# Show that $f(x,y,z)=x^2-y^2z$ is irreducible in $\mathbb{C}[x,y,z]$.

Let $$p\in\mathbb{C}[x,y,z]$$ be defined by $$p(x,y,z)=x^2-y^2z$$.

Goal: Prove that $$p$$ is irreducible.

Let $$I\subset\mathbb{C}[x,y,z]$$ be the ideal defined by

$$I:=(p).$$

My approach is to show that

$$\mathbb{C}[x,y,z]/I,$$ is an integral domain and, hence, $$I$$ prime ideal or equivalently here, $$p$$ is irreducible.

Let us consider a ring homomorphism

$$\varphi:\mathbb{C}[x,y,z]\to\mathbb{C}[t_1,t_2]$$ $$\begin{cases}x\mapsto t_1^2t_2 \\ y\mapsto t_1^2 \\ z \mapsto t_2^2. \end{cases}$$ Notice that, since $$\varphi$$ is a ring homomorphism, $$\varphi(p)=\varphi(x^2-y^2z)=\varphi(x^2)-\varphi(y^2)\varphi(z)=t_1^4t_2^2-t_1^4t_2^2=0.$$

Here is a claim, which I think is true (it atleast feels like it), but I don't know how to prove it. My argument relies on this:

Claim: Any polynomial $$f\in\mathbb{C}[x,y,z]$$ can be written as $$f(x,y,z)=f_0(y,z)+xf_1(y,z)+(x^2-y^2z)g(x,y,z).$$

I read on another link that, for any $$f(x,y)\in \mathbb{C}[x,y]$$, we can write

$$f(x,y)=f_0(x)+yf_1(x)+(x^3-y^2)g(x,y).$$

My approach is inspired by the above identity.

$$\text{ }$$

It feels like my claim should be true, since $$f_0(y,z)$$ takes care of all expressions of the form $$\sum_{k=0}^m\sum_{j=0}^nx^0y^jz^k$$ and $$xf_1(y,z)$$ takes care of everything of the form $$\sum_{k=0}^m\sum_{j=0}^nx^1y^jz^k$$.

Lastly, we have $$x^2g(x,y,z)$$ which feels like it should take care of any polynomial

$$\sum_{k=0}^m\sum_{j=0}^n\sum_{l=2}^o x^ly^jz^k\quad (o \text{ was quite ugly to use in the summation, but anyway)},$$

so I don't really understand what's so special with the

$$-y^2zg(x,y,z)-\text{part},$$

it feels like it doesn't really contribute, or restrict, the polynomial $$f$$ in any way. So it feels like you could do a similar construction for any polynomial. But probably not.

$$\text{ }$$

My questions now is:

• Is my claim true?
• If it is, could you please tell me why?

Continuing on the actual problem:

Thus, applying $$\varphi$$ on $$f\in\ker\varphi$$ gives us $$\varphi(f)=\varphi(f_0)+\varphi(x)\varphi(f_1)+\varphi(x^2-y^2z)\varphi(g)$$ $$=f_0(t_1^2,t_2^2)+t_1^2t_2f_1(t_1^2,t_2^2)=0$$ $$\Rightarrow f_0=f_1=0,$$ Where the last implication holds since $$f_0(t_1^2,t_2^2)$$ is of even degree while $$t_1^2t_2f_1(t_1^2,t_2^2)$$ is of odd degree. Hence the only case the sum can be $$0$$ is if $$f_0$$ and $$f_1$$ are identically $$0$$.

This shows us that $$\ker\varphi=I$$. By the first isomorphism theorem, we have $$\mathbb{C}[x,y,z]/I\cong \operatorname{im}(\varphi).$$

But this shows us that $$\mathbb{C}[x,y,z]/I$$ is a subring of the integral domain $$\mathbb{C}[t_1,t_2]$$. Hence $$I$$ is a prime ideal, so $$p$$ is irreducible, which is what we wanted to prove.

Questions:

• Does my approach work? In particular, is my claim true and if so, why?
• If this approach does not work, could you please help me with a better approach and solution?

Yes, your claim is true, for a very simple reason; considering $$f(x,yz)$$ as a polynomial in the integral domain $$\mathbf C[y,z][x]$$, you always can perform in this ring a Euclidean division by a monic (in $$x$$) polynomial. The remainder has degree $$\le 1$$ for a quadratic monic polynomial.
However, this polynomial is irreducible for a shorter reason: Eisenstein criterion. Indeed $$\mathbf C[y,z]$$ is a U.F.D. and the irreducible element $$z$$ divides all coefficients except the lead coefficient (of $$x^2),\,$$ and $$z^2$$ does not divide the "constant" coefficient $$y^2z$$ (of $$x^0)$$