So i'm trying to understand the following computation
$(x_1^2-2x_2^2)$ is a prime ideal in $\Bbb Q[x_1,x_2]$ but not in $\Bbb C[x_1,x_2]$
I only know that both polynomial rings are UFD. I wonder what are the standard strategies for such computations.
Note that in both $\mathbb{Q}[x_1,x_2]$ and $\mathbb{C}[x_1,x_2]$, the units are the constant polynomials.
Let $f=x_1^2-2x_2^2$.
In $\mathbb C[x_1,x_2]$ we have the factorization $$ x_1^2-2x_2^2=(x_1+\sqrt{2}x_2)(x_1-\sqrt{2}x_2) $$ hence $f$ is not irreducible in $\mathbb{C}[x_1,x_2]$.
Note also that the factors $x_1+\sqrt{2}x_2$ and $x_1-\sqrt{2}x_2$ are irreducible in $\mathbb{C}[x_1,x_2]$, so they are the unique irreducible factors of $f$ in $\mathbb{C}[x_1,x_2]$ (unique up to nonzero constant multiple).
Suppose $f$ could be expressed as $gh$ where $g,h\in\mathbb{Q}[x_1,x_2]$ are non-constant.
Then the factorization $f=gh$ would still work in $\mathbb{C}[x_1,x_2]$, hence, since $\mathbb{C}[x_1,x_2]$ is a $\text{UFD}$, it follows that $g=a(x_1+bx_2)$ for some nonzero $a\in \mathbb{C}$ and some $b\in\{\pm\sqrt{2}\}$.
Since $g\in\mathbb{Q}[x_1,x_2]$, we must have $a\in\mathbb{Q}$ (since $a$ is the coefficient of the $x_1$ term of $g$), and we must also have $ab\in\mathbb{Q}$ (since $ab$ is the coefficient of the $x_2$ term of $g$).
But then we would have ${\Large{\frac{ab}{a}}}=b\in\mathbb{Q}$, contradiction.
Therefore $f$ is irreducible in $\mathbb{Q}[x_1,x_2]$.