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This naive question came as the last problem in my homework. The author asked me to use linear relations of the discriminant like $\operatorname{disc}(ra_{1},a_{2}...a_{n})=r^{2}\operatorname{disc}(a_{1},..a_{n})$, $\operatorname{disc}(a_{1}+\beta,a_{2}...a_{n})=\operatorname{disc}(a_{1},..a_{n})$ for $\beta$ be a linear combination of $a_{i}$. I could not see how to make use of the hint or how to solve the problem in an easier way than taking the norm and evaluate, which is inpractical since 46$\times$23 terms are involved.

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Do you mean irreducible as an element in $\mathbf Z[\omega]$? –  Dylan Moreland Mar 8 '12 at 7:11
    
Remember that for a general number field $K$ with integers $\mathcal O$ and $\alpha \in \mathcal O$ we have $|N^K_{\mathbf Q}(\alpha)| = (\mathcal O : (\alpha))$. This last number is something you could get from relating the discriminants of $\mathcal O$ and $(\alpha)$. What's a good basis for the latter? –  Dylan Moreland Mar 8 '12 at 7:12
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You may find something useful in the links in my answer to math.stackexchange.com/questions/85520/… –  Gerry Myerson Mar 8 '12 at 11:20
    
@DylanMoreland: I could prove it is irreducible in $A\cap \mathbb{Q}(\sqrt{-23})$, but I could not prove it is irreducible in $A\cap \mathbb{Q}(\omega)$. –  Kerry Mar 8 '12 at 14:30
    
@DylanMoreland: I think it is a naive approach, but supoose $(2)\in (\alpha)$, and write $\alpha=\sum^{22}_{i=0} a_{i}\omega^{i}$. Then by your hint we have $|N(\alpha)|=(\mathbb{Z}[\omega]/\alpha)$. But this is not very clear - oh, do you mean using the integral basis of $O$ including $\alpha$? –  Kerry Mar 8 '12 at 17:01

3 Answers 3

A strategy of attacking this problem is this:

To show $2$ is irreducible in $\mathbb{Z}[\omega]$, we need to show that $F_{2}[\omega]$ is an integral domain. But this is equivalent to $F_{2}[x]/\Phi_{23}[x]$ is an integral domain. Thus we translated the problem to show $\Phi_{23}(x)$ is irreducible in $F_{2}[x]$. And this can be checked since $\Phi_{23}[x]=\sum^{22}_{i=0} x^{i}$, thus $\Phi_{23}[x]$ is irreducible in $F_{2^{n}}[x]$ if and only if $23|2^{n}-1$. In this case verify via brutal force give $n=22$. Thus $\Phi_{23}[x]$ is irreducible in $\mathbb{Z}[\omega]$.

I am slightly confused with the sublety since $\mathbb{Z}[\omega]$ is known to be not a UFD. Thus irreducible elements and primes does not coincide. I need to fix this.

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Dear Changwei, The fact that $23| 2^{22} -1 $ is just Fermat's little theorem. What is more to the point here is that $23$ also divides $2^{11} - 1$, and so $\Phi_{23}$ is not irreducible in $F_{2^{22}}[x]$. Consquently, $2$ does not generate a prime ideal in $\mathbb Z[\omega]$, and $\mathbb Z[\omega]/2\mathbb Z[\omega]$ is not an integral domain; rather, the principal ideal $2 \mathbb Z[\omega]$ factors as the product of two prime ideals. However, neither of them will be principal. Regards, –  Matt E Mar 13 '12 at 2:18
    
@MattE: Dear Matt: I obviously did not work right in the brutal force process. I have to re-think this through. Thank you for the suggestions. –  Kerry Mar 13 '12 at 17:03
    
@MattE I have posted a proof above to show that one of these ideals will not be principal. –  user38268 Dec 26 '12 at 0:00
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@BenjaLim: Dear Benjamin, Thanks for letting me know; I just read and upvoted it. Regards, –  Matt E Dec 26 '12 at 3:28

We recall that a non-zero non-unit of a ring $S$ is said to be irreducible if it cannot be written as the product of two non-units. Set $S = \Bbb{Z}[w]$ now that is the ring of integers of the algebraic number field $\Bbb{Q}(w)$. $w = e^{2\pi i/23}$ here.

