Why $y^2-x^3-x^2$ is irreducible? 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 $\mathbb{C}[x,y]$?
 A: 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 $x^3+x^2$.
A: Not quite an answer to your question, but if you graph the equation $p(x,y)=0$ in the case that $p=q_1q_2$, then your picture will be the union of $q_1=0$ and $q_2=0$. Look at the graph when your polynomial $p$ is set to zero, and you see that it contains no lines.
A: Separate into $p(x,y) = F_3(x,y) + F_2(x,y)$ where $F_3(x,y)= x^3$ and $F_2(x,y)= y^2-x^2$. There's a theorem that says if a polynomial in $x$, $y$ is the sum of two forms whose degree differs by 1, and the forms have no linear factor in common, then the polynomial is irreducible. Since the $F_2(x,y)=(y+ix)(y-ix)$, we see $p$ must be irreducible. 
A: Eisenstein's criterion with respect to $x+1$ should work.
