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I cannot prove that $x$ is irreducible on $\mathbb{Q}[x,y]$.

My attempt is here:

Assume that the following equation holds.
$$x=(a_{00}+a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+\dots)(b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\dots).$$
If $a_{00}\neq 0$ and $b_{00}\neq 0$, then the left hand side has no constant term, but the right hand side has a constant term $a_{00}b_{00}\neq 0$.
This is impossible.

So, $a_{00}=0$ or $b_{00}=0$.
If $a_{00}=b_{00}=0$, then $$x=(a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+\dots)(b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\dots).$$
The left hand side has a term whose degree is $1$, but the right hand side is $0$ or the degree of any term on the right hand side is greater than or equal to $2$.
This is impossible.

So, $a_{00}=0$ and $b_{00}\neq 0$ hold or $a_{00}\neq 0$ and $b_{00}=0$ hold.

If $a_{00}=0$ and $b_{00}\neq 0$, then $$x=(a_{10}x+a_{01}y+a_{20}x^2+a_{11}xy+a_{02}y^2+\dots)(b_{00}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\dots)\\=a_{10}b_{00}x+a_{01}b_{00}y+p(x,y).$$
If $p(x,y)$ is not $0$ then, the degree of any term of $p(x,y)$ is greater than or equal to $2$.
This is impossible since the degree of the right hand side is $1$ and the degree of $a_{10}b_{00}x$ is $1$ and the degree of $a_{01}b_{00}y$ is $1$.
So, $p(x,y)=0$.
So, $$x=a_{10}b_{00}x+a_{01}b_{00}y.$$
It is obvious that $b_{00}=\frac{1}{a_{10}}$ and $a_{01}=0$.
So, $$x=(a_{10}x+a_{20}x^2+a_{11}xy+a_{02}y^2+\dots)(\frac{1}{a_{10}}+b_{10}x+b_{01}y+b_{20}x^2+b_{11}xy+b_{02}y^2+\dots).$$

tchappy ha
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    You can use that $\mathbb Q[x,y]$ is a UFD, so prime and irreducible are equivalent, and $\mathbb Q[x,y]/(x)$ is a domain iff $x$ is prime. – Vercassivelaunos Apr 13 '22 at 12:02
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    Another approach is to use the fact that $\deg(fg) = \deg(f) + \deg(g)$ for all nonzero polynomials $f$ and $g$ (in any number of variables over any field). – Daniel Hast Apr 13 '22 at 12:56
  • @Vercassivelaunos Thank you very much for your comment. Your approach is difficult for me. I need to study more. Thank you. – tchappy ha Apr 14 '22 at 00:17
  • @DanielHast I didn't know that $\deg(fg)=\deg(f)+\deg(g)$ holds for multivariable polynomials. From this fact, it is easy to prove that $x$ is irreducible. Thank you very much for your comment. – tchappy ha Apr 14 '22 at 00:18

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