Re: list all maximal ideals

The key to all of these is the lattice (fourth) isomorphism theorem. Namely, you know that the maximal ideals of a quotient ring $\displaystyle R/I$ are the maximal ideals of the form $\displaystyle M/I$ for $\displaystyle M$ a maximal ideal of $\displaystyle R$ containing $\displaystyle I$. So, let's see here:

a) Yes, as you pointed out, the maximal ideals are the principal ones generated by $\displaystyle 2,3,5$. This follows from what I said since the maximal ideals of $\displaystyle \mathbb{Z}$ that contain $\displaystyle (90)$ are the principal ideals of the form $\displaystyle (p)$ where $\displaystyle p\mid 90$ is prime.

b) Exactly. Since $\displaystyle \mathbb{Q}$ is a field, you know that $\displaystyle \mathbb{Q}[x]$ is a PID, and since $\displaystyle x^2+1$ is irreducible we must have that $\displaystyle (x^2+1)$ is prime, and thus maximal. Thus, the only maximal ideals containing $\displaystyle (x^2+1)$ are the ideal itself!

c) Exactly. Since $\displaystyle \mathbb{C}$ is algebraically closed, you know that all the maximal ideals are the principal ones with linear generators. Since the only linear factors dividing $\displaystyle x^2+1$ are $\displaystyle x\pm i$ your answer is correct.

d) Once again, since $\displaystyle \mathbb{Q}[x]$ is a PID, the maximal ideals containing $\displaystyle (x^3+x^2)$ are the principal ideals generated by the irreducible factors of $\displaystyle x^3+x^2$. In other words, the maximal ideals above $\displaystyle (x^3+x^2)$ are the ideals $\displaystyle (x+1)$ and $\displaystyle (x)$. By the way, using this idea, you see that the maximal ideals of $\displaystyle \mathbb{Q}[x]/(x^2)$ is just the ideal $\displaystyle (x)$.

Best,

Alex