$Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial

Question

How to show that $Z^2-YZ-Y^2+X^2+2XY$ is an irreducible polynomial in $\Bbb{C}[X,Y,Z]$?

Answer 1

Suppose that $Z^2-YZ-Y^2+X^2+2XY$ is reducible. Since this is a degree two polynomial in $Z$ with coefficients in $\Bbb C[X,Y]$ it splits into a product of two polynomials of degree one in $Z$, that is, $Z^2-YZ-Y^2+X^2+2XY=(Z+f(X,Y))(Z+g(X,Y))$. We get $f(X,Y)+g(X,Y)=-Y$ and $f(X,Y)g(X,Y)=-Y^2+X^2+2XY$. Now plug in the second relation what you get from the first and find $f(X,Y)(f(X,Y)+Y)=Y^2-2XY-X^2$. This implies that $\deg_Yf(X,Y)=1$, so $f(X,Y)=a(X)+b(X)Y$. Now just calculate what you need and get $b(X)(b(X)+1)=1$, $2a(x)b(x)+a(X)=-2X$, and $a^2(X)=-X^2$. From $b(X)(b(X)+1)=1$ we deduce that $\deg b(X)=0$ and from now on write it $b$ instead of $b(X)$. We have $b(b+1)=1$. We also get $a(X)=\frac{-2}{2b+1}X$ and plugin this into the first relation we get $4=-(2b+1)^2$, so $4=-4(b^2+b)-1$ and from $b(b+1)=1$ we get $4=-4-1$, a contradiction.

Answer 2

You can see it is a cone. It is irreducible because its smooth part is connected and dense. Compute where it is smooth using the Jacobian.

Answer 3

Alternatively, it is a quadratic polynomial in $Z$. Complete the square in $Z$. You will something like $(Z-A)^2+B$. For it to be reducible $B$ must be a square (over the complex numbers). But $B$ is again a quadratic polynomial in the remaining variables. If it is a square, it is reducible. Repeat the same work as before with this polynomial using another one of the variables.

Answer 4

Yet another way: substitute some value for $Y,$ and check that the resulting polynomial is irreducible. My favorite value is $Y=2.$