Irreducibility of a polynomial in the ring $\mathbb{R}[x,y]$.

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

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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

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.

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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.

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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 $\rm\:g\:$

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