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I want to construct a cyclic division algebra of degree $3$ over some degree $3$ Galois extension $E$ of $\mathbb{Q}$. So the construction is as follows:

As a set $D=E\oplus uE \oplus u^2 E$ where $u$ is an indeterminate. Addition is defined component wise while multiplication is defined as $e.u=u\sigma(e)$ where $\sigma\in Gal(E/\mathbb{Q})$ is fixed and $u^3=a$ for some fixed $a\in \mathbb{Q}\setminus N_{E/\mathbb{Q}}(E)$ where $N_{E/\mathbb{Q}}$ is the norm map.

I want to find particular $a$ for some extension for some calculation purpose in $D^*$.

I started with $E=\frac{\mathbb{Q}[x]}{x^3+x^2-2x-1}$ and calculated that $N_{E/\mathbb{Q}}(a+bx+cx^2)= a^3-a^2b-2ab^2+b^3+5ac^2-abc-b^2c+6ac^2-2bc^2+c^3$. But it is difficult to find one element which is not norm of any element of the extension in this manner.

So here is my question:

Can you find explicitly one $a\in\mathbb{Q}$ such that $N_{E/\mathbb{Q}}(\alpha)\neq a$ for all $\alpha\in E$.

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    $\begingroup$ Your field $E$ in the example is $\mathbb{Q}(\mu_7)^{+}$; it has class group $1$. Primes that split are those congruent to $\pm 1$ modulo $7$. These primes, $7$, $-1$ and the cubes of all other primes will generate the image of the norm map in $\mathbb{Q}^{\times}$. For instance $a=2$ is not a norm. This can be done similar for other $E$. $\endgroup$
    – Chris Wuthrich
    Sep 15, 2018 at 10:14
  • $\begingroup$ @ChrisWuthrich : Ok. Can you give me a reference for the results, you used here? Thanks. $\endgroup$
    – user300
    Sep 15, 2018 at 10:19
  • $\begingroup$ In this case this is just basic algebraic number theory. If you really want to understand the image of the norm map in abelian extensions, read any text on global class field theory. $\endgroup$
    – Chris Wuthrich
    Sep 15, 2018 at 11:37
  • $\begingroup$ As an aged mathematician with failing eyesight, I ask you to avoid using $\alpha$ and $a$ in the same equation. $\endgroup$
    – Lubin
    Sep 15, 2018 at 16:08
  • $\begingroup$ @Lubin: haha ok :) $\endgroup$
    – user300
    Sep 15, 2018 at 20:09

1 Answer 1

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There is a general strategy to tackle such question. Since the norm is multiplicative, it make sense to look for a prime $p$ not in the image. Then, we can ask about the number of times $p$ divides a general norm. The norm map extends to prime ideals of the integer ring, and if the norm $N_{E/F}(q)$ is $p^3$ for every such ideal over $p$, then $p$ can not possibly be a norm, since its ideal is even not a norm of a fractional ideal of $\mathcal{O}_E$. This happens exactly for those primes $p$ which are inert in $E$, namely those primes modulo which the defining polynomial of $E$ remains irreducible. Indeed, the norm $N_{E/F}(q)$ coincide the the size of $\mathcal{O}_E/q$ which is $p^3$ for inert $q$, the quotient being $\mathbb{F}_{p^3}$.

In our case, for example, the polynomial $x^3+x^2-2x-1$ reduces mod $2$ to the polynomial $x^3+x^2+1$ which is irreducible, so $2$ is an inert prime and the number $2$ can not be realized as a norm. So you can choose $a=2$.

edit: Apparently a similar answer was put in the comments while I written this, sorry for the double answer.

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  • $\begingroup$ can you please suggest me some reference about the results, you used here? $\endgroup$
    – user300
    Sep 15, 2018 at 10:25
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    $\begingroup$ @user300 unfortunately I don't know any reference for that, but I guess that every standard textbook on algebraic number theory, e.g. Neukirch book should contain something along this lines. $\endgroup$
    – S. carmeli
    Sep 15, 2018 at 11:29
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    $\begingroup$ This example can be handled by more elementary methods too. Lam shows on pp. 238-240 of his book "A First Course in Noncommutative Rings" that $2$ and $4$ are not norms from this cubic field by proving in a simple way that an even integer that is a norm from that field has to be a multiple of $8$. He attributes the example (and perhaps implicitly the technique?) to Dickson. $\endgroup$
    – KConrad
    Sep 15, 2018 at 11:52

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