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

Why is $f(x,y)=x^{2}+y^{2}(y-1)^{2}$ is irreducible over $\mathbb{R}[x,y]$?

share|cite|improve this question
Think of it as a polynomial in the variable $x$ with coefficients in $\mathbb{R}[y]$ (or, if it helps, in $\mathbb{R}(y)$). If it were reducible, what would a factorization have to look like? – Qiaochu Yuan Mar 27 '12 at 18:20
Have you tried proving directly that any factorization involves one of the factors being a constant polynomial? This is a direct, albeit tedious, calculation. Alternatively, you could factor the polynomial in $\mathbb{C}[x,y]$ and deduce from that factorization that the polynomial is irreducible in $\mathbb{R}[x,y]$. – Michael Joyce Mar 27 '12 at 18:22
up vote 7 down vote accepted

Let me expand on my answer in the comments. Hopefully, you'll try the hints/suggestions provided in all of the comments before reading this answer.

In $\mathbb{C}[x,y]$ (a UFD), $$x^2 + y^2 (y-1)^2 = (x + iy(y-1))(x - iy(y-1))$$ so if $x^2 + y^2 (y-1)^2 = g(x,y) h(x,y)$ with $g(x,y), h(x,y) \in \mathbb{R}[x,y]$, then using that $\mathbb{C}[x,y]$ is a UFD, we must have one of the factors, say $g(x,y)$, associate to $x + iy(y-1)$ (and the other factor associate to $x - iy(y-1)$). This implies that there is a constant $\lambda \in \mathbb{C} \setminus \{0\}$ such that $\lambda \cdot (x + iy(y-1)) \in \mathbb{R}[x,y]$, which a simple calculation reveals is impossible.

share|cite|improve this answer
Thanks, a question: wouldn't we need to say "suppose $g,h$ are irreducible over $\mathbb{C}[x,y]$" to apply uniqueness of factorization? My question is: why are these polynomials also irreducible over $\mathbb{C}[x,y]$? don't we need this to use uniqueness? – user6495 Mar 27 '12 at 18:50
You should assume that $g(x,y)$ and $h(x,y)$ are not units in $\mathbb{R}[x,y]$. Then they are not units in $\mathbb{C}[x,y]$, so $g(x,y) h(x,y)$ is a non-trivial factorization. Since we already have a non-trivial factorization with just two factors, unique factorization implies that each of $g(x,y)$ and $h(x,y)$ must be associate to one of the linear factors in the original factorization. – Michael Joyce Mar 27 '12 at 18:58
ah, I see where I was going wrong, cheers! – user6495 Mar 27 '12 at 19:10

Modulo $\rm y-2$ it is $\rm x^2+4$, which is irreducible in $\rm \mathbb{R}[x,y]/(y-2)\cong\mathbb{R}[x]$.

Modulo $\rm x-1$ it is $\rm 1+y^2(y-1)^2$, which is irreducible in $\rm \mathbb{R}[x,y]/(x-1)\cong\mathbb{R}[y]$.

Therefore we must have $\rm x^2+y^2(y-1)^2=f(x-1)g(y-2)=p(x)q(y)$. Reducing this mod $\rm x,y$ tells us that $\rm p$ and $\rm q$ have zero constant terms, implying it is divisble by $\rm x$ and $\rm y$, a contradiction.

share|cite|improve this answer
You also have to rule the possibility that $f(x,y) = g(x,y) h(x,y)$ but either $g(x,2)$ or $h(x,2)$ is a constant polynomial. – Michael Joyce Mar 27 '12 at 18:32

Hint $\rm\ \ x - f(y)\ \ |\ \ x^2 + g(y)^2\: \Rightarrow \left(\dfrac{f(y)}{g(y)}\!\right)^{\!2} =\: -1\ \Rightarrow\: -1 \in \mathbb R^2\:$ via eval $\rm\:y\:$ at any nonroot of $\rm\:g\:$

share|cite|improve this answer

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.