Prove that the norm function $R \to \mathbb{Z}$ defined by $N(a+\sqrt db) = a^2 - db^2$ is a multiplicative integer-valued function

Question

Let $R = \mathbb{Z}[\sqrt d]$ where $d$ is a square-free integer. Prove that the norm function $R \to \mathbb{Z}$ defined by $N(a+\sqrt db) = a^2 - db^2$ is a multiplicative integer-valued function, with $N(\alpha) = 0 $ if only if $\alpha = 0$.

For this problem, should I simply show that for $\alpha, \beta \in R, N(\alpha\beta) = N(\alpha)N(\beta)$? Is that what the question is asking for?

you have to show that, and the fact that for every $\alpha\in R$, $N(\alpha)\in \Bbb{Z}$ (by the way, in your question you haven't defined $N$ yet), and also that $N(\alpha)=0$ if and only if $\alpha=0$. — peek-a-boo, Nov 27 '20 at 11:03
@Angelo but that's what the problem says. $N(a+\sqrt db) = a^2 - \sqrt db^2$. Is that a typo? Then there's only one definition of norm? I thought it was defined that way for purpose. — jun, Nov 27 '20 at 11:10

Angelo · Answer 1 · 2020-11-27T13:27:29.610

$\begin{align} &\text{Let }\;R=\mathbb{Z}[\sqrt d]\;\text{where }\;d\;\text{ is a square-free integer.}\\ &\text{Prove that the norm function }R\to\mathbb{Z}\text{ defined by }\\ &N(a+\sqrt db)=a^2-db^2\text{ is a multiplicative integer-}\\ &\text{valued function, with }N(\alpha)=0\text{ if only if }\alpha= 0\;. \end{align}$

For all $\alpha,\beta\in R$, it results that

$N(\alpha\beta)=N\big((a_1+\sqrt db_1)(a_2+\sqrt db_2)\big)=$

$=N\big(a_1a_2+db_1b_2+\sqrt d(a_1b_2+a_2b_1)\big)=$

$=(a_1a_2+db_1b_2)^2-d(a_1b_2+a_2b_1)^2=$

$=a_1^2a_2^2+d^2b_1^2b_2^2+\color{blue}{2da_1a_2b_1b_2}-da_1^2b_2^2-da_2^2b_1^2-\color{blue}{2da_1a_2b_1b_2}=$

$=\color{brown}{a_1^2a_2^2}+\color{green}{d^2b_1^2b_2^2}-\color{green}{da_1^2b_2^2}-\color{brown}{da_2^2b_1^2}=$

$=a_2^2(a_1^2-db_1^2)-db_2^2(a_1^2-db_1^2)=$

$=(a_1^2-db_1^2)(a_2^2-db_2^2)=$

$=N(a_1+\sqrt db_1)N(a_2+\sqrt db_2)=$

$=N(\alpha)N(\beta)\;.$

Moreover,

$N(\alpha)=0\iff N(a+\sqrt db)=0\iff a^2-db^2=0\;.$

Now we are going to prove by contradiction that $\;b=0\;.$

Let $\;p\;$ be a prime factor of $\;d\;,\;$ then

$d=pd_1\;$ and $\;p\nmid d_1\;$ (indeed $\;d\;$ is square-free),

$b=p^{n}b_1\;$ where $\;n\in\mathbb{N}\cup\left\{0\right\}\;$ and $\;p\nmid b_1\;.$

If $\;b\ne0\;,\;$ then $\;db^2=pd_1p^{2n}b_1^2=p^{2n+1}d_1b_1^2\;,\;$ consequently

$a^2=db^2=p^{2n+1}d_1b_1^2\;,$

but it is impossible for the unique-prime-factorization theorem, indeed $\;a^2\;$ has the prime factor $\;p\;$ with an even exponent, while $\;p^{2n+1}d_1b_1^2\;$ has the prime factor $\;p\;$ with an odd exponent.

Hence $\;b=0\;.$

Since $\;a^2=db^2\;$ and $\;b=0\;,\;$ it follows that $\;a=0\;.$

Consequently, $\;\alpha=a+\sqrt db=0\;.$

So far we have proved that $\;N(\alpha)=0\implies\alpha=0\;.$

Conversely,

if $\;\alpha=0\;,\;$ then $\;N(\alpha)=N(0+\sqrt d\cdot0)=0^2-d\cdot0^2=0\;.$

Hence, $\;\alpha=0\implies N(\alpha)=0\;.$

Prove that the norm function $R \to \mathbb{Z}$ defined by $N(a+\sqrt db) = a^2 - db^2$ is a multiplicative integer-valued function

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