Is $x_1^2-x_2^3$ irreducible?

Question

Is the polynomial $f=x_1^2-x_2^3 \in K[x_1,x_2]$, with $K$ field, irreducible? I think yes, because:

• $f=x_1^2-x_2^3$ has degree $3$. If suppose that $f=hk$ is a non trivial factorisation, than deg$h=1$ and deg$k=2$. Then i write $h=ax_1+bx_2+c$ and $k=a'x_1^2+b'x_2^2+c'$ (i don't write $k=a'x_1^2+b'x_2^2+c'+d'x_1x_2$ because in $f$ there isn't the mixed product $x_1x_2$). Then multiplyng and imposing the equality between $f$ and $hk$ i find that $aa'=0; \quad ab'=0; \quad ac'=0$, so i can conclude $a=0$ otherwise $k=0$ and that is absurd. Continuing, i have also $ba'=0; \quad bb'=-1; \quad bc'=0$, so i conclude that $b\neq0$ and $a'=c'=0$. But now in $hk$ doesn't compare the monomial $x_1^2$. Absurd.
• Is correct my reasoning?
• Are there other ways to prove this irreducibility? In genereal which are the "tricks" or observations one can use to prove irreducibility of polynomial in such cases?

Answer 1

To prove that $x^3-y^2\in K[x,y]$ is irreducible. We treat it as polynomial in $K(y)[x]$. Then it has degree $3$. Since there is no $f(y)\in K(y)$ such that $f(y)^3 - y^2=0$. We see that $x^3-y^2$ is irreducible in $K(y)[x]$, hence irreducible in $K[y][x]$.

Answer 2

If $A$ is a commutative ring the polynomial $y^2-a\in A[y]$ is irreducible iff $a$ is not a square in $A$.
Apply to $a=x^3\in A=K[x]$.