial $Y^2+X^2(X-1)^2$ is irreducible in $\mathbb R[X,Y]$. I tried to show that $\mathbb R[X,Y]$ modulo this ideal is an integral domain but I cannot find any homomorphism.

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It is helpful to think of this polynomial not as an element of $\mathbb{R}[X,Y]$, but as an element of $A[Y]$, where $A=\mathbb{R}[X]$. That is, we consider it as a polynomial only in $Y$, with polynomials in $X$ as coefficients. Now suppose we had a factorization $Y^2+X^2(X-1)^2=f(X,Y)g(X,Y)$. Then as polynomials in $Y$, the degrees of $f$ and $g$ must add to $2$, and their leading coefficients must multiply to $1$. The only units in $A$ are constants, so we may multiply $f$ and $g$ by constants to assume they are both monic. If either $f$ or $g$ has degree $0$, then it is just $1$, so we have the trivial factorization. The only other possibility is that they both have degree $1$. This means we have $f(X,Y)=Y+f_0(X)$ and $g(X,Y)=Y+g_0(X)$ for some $f_0(X),g_0(X)\in A$. So we must have $$Y^2+X^2(X-1)^2=(Y+f_0(X))(Y+g_0(X))=Y^2+(f_0(X)+g_0(X))Y+f_0(X)g_0(X).$$
Thus $g_0(X)=-f_0(X)$ and $-f_0(X)^2=X^2(X-1)^2$. But no such $f_0(X)$ exists (for instance, the leading coefficient of the left-hand side must be negative but the leading coefficient of the right-hand side is $1$).
Dear @Eric Wofsey. I have another polynomial $Y^2-X(X^2-1)$ in $\mathbb C[X,Y]$ and going the same way as above I see that $f_0(X)^2=X(X-1)^2$ which is not possible as in L.H.S. the degree of the polynomial is even.I don't see any role of $\mathbb C$ or $\mathbb R$ here or in the above question. Am I right? – 2015 Jan 17 at 7:24
The only place I used anything about $\mathbb{R}$ is in the very last step, where I said that $f_0(X)=-X^2(X-1)^2$ has no solution. In the case of $Y^2-X(X^2-1)$, your argument works over any field. – Eric Wofsey Jan 17 at 7:29
I think you can arrive at $-f_0(X)^2=X^2(X-1)^2$ immediately by noticing that one deals with a degree two polynomial and this is reducible iff it has a root in the base ring. – user26857 Jan 17 at 9:45