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Thread: list all maximal ideals

  1. #1
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    list all maximal ideals

    I seem to have difficulty with this type of problem - perhaps a lack of intuition about prime and maximal ideals. Here's the problem. I'm having trouble with part (d).

    For each of the following rings, list all maximal ideals.
    (a) $\displaystyle \mathbb{Z}/90\mathbb{Z}$
    (b) $\displaystyle \mathbb{Q}[x]/(x^2+1)$
    (c) $\displaystyle \mathbb{C}[x]/(x^2+1)$
    (d) $\displaystyle \mathbb{Q}[x]/(x^3+x^2)$

    My solutions for (a) through (c):

    (a) These are generated by the prime factors of 90: (2), (3), and (5).
    (b) Since $\displaystyle x^2+1$ is irreducible in $\displaystyle \mathbb{Q}[x]$, this is a field, so the only maximal ideal is (0).
    (c) $\displaystyle \mathbb{C}[x]/(x^2+1) = \mathbb{C}[x]/((x+i)(x-i)) \cong \mathbb{C}[x]/(x+i)\times\mathbb{C}[x]/(x-i) \cong \mathbb{C}\times\mathbb{C}$, so the maximal ideals are ((0,1)) and ((1,0)), which correspond to the ideals (x+i) and (x-i) in the original ring.

    I think that answer is correct, but I'm not entirely clear on how I got from the ideals in $\displaystyle \mathbb{C}\times\mathbb{C}$ to the ideals in the original ring.

    For (d), I think it's true that $\displaystyle \mathbb{Q}[x]/(x^3+x^2) \cong \mathbb{Q}[x]/(x^2) \times \mathbb{Q}[x]/(x+1)$, and the second is isomorphic to $\displaystyle \mathbb{Q}$. But what are the maximal ideals of $\displaystyle \mathbb{Q}[x]/(x^2)$?

    - Hollywood
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  2. #2
    MHF Contributor Drexel28's Avatar
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    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
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