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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?

confused
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    How did you conclude that $X\in P$? The result you linked to may leave other alternatives? Sure, $p=3$. But what about $f(X)$? – Jyrki Lahtonen Jul 07 '23 at 09:56
  • You are right! Now I see that $f(X)=X+3$ also works but I couldn't find a way of showing all such $f$'s. Can we classify all those $f$'s? – confused Jul 07 '23 at 10:31
  • We don't need to find all the $f$s, we only need to find all the ideals. And fortunately, we have $(X, 3)=(X+3, 3)$ so $X$ and $X+3$, while being different functions, yield the same ideal. On the other hand, $f(X)=X-1$ also works, and that genuinely gives a different ideal. – Arthur Jul 07 '23 at 10:51
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    I have a new approach: prime ideals of $R$ containing $\overline{3}$ are in one-to-one correspondence with the prime ideals of $(\mathbb{Z}/3\mathbb{Z})[X]$ containing $X^4-X$. Now if a prime ideal $P$ contains $X^4-X$. Then we have three options. Either $X\in P$, $X-1\in P$, or $X^2+X+1\in P$. Since the corresponding quotients of first two ideals are isomorphic to $\mathbb{Z}/3\mathbb{Z}$, they are maximal, in particular primes. However, $P$ cannot contain $X^2+X+1$ since in $\mathbb{Z}/3\mathbb{Z}$, $X^2+X+1=(X-1)(X+2)$. Thus, the ideals we were looking are $(3,X)$ and $(3,X-1)$. – confused Jul 07 '23 at 12:27
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    That's correct! Modulo three we have the factorization $$x^4-3x^2-x\equiv x^4-x\equiv x(x-1)^3,$$ and you can read the irreducible factors from there. – Jyrki Lahtonen Jul 07 '23 at 16:32
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    Note that $$x^2+x+1\equiv x^2-2x+1=(x-1)^2.$$ – Jyrki Lahtonen Jul 07 '23 at 16:32
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    So I want to encourage you to compile all your observations into an answer. It is very much ok to self-answer when you figure things out after posting the question. At least my comments were made with a view of getting you to the finish line. That way you may also get more feedback on your solution. And the question will be removed from the list of unanswered questions. – Jyrki Lahtonen Jul 07 '23 at 16:58
  • Thanks a lot! Sure will do. – confused Jul 08 '23 at 11:41

1 Answers1

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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 $(3,X-1)$.

user26857
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confused
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