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Inspired by this question: Need the norm of positive number be positive? The "Best Answer" for that question gives as examples six cubic rings that will perhaps be forever beyond my ability to comprehend.

Given $d \in \mathbb{Z}$, cubefree, not a perfect square, consider the ring $\mathcal{O}$ of algebraic integers of $\mathbb{Q}(\root 3 \of d)$. Is "simple" the appropriate terminology for such a ring? If it's not, you'll let me know (and how!).

A number $x \in \mathcal{O}$ is real, and if $x \neq 0$, we can say $x < 0$ or $x > 0$. Likewise, $N(x) < 0$ or $N(x) > 0$. It seems like $\textrm{sgn}(x) = \textrm{sgn}(N(x))$ always.

From this other question What is the norm of a number in a cubic integer ring? I see that $N(a + b \root 3 \of d + c \root 3 \of {d^2}) = a^3 + b^3 d + c^3 d^2 - 3abcd$ ($a, b, c$ are integers, or at least rational, depending on $d$).

I have played around with some primes in $\mathbb{Z}[\root 3 \of 2]$, $\mathbb{Z}[\root 3 \of 3]$ and $\mathbb{Z}[\root 3 \of 5]$, reasoning that if $x$ is a product of two primes exhibiting $\textrm{sgn}(x) \neq \textrm{sgn}(N(x))$, only one of the prime factors also shows this sign disagreement. (I would settle for irreducible numbers, but I've so far only looked at UFDs.)

On further reflection, I could be wrong on that last point, and in any case it might be easier to find a "composite" number exhibiting the sign disagreement.

My main question here is: in a cubic ring as described above, for which the norm of a number can be readily calculated by plugging $a, b, c$ in the formula given above, does the sign of a number always match the sign of the norm?

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A cubic field of the form $K=\mathbf Q(\sqrt [3]d)$ is usually called "pure". Its ring of integers is generally not $\mathbf Z[\sqrt [3]d]$, there are two cases depending on whether $d \equiv \pm 1$ mod $9$ or not (see e.g. Marcus' "Number Fields", end of chapter 2). But since your question is simply about signs, why bother with integers, we just have to work in the field $K$ itself. Since $3=1+2$, the embeddings of $K$ into $\mathbf C$ are exactly the three which are determined by $Id: \sqrt [3]d \to \sqrt [3]d$ , $\sigma: \sqrt [3]d \to j \sqrt [3]d$, $\tau: \sqrt [3]d \to j^2 \sqrt [3]d$, where $j$ denotes a primitive 3rd root of unity. The norm of any $x \in K$ is given by $N(x)=x.\sigma (x) .\tau (x)$. But by definition, $\sigma (x)$ and $\tau (x)$ are complex conjugates, hence their product is just the square of the module of $ \sigma (x)$, which shows that $x$ and $N(x)$ have the same sign.

Note that the above argument carries over to arbitrary cubic fields which are not totally real, plainly because $3=1+2$. Note also that it is not fundamentally different from @Will Jagy, except that we avoided to express explicitly the quotient $N(x)/x$, which allowed us not to appeal to the theory of ternary quadratic forms.

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  • $\begingroup$ Looks good, but, if we assume that KCd's answer to one of the linked questions is correct, wouldn't that mean your second paragraph is at least partially incorrect? $\endgroup$ – Robert Soupe Mar 2 '17 at 19:51
  • $\begingroup$ To add on to @Robert's comment above, this argument carries over to cubic fields which are not totally real. The statement is false in totally real cubic fields, as KCd's answer shows. $\endgroup$ – Brandon Carter Mar 2 '17 at 20:10
  • $\begingroup$ You are right, the argument does not carry on to totally real fields. I edit the beginning of my second paragraph. $\endgroup$ – nguyen quang do Mar 2 '17 at 20:25
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I'm assuming $d > 0$ and $\omega^3 = d,$ so that also $d \omega = \omega^4$ for example.

$$ \left( a + b \omega + c \omega^2 \right) \left( a^2 + b^2 \omega^2 + c^2 d \omega - bcd - ca \omega^2 - ab \omega \right) = a^3 + b^3 d + c^3 d^2 - 3abcd $$ so we just need to show, using $\omega$ real, that the quadratic form is positive. Better to write $$ a^2 x^2 + b^2 y^2 + c^2 z^2 -bcyz - ca zx - ab xy. $$ We do find $a^2$ and binary $a^2 x^2 -abxy + b^2 y^2$ positive. The discriminant (Watson and Lehman) is $$ \Delta = 4 a^2 b^2 c^2 +(bccaab) - a^2b^2 c^2 - b^2c^2 a^2 - c^2a^2 b^2, $$ $$ \Delta = 2 a^2 b^2 c^2.$$ So this is a positive definite form, take $$ x=1, \; \; y = \omega, \; \; z = \omega^2 $$ in $$ a^2 x^2 + b^2 y^2 + c^2 z^2 -bcyz - ca zx - ab xy $$ to show $$ a^2 + b^2 \omega^2 + c^2 d \omega - bcd - ca \omega^2 - ab \omega > 0. $$

I see how to factor the quadratic form. We will need $$ \sigma \neq 1, \; \; \sigma^3 = 1, $$ so that $$ \sigma^2 + \sigma = -1, $$ $$ \sigma^2 + \sigma^4 = -1. $$ Then $$ \left( a + b \omega \sigma + c \omega^2 \sigma^2 \right) \left( a + b \omega \sigma^2 + c \omega^2 \sigma \right) = a^2 + b^2 \omega^2 + c^2 d \omega - bcd - ca \omega^2 - ab \omega$$ Since we also have $$ \sigma^2 = \bar{\sigma},$$ our quadratic form is the product of complex conjugates and positive for that reason.

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