TITLE: Why $y^2-x^3-x^2$ is irreducible?
QUESTION [6 upvotes]: Let $p(x,y)=y^2-x^3-x^2$ a polynomial in $\mathbb{C}[x,y]$. How can I prove that $p(x,y)=y^2-x^3-x^2$ is irreducible as element of $\mathbb{C}[x,y]$?

REPLY [8 votes]: If a decomposition exists, it must be of the form $(y+f(x))(y+g(x))$, because surely no polynomial only in $x$ or $y$ divides $y^2-x^3-x^2$. It is not restrictive to take the coefficients of $y$ in the factors both equal to $1$. This gives the equations
\begin{gather}
f(x)+g(x)=0\\
f(x)g(x)=-x^3-x^2
\end{gather}
that is, $g(x)=-f(x)$ and
$$
f(x)^2=x^3+x^2
$$
which can't be satisfied, because the irreducible polynomial $x+1$ appears with exponent $1$ in the decomposition of $x^3+x^2$.