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Do units in $\mathbb{Z} [(1+\sqrt{d})/2] $ have norm $\pm1$? As I understood the norm of units of integer rings of $\mathbb{Q} (\sqrt{d} ) $ have norm $\pm 1$ and at the same time $N( \frac{a + b \sqrt{d}}{2} ) = \frac{1}{4} (a^2 - d b^2)$ in my understanding. Is that right?

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    $\begingroup$ Bear in mind that $\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right]$ is the ring of integers of $\mathbb{Q}(\sqrt{d})$ only when $d\equiv_41$, otherwise it is $\mathbb{Z}[\sqrt{d}]$. $\endgroup$ – j0equ1nn Nov 26 '17 at 23:36
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For every number field $K$, the units in the ring of integers $\mathcal O_K$ have norm $\pm 1$.

This follows from the multiplicativity of the norm. If $u\in \mathcal O_K$ is a unit, we have $uu^{-1}=1$, where $u^{-1} \in \mathcal O_K$. When we apply the norm to this equation, we get $N(u)N(u^{-1})=N(1)=1$. But the norm of an algebraic integer is always an integer, so $N(u)$ is a unit in $\Bbb Z$. The only units in $\Bbb Z$ are $\pm 1$.

Your computation of $N(\frac{a+b\sqrt{d}}{2})$ is correct.

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It's important to distinguish between $d$ positive and negative and $d \equiv 1 \bmod 4$ and $d \not\equiv 1 \bmod 4$. Maybe you do have those distinctions in mind, but didn't write them in your question here. If that's the case, I might be simple reiterating what you already know very well.

If $d \equiv 1 \bmod 4$, then the ring of integers $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$ includes numbers of the form $$\frac{a + b \sqrt{d}}{2}$$ with $a$ and $b$ both odd, in addition to numbers of the form $a + b \sqrt{d}$ with $a$ and $b$ of whatever parity combination. Then, with $a$ and $b$ both odd, we have $$N\left(\frac{a + b \sqrt{d}}{2}\right) = \frac{a^2 - b^2 d}{4}.$$ Since $a$ is odd, $a^2 \equiv 1 \bmod 4$, and since $b$ is odd, then $b^2 \equiv 1 \bmod 4$ as well. Then $a^2 - b^2 d$ is a multiple of $4$. Then that's why you care to find $N(x) = -4$ or $4$.

With two exceptions, if $d < 0$, then there are precisely only two units in $\mathcal{O}_{\mathbb{Q}(\sqrt{d})}$: $-1$ and $1$, and they both have norm $1$. In the two exceptions ($\mathbb{Z}[i]$ and $\mathbb{Z}[\omega]$), there are a few more units, but they all also have norm $1$. If $d > 0$, then there could be units of norm $-1$, but that's not guaranteed. Look up "fundamental unit" on this site for a more thorough discussion than I have time to write here.

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I only wish to supplement, not supplant Mathein's answer.

Every quadratic ring has units of norm 1. But some real rings have units of norm $-1$, and that's where things get interesting, in my opinion. It suffices to check the norm of the fundamental unit. Observe:

  • $N(\phi) = -1$
  • $N\left(\frac{3}{2} + \frac{\sqrt{13}}{2}\right) = -1$
  • $N(4 + \sqrt{17}) = -1$
  • $N\left(\frac{5}{2} + \frac{\sqrt{21}}{2}\right) = 1$... whoa!

It depends on $d$. We see that 5, 13 and 17 are primes congruent to 1 modulo 4, but 21 is composite, the product of primes congruent to 3 rather than 1 modulo 4. If you find that interesting, read Chapter 11 in Alaca & Williams's Introductory Algebraic Number Theory.

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