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Suppose $K$ is the splitting field of $x^3-2$ over $\mathbb{Q}$, and let $\mathcal{O}_K$ denote its ring of integers. I want to show that for any prime number $p$, $p\mathcal{O}_K$ is not a prime ideal of $\mathcal{O}_K$. The hint I was given was to consider the factorization of $x^3-2$ over $F_p$.

My question(s)

My first instinct was to use Kummer's criterion, which guarantees that $p\mathcal{O}_K$ is not prime as long as $x^3-2$ is reducible (has a root) in $F_p$ [and $p>3$]. But how to handle the case when $p=2,3$ (I suspect it ramifies, but I can't seem to show this).

My second instinct was that if $p\mathcal{O}_K$ is prime in $\mathcal{O}_K$, then $\mathcal{O}_K/p\mathcal{O}_K$ would be the splitting field of $x^3-2$ over $F_p$, and so I just need to check that the degree is always less than $[K:\mathbb{Q}]=6$. But is it always true in this case that $\mathcal{O}_K/p\mathcal{O}_K$ is the splitting field of $x^3-2$ over $F_p$?

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2 Answers

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Let's work inside $L=\mathbb{Q}(\sqrt[3]2)$. It really should be easy to see that $2$ ramifies (there an ideal $\mathfrak{p}$ with $\mathfrak{p}^3=2\mathfrak{O}_L$). Proving that $3$ ramifies is a little trickier. If we could find an element of $\mathfrak{O}_L$ whose minimal polynomial is Eisenstein at $3$, that would do. An Eisenstein polynomial at $3$ would be $$x^3+ax^2+bx+c$$ where $a$, $b$ and $c$ are divisible by $3$ and $c$ isn't divisible by $9$. Obviously $\sqrt[3]2$ doesn't work - but is there something similar which does?

Added I was implicitly arguing that since the primes $2$ and $3$ aren't inert in $L=\mathbb{Q}(\sqrt[3]2)$ then a fortiori they aren't inert in its Galois closure $K$. But no prime remains inert in a non-cyclic Galois extension of a number field. For a prime to remain inert, its decomposition group is the whole Galois group and its inertia group is trivial. But the quotient of the decomposition group by the inertia group must be cyclic (its the Galois group of the extension of residue fields)....

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I think you mean $\mathfrak{p}^3 = 2\mathcal{O}_K$ in the first line, not $\mathfrak{p}^3 = 2\mathfrak{p}$. –  Arturo Magidin Sep 8 '10 at 17:10
    
Dear Robin, I think that you mean $\mathfrak p^3 = 2\mathcal O_K$. –  Matt E Sep 8 '10 at 17:11
    
Thanks, corrected. –  Robin Chapman Sep 8 '10 at 17:21
    
Thanks so much for your last paragraph! I'm using Fröhlich and Taylor, and they do these type of things for complete fields. But after thinking about it, I see we can basically do the same thing for fields which aren't complete. –  PeteRR Sep 8 '10 at 21:46
    
Robin, it is easy to see 3 ramifies: look at the minimal polynomial of 2^(1/3) + 1, which is (x-1)^3 - 2. –  KCd Sep 9 '10 at 12:05
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In general, if the ring of integers of $\mathcal{O}_K$ is of the form $\mathbb{Z}[\alpha]$, and $q(x)$ is the irreducible polynomial of $\alpha$ over $\mathbb{Q}$, then there is a natural bijection between the primes in $\mathcal{O}_K$ that lie above the rational prime $p$ and the irreducible factors of $\overline{q}(x)$ in $\mathbb{F}_p$, where $\overline{q}$ is the image of $q(x)$ under reduction modulo $p$. The bijection is that a prime $\mathfrak{p}$ of $\mathcal{O}_K$ lies above $(p)$ corresponds to the factor $\overline{Q}(x)$ if and only if $\mathfrak{p}$ is the kernel of the map $\mathbb{Z}[\alpha]\to\mathbb{F}_p[\alpha]$. So the factorization of $(p)$ in $\mathcal{O}_K$ exactly mimicks the factorization of $\overline{q}(x)$ over $\mathbb{F}_p$: same number of factors, same repetitions, etc.

So you want to factor $x^3-2$ over $\mathbb{F}_p$; this suffices here because if $\alpha$ is a root of $x^3-2$, then $\mathcal{O}_K = \mathbb{Z}[\alpha]$ (see for example Problem 41(a)-(d) in Chapter 2 of Marcus's Number Fields). Edit: I was working in $\mathbb{Q}(\sqrt[3]{2})$ instead of the splitting field; sorry. But from the fact that they ramify in this subextension, you know they will ramify in $K$; the only question is whether they ramify as $\mathfrak{p}^6$ or as $\mathfrak{p_1}^3\mathfrak{p_2}^3$, depending on whether the prime above $p$ splits or ramifies, or is inert in $K$. Since you can go from $\mathbb{Q}(\sqrt[3]{2})$ to $K$ by adjoining $\omega$, and the ring of integers will be, if I'm not mistaken, $\mathbb{Z}[\alpha][i]$, you can use the same theorem above (which holds for arbitrary integral extensions, not just over $\mathbb{Z}$) to get that next step, looking at $x^2+x+1$.

With $p=2$, the polynomial factors over $\mathbb{F}_p$ as $x^3$, so $(2) = \mathfrak{p}^3$ for some prime $\mathfrak{p}$ of $\mathcal{O}_K$; for $p=3$, you have $x^3 + 2 = x^3 - 1 = (x-1)^3$, so again $(3)=\mathfrak{q}^3$ for some prime $\mathfrak{q}$ of $\mathcal{O}_K$.

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Thanks for your answer. I am working from Fröhlich and Taylor's book. They prove that you can factor $p\mathcal{O}_K$ as you factor $x^3-2$ in $F_p$, provided $p$ doesn't divide $[\mathcal{O}_K:\mathbb{Z}[\alpha]]$. Of course, you claim $\mathcal{O}_K=\mathbb{Z}[\alpha]$, but I don't think I've learned enough tools to see that. –  PeteRR Sep 8 '10 at 21:13
    
Note that I messed up and was working in $\mathbb{Q}[\sqrt[3]{2}]$ instead of $K$. What I quoted works in any integral extension $B$ over $A$ provided that $B=A[\alpha]$ for some $\alpha$; I don't think it applies directly to your $K$, but you can do it in two steps (I need to fix that $x^2+1$ and $i$ to $x^2+x+1$ and $\omega$, though). –  Arturo Magidin Sep 9 '10 at 3:43
    
Pete, look at Example 3.4 of math.uconn.edu/~kconrad/blurbs/gradnumthy/totram.pdf –  KCd Sep 9 '10 at 12:07
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