I saw the following question on a question paper:
Decide whether the following statements are true or false:
(1) The ring $\mathbb{C}[X]/(X^4-9X^3+18X^2-13X+3)$ has $4$ maximal ideals.
(2) In $\mathbb{Z}[X]$, there are $3$ maximal ideals that contain the ideal $(7, X^{147}+21X^{98}+12)$.
(3) There are $24$ monic irreducible polynomials in $\mathbb{R}[X]$ that divide $X^{48}+3X^{24}+4$.
For (1), I know that the ideals of our ring are of the form $I/(X^4-9X^3+18X^2-13X+3)$, where $I$ is an ideal in $\mathbb{C}[X]$ that contains $(X^4-9X^3+18X^2-13X+3)$. Via one of the isomorphism theorems (I don't know exactly its number, because different people number them differently), our ideal $I/(X^4-9X^3+18X^2-13X+3)$ is maximal in our factor ring iff $I$ is maximal in $\mathbb{C}[X]$. We know that an ideal in $\mathbb{C}[X]$ is maximal iff it is generated by a polynomial of degree $1$. This combined with the fact that this polynomial should divide $X^4-9X^3+18X^2-13X+3$ and this polynomial has $3$ distinct roots, we get that $\mathbb{C}[X]/(X^4-9X^3+18X^2-13X+3)$ has precisely $3$ maximal ideals.
For (3), our polynomial has no real roots. All of its roots are complex and come in conjugate pairs. As a result, $X^{48}+3X^{24}+4$ is the product of $24$ monic irreducible polynomials of degree $2$ in $\mathbb{R}[X]$, so (3) is true.
For (2), I know that the maximal ideals in $\mathbb{Z}[X]$ are of the form $(p, f)$, where $p$ is a prime number and $f$ is a polynomial that is irreducible in $\mathbb{Z}_p[X]$. Since we want to have $(7, X^{147}+21X^{98}+12)\subset (p, f)$, we get that $p=7$ and $f$ divides $X^{147}+21X^{98}+12 $. But via Eisenstein's Criterion, $X^{147}+21X^{98}+12 $ is irreducible so I either have $f=1$ or $f=X^{147}+21X^{98}+12 $, but this doesn't look right because I would get that there is no maximal ideal that contains my ideal and this is wrong.
So, my reasoning (at least) for (2) is flawed. How should I approach this?
EDIT: I will add the exact wording of the question so it is relevant why I ask all these questions together. We need to choose between the options:
a) (1), (2), and (3) are true.
b) (1) and (2) are false.
c) (1) and (3) are false.
d) (2) and (3) are true.
Your proof of (1) is good. You might consider mentioning explicitly that you're using that $\mathbb C$ is algebraically closed, as well as showing why there are exactly 3 distinct roots.
For (3) your proof is mostly there, but again I think there are some details to be added. I interpret the question as saying that there are precisely 24 monic irreducibles dividing. How do you know that the 24 polynomials you've found are all distinct? How do you know there aren't any more monic irreducibles dividing it? These questions aren't as hard as the work you've already done in my opinion, but it's worth adding the detail.
For (2) I must point out that $(q, g) \subseteq (p, f)$ does not mean $q = p$ and $f \mid g$. You can deduce $p=q$ by the existence of a map $\mathbb Z[x]/(q, g) \longrightarrow \mathbb Z[x]/(p, f)$ and considering characteristic. However, $f \nmid g$ is possible. You could have $f = x$ and $g = x + p$, for instance. What you actually have is divisibility in $\mathbb Z/p$.
Note that $x^{147} + 21x^{98} + 12 = x^{147} + 5$ in $\mathbb Z/7 [x]$. This problem reduces to factoring this polynomial. We have $147 + 3 * 7^2$, so $x^{147} + 5 = (x^3 + 5)^{49}$. The equation $x^3 + 5 = 0$ has no solutions in $\mathbb Z/7$ so this cubic is irreducible. Hence, the only maximal ideal containing $(7, x^{147} + 21x^{98} + 12)$ should be $(7, x^3 + 5)$.