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I need to prove that $x_1^2 + 2x_2^2 + 1$ is irreducible over $\mathbb{C}$. And after prove that algebraic curve $X$ on affine plane $\mathbb{A}^2$ given by the equation $f = 0$ is also irreducible. So with this proof I am hoping to that polynomial equation is irreducible, so the second part of the task is easy.

What I have stopped on. It is proof by contradiction. Let's say that $f$ is reducible over $\mathbb{C}$, then it could be easily written as:

  1. $f = x_1^2 + 2x_2^2 + 1 = 4 = g \cdot t$, where $g = \alpha x_1 + \beta x_2 + \gamma$ and $t = \theta x_1 + \tau x_2 + \phi$ then by multiplying the two we would get: $$f = g \cdot t = \underbrace{\alpha \theta}_{1} x_1^2 + \underbrace{\ \beta \theta}_{2} x_2^2 + (\alpha \tau + \beta \theta)x_1 x_2 + (\alpha \phi + \gamma\theta)x_1 + (\beta\phi + \gamma \tau)x_2 + \underbrace{\gamma \phi}_{1}$$
  2. According to that, I have written a system of equations:

$\begin{cases} \alpha \theta = 1 \\ \beta \tau = 2 \\ \gamma \phi = 1 \\ (\alpha \tau + \beta \theta) = 0 \\ (\alpha \phi + \gamma\theta) = 0 \\ (\beta\phi + \gamma \tau) = 0 \end{cases}$ $\Leftrightarrow$ $\begin{cases} \theta = \alpha^{-1} \\ \tau = 2 \beta^{-1} \\ \phi = \gamma^{-1} \\ (\alpha \beta^{-1} + \beta \alpha^{-1}) = 0 \quad * \alpha\beta\\ (\alpha \gamma^{-1} + \gamma\alpha^{-1}) = 0 \quad * \alpha\gamma \\ (\beta\gamma^{-1} + \gamma \beta^{-1}) = 0 \quad * \beta\gamma \end{cases}$ $\Leftrightarrow$ $\begin{cases} \theta = \alpha^{-1} \\ \tau = 2 \beta^{-1} \\ \phi = \gamma^{-1} \\ \alpha^2 = - \beta^{2} \\ \alpha^2 = -\gamma^{2} \\ \beta^2 = -\gamma^{2} \end{cases}$

And from the three last equations, I got $\beta^2 = - \gamma^2, \ \alpha^2 = -\gamma^2, \ \alpha^2 = -\beta^2 \rightarrow -\gamma^2 = -(-\gamma^2)$. And I do not see where this is getting me.

Any advice would be appreciated, honestly with both proof 1 and proof 2!

Charlotte
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  • You are overcomplicating it way too much. Do you know about Eisenstein's criterion? – Mark Nov 19 '24 at 18:36
  • From the last three equations you get $\beta^2=\gamma^2$ but also $\beta^2=-\gamma^2$, meaning that $\alpha=\beta=\gamma=0$. – lisyarus Nov 19 '24 at 18:40
  • @Mark we haven't used it during seminars, I will do some reading on it, because from what I see, this is an indirect case, requiring some modifying to the equation? – Charlotte Nov 19 '24 at 18:44
  • @CharlotteCorrin You can think of it as a polynomial in $\mathbb{C}[x_2][x_1]$, then you can apply it. Or as a polynomial in $\mathbb{C}[x_1][x_2]$. – Mark Nov 19 '24 at 18:47
  • @lisyarus yep, which leads to everything being 0, but $g \cdot t$ can not give us a polynomial equation when all the coefficients are zeroes, that's the contradiction? – Charlotte Nov 19 '24 at 18:48
  • @CharlotteCorrin I'd say so, yes! – lisyarus Nov 19 '24 at 18:48
  • @lisyarus thanks!! – Charlotte Nov 19 '24 at 18:51
  • @Mark thanks, will look into it for future refereces! – Charlotte Nov 19 '24 at 18:51

2 Answers2

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You know that $p(x,y):=x^2 + y^2 + 1 \in \mathbb{C}[x,y]$ is irreducible by for example this post $f(x)=x^2+y^2+1$ is irreducible in $\mathbb{C}[x,y]$. Now it is $f(x_1,x_2) = p(x_1,\sqrt{2}x_2)$. Therefore $f$ is irreducible.

psl2Z
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Thank you everyone, you had some new ideas I haven’t considered.

I was one step away from finding a solution.

So $\alpha^2 = -\gamma^2, \ \beta^2 = - \gamma^2, \ \alpha^2 = -\beta^2 \rightarrow \gamma^2 = -\gamma^2$. That’s how we got $\gamma = 0$.

We also have first three equations where $\phi = \gamma^{-1}$, but there’s no inverse element for 0 when considering multiplication.

That’s the contradiction. It means that $f$ is in fact irreducible over $\mathbb{C}$.

Charlotte
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