1
$\begingroup$

I heard that the discriminant of an extension is equal to the product of local discriminants. To better understand this I tried it out on some quadratic number fields, but ended up getting stuck.

For example I considered the number field $\mathbb{Q}(\sqrt{5})$. Since $$disc(\mathbb{Q}(\sqrt{5}))=-20$$ we have that the only ramified primes are 2 and 5 and so we should have $$-20=disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)\cdot disc(\mathbb{Q}_5(\sqrt{5})/\mathbb{Q}_5).$$ (I suppose knowing the discriminant before hand is kind of cheating...)

Anyway, following the answer here (and also with the help of Gouvea's book p-adic Numbers) I was able to show that $\mathbb{Q}_5(\sqrt{5})/\mathbb{Q}_5$ is ramified and thus divides $disc(\mathbb{Q}(\sqrt{5}))$. I am not sure about what to do about $disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)$. I do know that $\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2$ is unramified, but this is part of my confusion (I guess things get weird in characteristic 2). In particular I am curios as to how $disc(\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2)$ can be computed? Also, can this be done without finding an explicit basis for the ring of integers of $\mathbb{Q}_2(\sqrt{5})/\mathbb{Q}_2$?

$\endgroup$
1
$\begingroup$

Maybe better for a full-fledged answer than a comment:

You have it wrong: though $\text{disc}(\Bbb Q(\sqrt{-5\,})=-20$, the fact is that $\text{disc}(\Bbb Q(\sqrt{5\,})=5$. This alone may dissipate all your confusions. But let me go into more detail.

Over $\Bbb Z$, an integral basis for (the integers of) $\Bbb Q(\sqrt{5\,})$ is $\{1,\frac{1+\sqrt5}2\}$. Since the number $\frac{1+\sqrt5}2$ has for its minimal polynomial $X^2-X-1$, you readily calculate the (global) discriminant to be $5$. I don’t know what general definition of the discriminant of a ring $R\supset\Bbb Z$ you’re using, but the most primitive one is to take your integral basis $\{a_1,\cdots,a_n\}$ and then calculate the determinant of the matrix whose $(i,j)$-entry is the trace of $a_ia_j$, the “trace” being the field-theoretic trace from the fraction-field of $R$ down to $\Bbb Q$.

To go locally, you look, as you did, at $\Bbb Q_5(\sqrt{5\,})\supset\Bbb Q_5$, for which a good basis is $\{1,\sqrt5\}$ or the global one I mentioned above. The discriminant ideal here is $(5)$. And at the prime $2$, you look at $\Bbb Q_2(\sqrt{5\,})\supset\Bbb Q_2$ and you have to use the global basis above; $\{1,\sqrt5\}$ is not an integral basis, even at $2$. And the discriminant ideal is $(1)\,$: that is, the extension is unramified at $2$.

If you have further questions, I can edit this answer or respond in a comment.

$\endgroup$
  • $\begingroup$ Thanks, that cleared up my confusion. This might be a different question all together, but I was wondering what can be said in the case where we have a suspected basis but do not know for sure that it actually is a basis (such as in your answer here:link). For example if $\alpha$ is a root of $x^6+5x^4+6x^2+1$ and I suspect that $\{1,\alpha,\dots,\alpha^5\}$ is an basis for the ring of integers, then are there techniques (newton polygons?) which would allow me to confirm that this is infact a basis for the ring of integer in $\mathbb{Q}(\alpha)$? $\endgroup$ – Tristan Phillips Oct 2 '17 at 3:49
  • $\begingroup$ Other people may have a systematic method, but I just use the Method of Ingenious Devices. First I would calculate the discriminant of your suspected basis, see whether it’s divisible by any squares. If not, voilà. Then I might look at the associated polynomial $X^3+5X^2+6X+1$ and see what I could find out there. knowing the story there might help with the main story. I just thrash around, not knowing whether I’ll come up with anything. $\endgroup$ – Lubin Oct 2 '17 at 14:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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