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 $\mathbb{R}[x,y]$?
 A: 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.
A: 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.
A: 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 $\rm\:g\:$
