2
$\begingroup$

Let $d$ be an integer that is not the square of an integer, and defined the subfield of $\mathbb{C}$ as

$\mathbb{Q}( \sqrt{d} ) = \{ a + b \sqrt{d} \ | \ a,b \in \mathbb{Q} \}$.

I need to prove that

$\mathbb{Q}(\sqrt{d}) \cong \mathbb{Q}[t]/(t^2 - d)$

I tried to prove it the following way. I defined a ring homomorphism $\phi: \mathbb{Q}[t] \to \mathbb{Q}( \sqrt{d} )$ by setting $\phi(f(t)) = f( \sqrt{d} )$. Now all that if left there to prove is that $\ker \psi = (t^2 - d)$

Obviously, $(t^2 - d) \subseteq \ker \psi$.

I also tried to use the division with remained in $\mathbb{C}[t]$: for a $f(t) \in \mathbb{Q}[t]$, $f( \sqrt{d} ) = 0 \Leftrightarrow f(t) \in (t - \sqrt{d}) \subseteq \mathbb{C}[t]$, so in $\mathbb{C}[t]$ we have

$(t^2 - d) \subseteq \ker \psi \subseteq (t - \sqrt{d})$

$\endgroup$
  • $\begingroup$ Does "$d$" stand for "$2$" or does "$2$" stand for "$d$" ? :) $\endgroup$ – user228113 Aug 31 '16 at 15:29
  • 1
    $\begingroup$ Try to use what you said, that is, the division algorithm, but in $\mathbb Q[t]$. Then $f(t)=(x^2-d)g(t)+a+bt$ and then $a+b\sqrt d=0$. Conclusion? $\endgroup$ – user26857 Aug 31 '16 at 15:34
  • 1
    $\begingroup$ $t^2-d$ is irreducible because it has not roots. In particular, the ideal $(t^2-d)$ is a maximal ideal. By maximality, $\ker \psi = (t^2-d)$. $\endgroup$ – Crostul Aug 31 '16 at 15:42
  • $\begingroup$ @G.Sassatelli It's $d$. I edited the question. $\endgroup$ – Jxt921 Sep 1 '16 at 11:23
  • $\begingroup$ @user26857 What a wonderful idea! Silly that I hadn't thought of this. Of course, since $d \in \mathbb{Z}, d \neq k^2, k \in \mathbb{Z}$, then if $a,b \neq 0$, then $d = {a^2}/{b^2}$, so $d$ would be either the square of integer $a/b$ or non integer, which is not true. Then $a = b = 0$, and $f(t) \in (t^2 - d)$. $\endgroup$ – Jxt921 Sep 1 '16 at 11:33
0
$\begingroup$

Consider $f:Q[t]\rightarrow Q(\sqrt d)$ defined by $f(t)=\sqrt d$, $(t^2-d)\subset Ker f$ and $f$ is surjective, it induces a morphism $\bar f:Q[t]/(t^2-d)\rightarrow Q(\sqrt d)$, since $d$ is not a square, $(t^2-d)$ is a maximal ideal and $Q[t]/(t^2-d)$ is a field, this implies that $Ker \bar f=0$ or $Q[t]/(t^2-d)$, it is $0$ since $\bar f$ is surjective.

$\endgroup$
  • $\begingroup$ "since $d$ is not a square, $(t^2-d)$ is a maximal ideal" this is exactly what OP had reduced the problem to, and you just asserted it without proof. It's not clear that you read the question (as opposed to just the title). $\endgroup$ – Najib Idrissi Sep 1 '16 at 11:44
  • $\begingroup$ I read the question (perhaps not as well as you), but I don't think that OP has reduced its question to that fact indeed he is using division remaining. OP want to determine the kernel of $\phi:Q[t]\rightarrow Q(\sqrt d)$. If you need more details to understand: the induced map $Q[t]/(t^2-d)\rightarrow Q(\sqrt d)$ is a surjective morphism of fields, so an isomorphism. The fact that $Q[t]/(t^2-d)$ is maxima results from the fact $t^2-d$ does not have roots and this implies that $Q[t]/(t^2-d)$ is a field. $\endgroup$ – Tsemo Aristide Sep 1 '16 at 11:54

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.