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Let $d\in \Bbb{Z}$ be a square free in, I need to find an isomorphism between $\Bbb{Z}[\sqrt{d}]$ and a quotient of $\Bbb{Z}[X].$ I Would guess it's $$\Bbb{Z}[\sqrt{d}]\simeq\Bbb{Z}[X]/\langle X^2-d\rangle.$$ But I need to prove it. So, as $X^2+1$ is monic, I can perform the euclidean algorithm: all polynomial $\Bbb{Z}[X]$ can be written $$P=(X^2-d)Q+(aX+b)$$ for $Q\in\Bbb{Z}[X].$ We have $P(\sqrt{d})=0$ if and only if $a+\sqrt{d}b=0.$

If $d<0$ then $\sqrt{d}=i\sqrt{-d}$ so $a=b=0.$

If $d>0$ I am not sure how can I prove that $a=b=0$, I can do it for some special ("easy") case.

More than a solution, I need to understand what's going on to get the idea of the proof. I can conclude, but

JeSuis
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    The title is misleading. You may want to say "an isomorphism between $\Bbb Z[\sqrt d]$ and a quotient of $\Bbb Z[X]$". – Pedro Nov 26 '15 at 16:05
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    Hint: Suppose that $a+\sqrt{d}b=0$. Then $\sqrt{d}=\frac{-a}{b}$ (and we can assume $a$ and $b$ are relatively prime). So $d=\frac{a^2}{b^2}$. Since $d$ is an integer, it must be that $b^2=1$ and so $d$ is not square free. – Michael Burr Nov 26 '15 at 16:07
  • @PedroTamaroff Right, fixed. – JeSuis Nov 26 '15 at 16:07
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    I would also suggest that you be a little more formal and start with a ring homomorphism $\mathbb{Z}[X]\to \mathbb{Z}[\sqrt{d}]$ constructed so that it is surjective. Then determine its kernel. Your method for doing that is correct, completed using Michael Burr's comment above. – rogerl Nov 26 '15 at 16:09
  • @rogerl I Know the formal method but I am trying this one today ;). – JeSuis Nov 26 '15 at 16:10
  • @MichaelBurr pff, I did this too but I didn't think about $d$ being an integer... If you can post this (more than a ) hint, it will be great. – JeSuis Nov 26 '15 at 16:13

2 Answers2

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Give the explicit application $$\phi: \Bbb{Z}[x]\to\Bbb{Z}[\sqrt{d}]$$ $$\; \; x \mapsto \sqrt{d}$$

Now check that, if you quotient left side by $x^2 - d$, the application is well-defined on the quotient (that is, $\phi(x^2 -d) = 0$ ) This gives you one side.

For the other one you have to give an explicit inverse, which is pretty straightforward (let me know if you can't do it!)

AnalysisStudent0414
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  • I got it using onto function and injectivity, but it's the same as proving $a+b\sqrt{d}$ implies $a=b=0$ isn't it? Or am I missing what you are doing? – JeSuis Nov 26 '15 at 16:41
  • If you mean $a+b\sqrt{d}=0$ then yes – AnalysisStudent0414 Nov 26 '15 at 16:50
  • So what is difference between my try ? I always need to prove $a=b=0$, or perhaps I did not understand your method. – JeSuis Nov 26 '15 at 16:52
  • Well, suppose $a\neq 0 \neq b$, and $a$ and $b$ such that $gcd(a,b)=1$, then $\sqrt{d} = \frac{a}{b}$ so $d=\frac{a^2}{b^2}$. But the latter can only be an integer if $b^2 =1$, which implies $d=a^2$ which is absurd for your initial hypothesis. – AnalysisStudent0414 Nov 26 '15 at 17:01
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@AnalysisStudent0414 's answer is much cleaner and more direct, but you can follow the idea in the question as follows:

Suppose that $a+b\sqrt{d}=$ (we may safely assume that $a$ and $b$ are relatively prime). Then, we know that $\sqrt{d}=-\frac{a}{b}$. Then, $d=\frac{a^2}{b^2}$ by squaring both sides. Since $d$ is an integer and $a$ and $b$ are relatively prime, it must be that $b^2=1$ (since otherwise the RHS is a rational which is not an integer), so $d=a^2$, which is impossible because then $d$ is not square-free.

Michael Burr
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  • In AnalysisStudento414's answer we also need to prove that $a=b=0$ so it does not answer my question, because I was stuck at this part of the proof. – JeSuis Nov 26 '15 at 17:38