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We know that $(\mathbb{Z} [\sqrt{2}],+,\cdot)$ is an integral domain.

Someone can prove it easily if he says that is a subring of $(\mathbb{R} ,+,\cdot)$ .

Can we find a different proof, more analytical? How can we show that $$\forall x,y\in \mathbb{Z} [\sqrt{2}], x\ne 0, y\ne 0\implies xy\neq 0$$

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3 Answers 3

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In these cases, considering the conjugate can help. The conjugate of $a+b\sqrt{2}$ is $a-b\sqrt{2}$. Now, if $(a+b\sqrt{2})(c+d\sqrt{2})=0$, also $$ (a+b\sqrt{2})(a-b\sqrt{2})(c+d\sqrt{2})(c-d\sqrt{2})=0 $$ and therefore $$ (a^2-2b^2)(c^2-2d^2)=0 $$ Since the integers form a domain, we conclude $a^2-2b^2=0$ or $c^2-2d^2=0$. The irrationality of $\sqrt{2}$ tells us that either $a=b=0$ or $c=d=0$.

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  • $\begingroup$ Nice proof, thank you. So we multiply with the conjugate, and we say that $$ (a/b)=2^{(1/2)}$$, false. Right? $\endgroup$
    – Chris
    Apr 29, 2016 at 23:08
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    $\begingroup$ @Chris Suppose $a^2-2b^2=0$. If $b=0$, then $a=0$; if $b\ne0$, then $\frac{a^2}{b^2}=2$. The latter case is impossible. $\endgroup$
    – egreg
    Apr 29, 2016 at 23:09
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The main proof you mention is the easiest and the best since it generalizes very well. If you really want something more intrinsic, note

$$(a+b\sqrt 2)(c+d\sqrt 2)=0\implies (a^2-2b^2)(c^2-2d^2)=0$$

But then if so, either $a^2=2b^2$ or $c^2=2d^2$, WLOG assume the former.

Then $a$ is even, but then if the prime factorization of $a$ is $2^kp_1^{e_1}\ldots p_r^{e_r}$ we have

$$a^2=2^{2k}p_1^{2e_1}\ldots p_r^{2e_r}$$

So the exponent of $2$ in $a$ is even, however since $b^2$ has an even power of $2$, $2b^2$ has an odd power, a contradiction, unless that power is infinity, but the only number infinitely divisible by $2$ is $0$, so $a=b=0$ holds

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  • $\begingroup$ Once we've eliminated the possibility that $b = 0$ (which is almost immediate), we can conclude that if $a^2 = 2 b^2$ then $\sqrt{2}$ is rational, and likewise for $c^2 = 2 d^2$. $\endgroup$ Apr 29, 2016 at 22:44
  • $\begingroup$ @Travis yes, but why bother? That's a corollary of this approach in general, may as well avoid saying it for a direct proof that takes place completely within the realm of $\Bbb Z$. $\endgroup$ Apr 30, 2016 at 18:00
  • $\begingroup$ I remarked it simply because it cuts the length of the (already efficient) presentation in half. $\endgroup$ Apr 30, 2016 at 19:34
  • $\begingroup$ @Travis I agree, assuredly, I chose this presentation for pedagogical reasons more than anything. I always find appealing to lower-level theorems is best when it doesn't make the proof unwieldy. $\endgroup$ Apr 30, 2016 at 19:43
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    $\begingroup$ @Travis no worries, that came across clearly, I was just responding to explain my thought process. Thanks very much for your comments. Cheers! $\endgroup$ Apr 30, 2016 at 19:54
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Hint Suppose we have $$(a + b \sqrt{2})(c + d \sqrt{2}) = 0$$ from some $a, b, c, d \in \Bbb Z$. Expanding gives $$(ac + 2 bd) + (ad + bc) \sqrt{2} = 0,$$ and since $\sqrt{2}$ is irrational, the coefficients must vanish separately $$ac + 2bd = ad + bc = 0 .$$ Substituting $0$ for any of the four parameters gives quickly that $a = b = c = d = 0$, so we may assume that none is zero. Then, it follows from the second equation that $c = -\lambda a$ and $d = \lambda b$ for some $\lambda \in \Bbb Q - \{ 0 \}$.

Substituting gives $$0 = a c + 2 b d = a(-\lambda a) + 2 b (\lambda b) = \lambda (-a^2 + 2b^2) .$$ Clearing $\lambda$ and rearranging gives $\left(\frac{a}{b}\right)^2 = 2$, but this contradicts the irrationality of $\sqrt{2}$.

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    $\begingroup$ I think you mean "this implies $\sqrt{2}$ is rational" $\endgroup$ Apr 29, 2016 at 22:57
  • $\begingroup$ Thank you for your answer. How can we say with certainty that exists $$λ\in \mathbb{Q^*} : c=-λa, d=λb$$ $\endgroup$
    – Chris
    Apr 29, 2016 at 23:01
  • $\begingroup$ Thanks, @StellaBiderman, you're right of course. I've fixed the statement. $\endgroup$ Apr 29, 2016 at 23:20
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    $\begingroup$ We can rearrange $ad + bc = 0$ and write $\lambda := -\frac{c}{a} = \frac{d}{b}$. $\endgroup$ Apr 29, 2016 at 23:28
  • $\begingroup$ You're welcome, I hope you found it useful. $\endgroup$ Apr 29, 2016 at 23:51

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