Ring Extension: Mapping: $ \mathbb Q[\sqrt d] \rightarrow \mathbb Q$

Question

Show that the Norm: $\mathbb Q[\sqrt d] \rightarrow \mathbb Q, (r+s\sqrt d) (r-s\sqrt d) = r^2 - ds^2$ is multiplicative, i.d. $N(xy) = N(x)N(y)$

How to show it without computing?

(I tried to do it by computing and went astray. If you know how to compute something like this compactly, please let me know.)

The following is known/given.

$d \in \mathbb Z, \sqrt d \in \mathbb C$ is a zero of $x^2 - d$

$ \mathbb Q[\sqrt d] = \{r + s\sqrt d | r,s \in \mathbb Q \} $

$ \mathbb Z[\sqrt d] = \{m + n\sqrt d | m,n \in \mathbb Z \} $

$\mathbb Q[\sqrt d]$ and $\mathbb Z[\sqrt d]$ are integrity domains, subrings of $\mathbb C$, $\mathbb Q[\sqrt d]$ is a field...

What if it could be shown that $\mathbb Q[\sqrt d] \cong Quot(\mathbb Z[\sqrt d])$ and then use the fact that to say that $\mathbb Q[\sqrt d] \rightarrow \mathbb Q$ is a ring homomorphisms or isomorphism.

How to show $\mathbb Q[\sqrt d] \cong Quot(\mathbb Z[\sqrt d])$?

Thank you very much in advance!

Answer 1

You indicate you would like to know about directly computing this...

Let $x = s+t\sqrt{d}$ and $y = u+v\sqrt{d}$ be elements of $\Bbb{Q}[\sqrt{d}]$. Then we compute \begin{align*} xy &= su + dtv + (sv+tu)\sqrt{d} \text{,} \\ N(xy) &= (su + dtv)^2 - d((sv+tu))^2 \\ &= s^2u^2 + 2dstuv + d^2t^2v^2 - ds^2v^2 - 2dstuv - dt^2u^2 \\ &= s^2u^2 + d^2t^2v^2 - ds^2v^2 - dt^2u^2 \text{, and} \\ N(x) N(y) &= (s^2 - dt^2)(u^2 - dv^2) \\ &= s^2u^2 - ds^2v^2 - dt^2u^2 + d^2t^2v^2 \text{,} \end{align*} and with a trifling rearrangement, we see $N(xy) = N(x) N(y)$.

Answer 2

Consider $\mathbb{Q}[\sqrt{d}]$ as $\mathbb{Q}$-vector space. For any $a \in \mathbb{Q}[\sqrt{d}]$ consider map $M_a: \mathbb{Q}[\sqrt{d}] \to \mathbb{Q}[\sqrt{d}]$ defined by $M_a(x) = ax$. This is $\mathbb{Q}$-linear, and easy calculation shows that for $a = r + s \sqrt{d}$, $\det(M_a) = N(a) = r^2 - ds^2$. Now, clearly $M_a \circ M_b = M_{ab}$, so $N(ab) = \det(M_{ab}) = \det(M_a \circ M_b) = \det(M_a)\det(M_b) = N(a)N(b)$.

To elaborate why $\det(M_a) = r^2 - ds^2$: note that $a$ satisfies a polynomial equation $a^2 - 2 r a + (r^2 - d s^2) = 0$ -- in fact, this is miniml polynomial of $a$ over $\mathbb{Q}$. It follows that $M_a^2 - 2r M_a + (r^2 - d s^2)$ is a zero linear map, so $x^2 - 2rx + (r^2 - d s^2)$ is a characteristic polynomial of $M_a$. On the other hand, we know that for 2x2 matrix $M$, its characteristic polynomial is $x^2 - 2 \mathrm{tr}(M) + \det M$.

Answer 3

You ask how to show $\Bbb{Q}(\sqrt{d}) \cong \mathrm{Frac}(\Bbb{Z}[\sqrt{d}])$, where I have altered your notation to indicate that the object on the left is a field extension (which is necessary, because the object on the right contains an element that it is not unreasonable to write as $1/\sqrt{d}$ (and which it is rigorous to write $\frac{1 + 0\sqrt{d}}{0+1\sqrt{d}}$).

Observe that $(\sqrt{d})^2 \in \Bbb{Z}$, so an arbitrary element of $\Bbb{Q}(\sqrt{d})$ is $a + b \sqrt{d} + c/\sqrt{d}$ and an arbitrary element of $\mathrm{Frac}(\Bbb{Z}[\sqrt{d}])$ is $\frac{s+t\sqrt{d}}{u+v\sqrt{d}}$. We find \begin{align*} a+b\sqrt{d} + c/\sqrt{d} &= \frac{(bd + c) + a \sqrt{d}}{0+1\sqrt{d}} \text{ and} \\ \frac{s+t\sqrt{d}}{u+v\sqrt{d}} &= \frac{s+t\sqrt{d}}{u+v\sqrt{d}} \frac{u-v\sqrt{d}}{u-v\sqrt{d}} \\ &= \frac{su - dtv + (tu-sv) \sqrt{d}}{u^2-d v^2} \\ &= \frac{su - dtv}{u^2-d v^2} + \frac{(tu-sv) }{u^2-d v^2} \sqrt{d} + 0/\sqrt{d} \text{.} \end{align*} If the maps $$ f:\Bbb{Q}(\sqrt{d}) \rightarrow \mathrm{Frac}(\Bbb{Z}[\sqrt{d}]): a+b\sqrt{d} + c/\sqrt{d} \mapsto \frac{(bd + c) + a \sqrt{d}}{0+1\sqrt{d}} $$ and $$ g: \mathrm{Frac}(\Bbb{Z}[\sqrt{d}]) \rightarrow \Bbb{Q}(\sqrt{d}): \frac{s+t\sqrt{d}}{u+v\sqrt{d}} \mapsto \frac{su - dtv}{u^2-d v^2} + \frac{(tu-sv) }{u^2-d v^2} \sqrt{d} + 0/\sqrt{d} $$ are inverses, we have the isomorphism you want. So we compute \begin{align*} f\left(\frac{su - dtv}{u^2-d v^2} + \frac{(tu-sv) }{u^2-d v^2} \sqrt{d} + 0/\sqrt{d} \right) &= \frac{\left( \frac{(tu-sv) }{u^2-d v^2}d + 0 \right) + \frac{su - dtv}{u^2-d v^2} \sqrt{d}}{\sqrt{d}} \\ &= \frac{su - t\sqrt{d} \, \sqrt{d}v + (tu-sv)\sqrt{d}}{u^2- (v\sqrt{d})^2} \\ &= \frac{s + t\sqrt{d} }{u + v\sqrt{d}} \text{ and} \\ g \left( \frac{(bd + c) + a \sqrt{d}}{0+1\sqrt{d}} \right) &= \frac{(bd + c)0 - da1}{0^2-d 1^2} + \frac{(a0-(bd + c)1) }{0^2-d 1^2} \sqrt{d} + 0/\sqrt{d} \\ &= a+b\sqrt{d} + c/\sqrt{d} \text{.} \end{align*} So these maps are inverses and $\Bbb{Q}(\sqrt{d}) \cong \mathrm{Frac}(\Bbb{Z}[\sqrt{d}])$ as fields.