Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A few weeks ago I asked a question about finding the number of maximal ideals lying above $3\mathbb{Z}$ in $B$, where $B$ is the integral closure of $\mathbb{Z}$ in a splitting extension $E\supset\mathbb{Q}$ for the polynomial $f(x)=x^3+x+1$.

I'm reading over the following solution, but would like to ask a few follow up questions.

Since the discriminant of $f(x)$ is negative, two of the roots are complex conjugates, and since conjugation has order 2, the degree of the extension must be 6. Hence we may think of $E$ as adjoining two roots $a$ and $b$ of $f(x)$ to $\mathbb{Q}$, where $a$ satisfies $a^3+a+1$ and $b$ satisfies the polynomial $(x^3+x+1)/(x-a)=x^2+ax+a^2+1$.

So factor $(3)$ in $\mathbb{Z}[a]$ by factoring $x^3+x+1\pmod{3}$. Since there is one root modulo 3, it follows that $$ x^3+x+1=(x-1)(x^2+x-1)\pmod{3} $$ and so as we have seen, $(3)$ factors in $\mathbb{Z}[a]$ as $(3)=(3,a-1)(3,a^2+a-1)$.

Now since $a\equiv 1\mod (3,a-1)$, the polynomial $x^2+ax+a^2+1$ becomes $x^2+x+2$, which is irreducible modulo 3, and thus does not reduce further. Also, since $a^2=-a+1$ modulo this ideal, the polynomial can be further simplified to $x^2+ax-a+2$, so any root has for $ra+s$ for $r$ and $s$ integers modulo 3. This gives the set of equations $$ \begin{align*} (ra + s)^2 + a(ra + s) - a + 2 = 0 &\iff r^2a^2 + 2ras + s^2 + ra^2 + sa - a + 2 = 0,\\ &\iff r^2(-a + 1) + 2ras + s^2 + r(-a + 1) + sa - a + 2 = 0. \end{align*} $$ and so $-r^2 + r + s - 1 = 0$, and $r^2 + s^2 + r + 2 = 0 \pmod{3}$. This former equation implies $s\equiv (r+1)^2\pmod{3}$, and substituting into the second gives $r^2+(r+1)^4+r+2\equiv 0\pmod{3}$. This is true when $r=0$ and $s=1$, so $x=1$ is a root of $x^2+ax-a+2\mod(3,a^2+a-1)$. This in turn implies it factors as $x^2+ax-a+2=(x-1)(x+a+1)$.

Hence $(3, a^2 + a - 1) = (3, a^2 + a - 1, b - 1)(3, a^2 + a -1, b + a + 1)$, and hence there are three prime factors, and hence maximal ideals since $3\mathbb{Z}$ is maximal in $\mathbb{Z}$, lying above $3\mathbb{Z}$ in $\mathbb{Z}$.

I see that the $3\mathbb{Z}$ is the product of the $3$ factors found, but is there a more detailed explanation about why this immediately implies that there are $3$ maximal ideals lying above $3\mathbb{Z}$? I don't see how the conclusion follows so quickly. Would there need to be some uniqueness about this factoring for it to work as it does? Thank you kindly for your responses.

share|improve this question
    
Many thanks @BrandonCarter Do you have a proof or reference of these facts, that i) the integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$ is a Dedekind domain, and that ii) each nonzero ideal factors uniquely? If you post it as an answer, I would be more than happy to accept. –  Sofie Feb 1 '12 at 19:17
    
You might as well link to the older question. Actually, links both ways would be great. I would do this, but it appears that the older question was created on a separate account. –  Dylan Moreland Feb 2 '12 at 1:59

1 Answer 1

up vote 1 down vote accepted

Yes. The integral closure of $\mathbb{Z}$ in a finite extension of $\mathbb{Q}$ is a Dedekind domain, where each (nonzero) ideal factors uniquely as a product of (maximal) prime ideals. For proofs of these facts (and much more), see Robert Ash's notes here.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.