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Prove that $\Bbb Z [\sqrt 2]\ncong \Bbb Z [\sqrt 3].$ ($\cong$ denotes an isomorphic relation)

My solution goes like this:

(Here, $I(\phi)$ denotes the kernel of a homomorphism $\phi.$)

We have $$\Bbb Z [\sqrt 2]=\{a+b\sqrt 2:a,b\in \Bbb Z\}$$ and $\Bbb Z[\sqrt 3]=\{a+b\sqrt 3:a,b\in\Bbb Z\}.$

Let $\phi:\Bbb Z[\sqrt 2]\to \Bbb Z [\sqrt 3]$ be an isomorphism.

We have, $\phi(1)^2=\phi(1.1)=\phi(1)\implies \phi(1)=1$ or $\phi(1)=0.$

But $\phi(1)=1$ as $\phi$ is onto.

Let $a\in \Bbb Z^+$

We note that, $\phi(a)=a\phi(1)=a.$

Again, let $a\in\Bbb Z^-$

We observe that, $\phi(a)=\phi(-1-1-\cdots-1)=|a|\phi(-1)=-|a|\phi(1)=a\phi(1)=a.$

Thus, $\phi(x)=x, \forall x\in\Bbb Z$ as $\phi(0)=0.$

Therefore, $\phi(a+b\sqrt 2)=a+b\phi(\sqrt 2),\forall a,b\in\Bbb Z.$

Now, $\phi(2)=\phi((\sqrt 2)^2)=\phi(\sqrt 2)^2\implies \phi(\sqrt 2)=\pm\sqrt {\phi(2)}=\pm \sqrt 2.$

Let $\phi(\sqrt 2)=-\sqrt 2$ then, $$\phi(a+\sqrt 2b)=a-\sqrt 2b\implies \phi(\sqrt 2)=\phi(\sqrt 2+\sqrt 2.0)=\sqrt 2, $$ a contradiction.

Thus, $\phi(\sqrt 2)=\sqrt 2$ and so, $\phi(a+b\sqrt 2)=a+b\sqrt 2.$

But then, $\phi$ is not onto.

For if, $ a,b\in\Bbb Z$ such that $\phi(a+b\sqrt 2)=a+b\sqrt 2=1-\sqrt 3$ then, $a$ and $b$ have no solution in $\Bbb Z.$

So, $\phi$ cannot be an isomorphism. This means $\Bbb Z [\sqrt 2]\ncong \Bbb Z [\sqrt 3].$

I feep the part where I wrote,

For if, $ a,b\in\Bbb Z$ such that $\phi(a+b\sqrt 2)=a+b\sqrt 2=1-\sqrt 3$ then, $a$ and $b$ have no solution in $\Bbb Z$

needs a little bit more justification. I can't find a way to rigorously show that indeed, no solution of $a,b\in\Bbb Z$ exists. Is there any way to preswnt this statement in a more justified way? Any help regarding this will be greatly appreciated. Lastly, if there is something wrong with my solution, please do point it out.

Thomas Finley
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  • try to square the expression and see if you get a contradiction – Sine of the Time Dec 17 '23 at 14:19
  • @SineoftheTime Squaring the expression gives us, $a^2+2b^2+2\sqrt ab=1+3-2\sqrt 3=4-2\sqrt 3.$ But I don't notice any contradictions. – Thomas Finley Dec 17 '23 at 14:26
  • The two rings both have the integer $1$ as the multiplication identity. In the first ring, exists $x$ s.t. $x^2=2\cdot 1$, while the second does not. – Asigan Dec 17 '23 at 14:27
  • $3$ is irreducible in $\Bbb Z [\sqrt 2]$ but not in $\Bbb Z [\sqrt 3]$. – lhf Dec 17 '23 at 14:38
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    You overcomplicate it. It is sufficient to show that there is no element in $\mathbb{Z}[\sqrt{3}]$ whose square is equal to $2$. If you know how a general element in this ring looks like, then it should be easy. – Mark Dec 17 '23 at 14:38
  • For $,a,b,c\in\Bbb Q!:\ a+b\sqrt 2+c\sqrt 3 = 0 \Rightarrow a,b,c=0,$ follows immediately from the well-known facts that the square roots of $:!2,:!3:!$ and $:!2\cdot 3:!$ are all irrational, see the Lemma in the linked dupe. The common direct proofs are special cases of the proof of that Lemma - which is the key idea (inductive step) needed to generalize the above implication to any number of square roots. – Bill Dubuque Dec 17 '23 at 19:27
  • Your argument that $\phi(\sqrt{2})\ne -\sqrt{2}$ is wrong. – Allen Bell Dec 18 '23 at 05:24
  • @AllenBell But why? – Thomas Finley Dec 18 '23 at 05:43
  • @Thomas Finley You let $a=\sqrt{2}$, but $a$ has to be an integer. – Allen Bell Dec 19 '23 at 22:59

2 Answers2

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Hint : If we assume that there exists an isomorphism between $\mathbb{Z}[\sqrt2]$ and $\mathbb{Z}[\sqrt 3]$ and call this map $\phi $ as you did. Then there exists $x,y \in \mathbb{Z}$ such that $\phi(\sqrt2)=x+y\sqrt3$. 2 is a square in $\mathbb{Z}[\sqrt2]$, so it must also be a square in $\mathbb{Z}\sqrt3$. But then we have : $$2=\phi(2)=\phi(\sqrt2^2)=\phi(\sqrt2)^2=(x+y\sqrt3)^2=(x^2+3y^2)+2xy\sqrt3$$

What are the possible solutions of x and y?

muhammed gunes
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Expanding my comment, you can try to square both sides to see if you arrive to a contradiction: $$a+b\sqrt 2=1-\sqrt 3\implies b\sqrt2+\sqrt3=1-a\implies2b^2+3+2b\sqrt6=1-2a+a^2$$ Now rewrite the expression as follows $2b\sqrt6=-2-2b^2-2a+a^2$, and consider two different cases: $2b\neq 0$ and $2b=0$.

Sine of the Time
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