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?

Thanks for your time!

up vote 3 down vote accepted

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 $x^0)$

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