I want to show that the polynomial $f=X^3+2Y^2+3\in \mathbb{Q}[X,Y]$ is prime. Since, however, $\mathbb{Q}[X,Y]$ is not a principal integral domain (as $\mathbb{Q}$ is not a field) it does not suffice to show that $f$ is irreducible. So, what I have been thinking is the following: $f$ would be prime if the ideal $(f)$ is prime. This is the case, if and only if the factor ring $R=\mathbb{Q}[X,Y]/(f)$ is an integral domain, i.e. it does not have any non-trivial zero-divisors. So I have been looking at $g=g_1+g_2(X^3+2Y^2+3)$, $h=h_1+h_2(X^3+2Y^2+3)\in R$ with $g_1,g_2,h_1,h_2\in \mathbb{Q}[X,Y]$. Now suppose, $gh=0\in R$, then it followings \begin{align} gh&=g_{1}h_1+g_{1}h_2(X^3+2Y^2+3)+g_{2}h_1(X^3+2Y^2+3)+h_{2} g_2(X^3+2Y^2+3)^2\\ &=g_{1}h_1 \end{align} Thus, $g_{1}h_2$ must be a multiple of f and, since f is irreducible, this only happens for either $g_1$ or $h_2$ a multiple of f, i.e. $g=0$ or $h=0$. I am a bit unsure. Is this correct, I am unsure about the last step, concluding that "$g_{1}h_2$ must be a multiple of f and, since f is irreducible". Because this would hold completely unrelated of the particular polynomials and, thus, irreducible and prime elements would be the same which they are only in principal integral domains which we do not have here.
Thanks a lot in advance
Let $R:=\Bbb Q[X]$. This is UFD, as is $R[Y]$. We have to check if $Y^2+\frac{1}{2}(X^3+3)$ is an irreducible polynomial of $R[Y]$; considering degree, necessarily it would reduce into linear factors and so the question really becomes: does $-\frac{1}{2}(X^3+3)$ have a square root in $R$?
Considering degree again, we realise that this cannot happen since $3$ is not even. We are done.