Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 8 down vote accepted

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 $1$).

share|cite|improve this answer
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
Yes! I missed that. Thank you for the above answer. And I suppose it is difficult to show the first polynomial irreducible by the way I stated in the question. – 2015 Jan 17 at 7:54
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.