Now suppose that $2 = \alpha\beta$. We want to show that one of $\alpha$ or $\beta$ is a unit. Suppose that the both of them are not units. Then we have $2S = (\alpha\beta) = (\alpha)(\beta)$ with the ideals $(\alpha),(\beta)$ both properly contained in $\Bbb{Z}[w]$. Now by Proposition 10.3 (Neukirch) s, we know that the ideal $2\Bbb{Z}$ is unramified in $S$; in fact it is the product of $22/11 = 2$ prime ideals in $S$. I leave it to you to check that

$$2S = \left(2, \frac{1 + \sqrt{-23}}{2}\right)\left(2, \frac{1 - \sqrt{-23}}{2}\right).$$

By unique factorisation, we must have $(\alpha) = \left(2, \frac{1 + \sqrt{-23}}{2}\right) = Q$ say. However this contradicts $Q$ not being principal and so at least one of $\alpha$ or $\beta$ must be a unit in $\Bbb{Z}[w]$.

Proof that $Q$ is not principal, à la Marcus Problem 3.17:

Let $K = \Bbb{Q}(\sqrt{-23})$, $L = \Bbb{Q}(w)$. By my answer here we know that $K\subseteq L$. Set $R = \mathcal{O}_K$ and $S = \mathcal{O}_L$.

Part (a): Let $P$ be the prime ideal $(2, \theta)$ in $R$ that lies over $(2)$ in $\Bbb{Z}$ where $\theta = \frac{1 + \sqrt{-23}}{2}$. Now if $Q$ is a prime of $S$ that lies over $P$ then it also lies over $(2)$ in $\Bbb{Z}$. By Proposition 10.3 (Neukirch) we have $$e(Q|(2)) = \varphi(2^0) = \varphi(1) = 1$$ and $f(Q| (2)) = 11$ because $2^{11} = 1 + 23(89)$ and one easily sees that $11$ is the multiplicative order of $2 \pmod{23}$. Since the inertia degree is multiplicative in towers, we get that

$$f(Q|P) = \frac{f(Q|(2))}{f(P|(2))} = \frac{11}{{f(P|(2))}}.$$

Now since $-23 \equiv 1 \pmod{8}$ by knowledge of how primes split in quadratic extensions we know that $2R = (2,\theta)(2,\bar{\theta})$ where $\bar{\theta}$ is the other Galois conjugate of $\theta$ over $\Bbb{Q}$. It will now follow that $f(P|(2)) = 1$ and so we conclude that $f(Q|P) = 11/1 = 11$. For the last part of this problem, we know already that $S/PS$ is an $11$ - dimensional $R/P$ vector space. Since this is equal to $f(Q|P)$, we see that the map $f$ in the diagram

                               enter image description here

is a surjective map between two finite sets of the same size and hence is injective. It follows that $PS = Q$, i.e. that $(2,\theta) = Q$ in the ring $S = \Bbb{Z}[w]$.

(b): Now it is not hard to see that $P^3 = (\theta - 2)$. If $P$ is a principal ideal generated by some $\alpha \in R$ then since $||P||^3 = ||(\theta - 2)|| = 8$ and so we would have $N_{K/\Bbb{Q}}(\alpha)$ being either $2$ or $-2$. We now show that there is no element in $R$ of norm $2$ or $-2$. Consider a general element $x$ in $R$ which is of the form $x = a + b \theta$ for $a,b \in \Bbb{Z}$. Then $N_{K/\Bbb{Q}} (x) = (a + b\theta)(a + b\bar{\theta}) = a^2 + ab + 12b^2$. If this is equal to $2$ or $ -2$, by the quadratic formula there must exist an integer $b$ such that $b^2 - 4(12b^2 -2) \geq 0$ or $b^2 -4(12b^2 +2) \geq 0$ which is a contradiction. It follows that $P$ is not principal.

(c) - Final part of the proof to show that $Q$ is not principal: Let $G(S)$ and $G(R)$ denote the ideal class gorups of $R$ and $S$. Let $d_Q$ denote the order of the class of $Q$, let $d_P$ denote the order of the class of $P$. If $Q$ is a principal ideal then we would have $d_Q = 1$. By my answer here and by part (a) of this problem we would have $d_P|11$. However from (b) we have $P^3 = (\theta - 2)$ and taking ideal classes gives $[P]^3 = [(\theta - 2)]$ where the latter is the class of the identity. We conclude that the order of $[P]$ is exactly $3$ because $P$ is not principal. However we now get that $3|11$ which is a contradiction from which it follows that $Q$ cannot possibly be principal.

$$\hspace{6in} \square$$

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"Number Fields" by Daniel A. Marcus has a guided exercise that proves your question - see chapter 3, exercise 17 (page 86):enter image description here

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