Note: I will abuse of notation and write $n$ instead of $n\cdot1$, for any $n \in \mathbf{Z}$. Obviously, $1$ being the identity of $K$.
Applying the first isomorphism theorem and the correspondence theorem, I have arrived to the following conclusion: $\mathfrak{a}$ is prime if and only if $\mathfrak{b}=(T^5-X^4+2X^2+3X^3)$ is prime. As $K[X,T]$ is a UFD, $\mathfrak{b}$ is prime if and only if $f(X,T)=T^5-X^4+ 3X^3 +2X^2$ is irreducible. If I find an irreducible element in $K[X]$, $p(X)$, such that $p(X)\mid -X^4+ 3X^3 +2X^2 $ and $p(x)^2 \not\mid -X^4+ 3X^3 +2X^2 $, I can apply the Eisenstein Criterion and conclude that $f$ is irreducible. But how can I get one $p(X)$ that satisfies those conditions?
I have tried the following. Note that: $-X^4+ 3X^3 +2X^2 = X^2(-X^2+3X+2)$. Unfortunately, $g(X)=-X^2+3X+2$ is not irreducible in some fields (for instance $\mathbf C$). So if $g(X)$ is irreducible in $K$, I take $p(X):= g(X)$. Otherwise, as $\deg g = 2$ and it is not irreducible, $g$ has two roots in $K$, $r_1$ and $r_2$. If I prove that $r_1 \neq r_2$, I take $p(X):= X-r_1$ (or $X-r_2$, it does not matter) and the conditions are satisfied for $p$. The sensible way to prove that $r_1 \neq r_2$ is to check the roots of the derivative of $g$. $g’(X)=-2X+3=0$. So, $X=3\cdot 2^{-1}$ (if $\text{char} \, K \neq 2$). Evaluating $g(3\cdot 2^{-1})=-9\cdot 2^{-2}+9\cdot 2^{-1}+2$ and this is different from zero. Hence, $r_1 \neq r_2$.
I conclude that $\mathfrak{a}$ is prime if $\text{char} \, K \neq2$. If $\text{char} \, K = 2$, I would have to check if $T^5+X^4+X^3$ is irreducible in $K$. Applying the Eisenstein Criterion to the irreducible $X+1$ it is straightforward.
Therefore, $\mathfrak{a}$ is prime in any field $K$.
Is there a better way to prove that $\mathfrak{a}$ is prime? What I have written is correct?
One could also use some field theory to prove that the polynomial $f:=T^5-X^4+2X^2+3X^3$ is irreducible: by the Lemma of Gauss it suffices to show that $f$ is irreducible as an element of $K(X)[T]$. If the characteristic of $K$ is different from $5$ and if $K$ contains the 5th roots of unity, then $f$ is either irreducible or splits completely into linear factors, the roots being of the form $\zeta^k\alpha$, where $\alpha\in K(X)$ is one root and $\zeta$ is a primitive 5th root of unity. This shows that actually $\alpha\in K[X]$, which is impossible since $X^4+2X^2+3X^3$ has degree $4$.
If $K$ does not contain the 5th roots of unity one can replace $K$ by $L=K(\zeta)$ thus getting the irreducibility of $f$ in $L(X)[T]$, which suffices.
If the characteristic of $K$ is $5$, then $f$ either is irreducible or has a root of multiplicity $5$ in $K(X)$. The latter is obviously not the case.