Question:
Let $R$ be the ring $\mathbb{Z}[X]/(X^4-3X^2-X)$. I am asked to find all prime ideals of $R$ containing $3+(X^4-3X^2-X)$.
Answer
We know that there is a one-to-one correspondence between the prime ideals of $R$ and the prime ideals $P$ of $\mathbb{Z}[X]$ containing $(X^4-3X^2-X)$. Thus, if we also add the condition that $3\in P$, the projection map yields the all desired prime ideals. To sum up, we want to find prime ideals $P$ of $\mathbb{Z}[X]$ such that $(X^4-3X^2-X)\subseteq P$ and $3\in P$. From the question here, we know that $P$ can only be $(3,X)$. Therefore, its projection gives us the all prime ideals of $R$ that contains $3+(X^4-3X^2-X)$. I just want to as if there is any mistake in my proof?
By the helps of the above comments, here is the answer to the question:
By the correspondence theorems, the prime ideals of $R$ containing $\overline{3}$ are in one-to-one correspondence with the prime ideals of $\mathbb{Z}[X]$ containing $(X^4-3X^2-X)$ and $3$. Also, observe that the latter prime ideals are in one-to-one correspondence with the prime ideals $P$ of $(\mathbb{Z}/3\mathbb{Z})[X]$ containing $X^4-3X^2-X$ but observe that $X^4-3X^2-X\equiv X^4-X=X(X-1)(X^2+X+1)$ in $(\mathbb{Z}/3\mathbb{Z})[X]$. Now since $P$ is a prime ideal, $X^4-X\in P$ implies that either $X\in P$, $X-1\in P$, or $X^2+X+1\in P$.
First suppose $X\in P$, whence $(X)\subseteq P$. Observe that $(\mathbb{Z}/3\mathbb{Z})[X]/(X)\cong\mathbb{Z}/3\mathbb{Z}$, so $(X)$ is a maximal ideal. Thus, $P=(X)$.
Similarly, suppose $X-1\in P$, whence $(X-1)\subseteq P$. Again, $(\mathbb{Z}/3\mathbb{Z})[X]/(X-1)\cong \mathbb{Z}/3\mathbb{Z}$, so $(X-1)$ is maximal ideal. Thus, $P=(X-1)$.
Now suppose $X^2+X+1\in P$. Observe that $X^2+X+1\equiv X^2-2X+1=(X-1)^2$. Thus, in $(\mathbb{Z}/3\mathbb{Z})[X]$, $X^2+X+1$ is not prime implying that the ideal generated by $X^2+X+1$ is not prime either. Thus, $P$ is not prime.
Therefore, combining the above information, we obtain that the prime ideals we were looking for were just $(3,X)$ and $(3,X-1)$.