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In $\mathbb{Z} \sqrt{- 5}$, $2$ is irreducible, but the ideal $(2)$ factors into non-units:

$(2)$ = $(2, 1 + \sqrt{- 5})(2, 1 - \sqrt{- 5})$.

In general, what gives one the intuitive motivation (or maybe a priori confidence) to attempt to factor an ideal?

I could say in the specific case above that $(1 + \sqrt{- 5})$ and $(1 - \sqrt{- 5})$ are easy combinations of the elements of the basis of $\mathbb{Z} \sqrt{- 5}$, {$1, \sqrt{- 5}$} and their product is divisible by $2$, so it's pretty easy to give them a try along with $2$.

But I was wondering if there is something that gives a clue that an ideal can be further factored.

Thanks

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1  
in quadratic fields you can tell which primes split by congruence conditions. –  user58512 Feb 28 '13 at 22:18
    
In a dedekind domain, every ideal can be factored into prime ideals in a unique way. So, if your ideal is prime it can't be facotored anymore. –  Mohan Feb 28 '13 at 22:19
    
@user58512 Do you think you could elaborate as an answer and give an example? Thanks –  Andrew Feb 28 '13 at 22:20
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There is a lot of general theory that gives clues. One of the immediately available ones is that $(2)$ is not a maximal ideal, because the quotient ring is not a field. Here $\mathbb{Z}[\sqrt{-5}]/(2)$ has four elements, the cosets of $0,1$,$\sqrt{-5}$ and $1+\sqrt{-5}$. But this quotient ring is not a field. In fact the last element is nilpotent, which is an indication the $(2)$ ramifies (as opposed to splits). The square of $(1+\sqrt{-5})$ is in the ideal $(2)$, so the latter is not a prime idea. –  Jyrki Lahtonen Feb 28 '13 at 22:20
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@Andrew, it's a consequence of quadratic reciprocity –  user58512 Feb 28 '13 at 22:20

1 Answer 1

up vote 5 down vote accepted

The key insight into all of this (at least for quadratic fields) is something called the absolute ideal norm. Let $K$ be a quadratic extension of $\Bbb{Q}$ and let $\mathcal{O}_K$ denote the integral closure of $\Bbb{Z}$ in $K$. For any ideal $\mathfrak{a} \in \mathcal{O}_K$, we define $||\mathfrak{a}||$ to be the number of elements in the finite group $\mathcal{O}_K/\mathfrak{a}$. To see why $\mathcal{O}_K/\mathfrak{a}$ is finite consider the ses

$$0 \to \mathfrak{a} \to \mathcal{O}_K \to \mathcal{O}_K/\mathfrak{a} \to 0$$

and apply the exact functor $-\otimes_{\Bbb{Z}} \Bbb{Q}$ to the conclude that the free part of $\mathcal{O}_K/\mathfrak{a}$ is zero. Right back to our problem. Suppose now that $\mathfrak{a}$ is a principal ideal. For a concrete example let's take $K = \Bbb{Q}(\sqrt{-21})$ and $\mathfrak{a} = (3 + \sqrt{-21})$.

Now factor $\mathfrak{a}$ into primes $p_1\ldots p_n$ (including multiplicities). Then using the fact that the absolute norm is multiplicative and that the absolute norm of a principal ideal is the field norm of the generator we get that

$$30 = N_{K/\Bbb{Q}}(3 + \sqrt{-21})) = ||p_1|| \ldots ||p_n||.$$

Here is now the killer blow: The right hand side is just a plain old product of integers!!!

Thus it must be the case that upto rearrangement, $||p_1|| = 2$, $||p_2||= 3$ and $||p_3||| =5$. Thus we now ask ourselves: What are the possibilities for a prime $p_1$ of norm $2$? Well it has to lie over $2$ of course! Why? Because we recall that for $P$ a prime in $\mathcal{O}_K$ lying over a prime $p \in \Bbb{Z}$, the number of elements in $\mathcal{O}_K$ is just

$$|\Bbb{Z}/p\Bbb{Z}|^{f(P|p)}$$

namely a power of a prime. Thus we see that it will suffice to factor $2\mathcal{O}_K$,$3\mathcal{O}_K$ and $5\mathcal{O}_K$. I get that

$$\begin{eqnarray*} 2\mathcal{O}_K &=& (2, 1 + \sqrt{-21})^2\\ 3\mathcal{O}_K &=& (3,\sqrt{-21})^2 \\ 5\mathcal{O}_K &=& (5,2+\sqrt{-21})(5,2-\sqrt{-21}). \end{eqnarray*}$$

and so $(3+\sqrt{-21})$ must necessarily factor as

$$(3+\sqrt{-21}) = (2, 1 + \sqrt{-21})(3,\sqrt{-21})\mathfrak{p}_5$$

where $\mathfrak{p}_5$ is one of the primes lying over $5$ of which there are two possibilities. To determine exactly which one it is takes some bashing, but at least you get the general idea of how we factor principal ideals. Lastly, I should say that it is not so hard to determine how a prime splits in a quadratic extension. Let me give an example. Consider $\Bbb{Q}(\sqrt{-21})$ as before and the prime $p = 5$ in $\Bbb{Z}$. We know that $\mathcal{O}_K = \Bbb{Z}[\sqrt{-21}]$ and so

$$\begin{eqnarray*}\Bbb{Z}[\sqrt{-21}]/(5) &\cong& \Bbb{Z}[x]/(x^2 + 21)/(5,x^2+ 21)/(x^2 + 21) \\ &\cong& \Bbb{Z}[x]/(5,x^2+21)\\ &\cong& \left(\Bbb{Z}/5\Bbb{Z}\right)[x]/(x^2 + 1) \\ &\cong& \left(\Bbb{Z}/5\Bbb{Z}\right)[x]/\left((x+2)(x+3)\right) \\ &\cong& \Bbb{Z}/5\Bbb{Z} \times \Bbb{Z}/5\Bbb{Z} \end{eqnarray*}$$

by the Chinese remainder theorem. This tells you that $5$ must split into two prime ideals, namely $(5,2+\sqrt{-21})$ and $(5,3+\sqrt{-21})$. The latter is of course the same as $(5,2-\sqrt{-21})$, which is exactly like what I wrote down above.

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Thanks BenjaLim. I appreciate it. To be honest, I get some of it, yet it is apparent to me that I need a great deal more background studying - which I am very willing and would enthusiastically do. It's ironic (e.g., for this particular question) that I was reading some notes that were going well, but presented such examples as a "done deal," leading me wonder, and post my question. Maybe there are books, notes you might recommend that would be beneficial to get to the level to derive the fullest out of your answer and related material, no matter how much work it entails. Best regards, Andrew –  Andrew Feb 28 '13 at 23:47
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Hi again. I posted my email address if you are so inclined. It would be nice to stay in touch. Best, Andrew –  Andrew Mar 1 '13 at 0:10
    
@Andrew Thanks for your kind words! For references about my answer, you can read about absolute norms in chapter 3 of Marcus' Number Fields. In that chapter is also Theorem 25 which is basically a "formula" for how primes split in quadratic extensions. The specific example of $\Bbb{Q}(\sqrt{-21})$ in my answer above I thought of while trying to compute some class group. For the last part of my answer, well I kinda learnt it from experience. You can read some other answers of mine on algebraic number theory to see how this kinda quotiening is done. –  user38268 Mar 1 '13 at 2:19
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Thanks, as always. –  Andrew Mar 1 '13 at 2:33

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