Are $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ isomorphic?

Question

I would like to show whether $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ are isomorphic.

I tried noting that $\pm\sqrt2$ and $\pm i\sqrt2$ are the roots of $x^4-4=\left(x^2+2\right)\left(x^2-2\right)$ and that if $x^4-4$ were irreducible over $\mathbb Q$, then

$$\mathbb Q\left[x\right]/\left(x^4-4\right)\cong\mathbb Q\left(\sqrt2\right)\cong\mathbb Q\left(i\sqrt2\right),$$

i.e., $\sqrt2$ and $i\sqrt2$ would be algebraically indistinguishable. Therefore, $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ are not isomorphic since $x^4-4$ is not irreducible over $\mathbb Q$.

Is this approach valid? Is there a better one? Thanks in advance.

@Mathmo123, I assumed that it is irreducible to show the isomorphism, implying that there is no isomorphism since it is actually reducible, but I am not sure if this logic is valid. — wjmolina, Apr 01 '16 at 05:09
I'm not convinced by your approach. For example, $\pm\sqrt2$ are both roots of the reducible polynomial $x^4-4$. Your logic would imply that $\mathbb Q(\sqrt2)\not\cong\mathbb Q(-\sqrt 2)$. What you need to show is that $\sqrt{\pm2}$ are not the roots of any irreducible polynomial. You've only shown this for one polynomial. — Mathmo123, Apr 01 '16 at 15:31

score 13 · Answer 1 · answered Apr 01 '16 at 05:02

13

$\mathbb{Q}(i\sqrt{2})$ contains a number whose square is $-2$.

But every element of $\mathbb{Q}(\sqrt{2})$ is a real number, and therefore has a nonnegative square.

answered Apr 01 '16 at 05:02

Andrew Dudzik

30,074

very clear shot – jlandercy Apr 01 '16 at 12:23
Concise and nice [: – Evangeline A. K. McDowell Apr 01 '16 at 15:04

Evangeline A. K. McDowell · Accepted Answer · 2016-04-01T04:05:16.993

3

The answer is no. Note that for this isomorphism each rational number must be mapped to itself: let $\phi$ be the required field isomorphism. Then a property of field homomorphism is that $\phi(q)=q$ $\forall q\in\mathbb{Q}$. More explicitly, given a field $F$, a field homomorphism must satisfy $\phi(ax+b)=\phi(a)\phi(x)+\phi(b)$ for $a,x,b\in F$. Therefore, any rational number $m/n$ ends up \begin{equation}\phi\left(\frac{m}{n}\right)=\frac{\phi(m)}{\phi(n)}=\frac{m\phi(1)}{n\phi(1)}=\frac{m}{n}\end{equation}

Therefore, we are left with $\phi(\pm\sqrt{2})=\pm i\sqrt{2}$. For example, suppose $\phi(\sqrt{2})=i\sqrt{2}$. Then \begin{equation}\phi(\sqrt{2})^{2}=\phi(2)=2=(i\sqrt{2})^{2}=-2\end{equation} a contradiction. You can easily show that any combination of $\pm$ signs lead to similar contradiction. Thus the supposed isomorphism cannot exist.

The reason I show this explicitly is because this is actually a standard method of eliminating certain isomorphisms, which become important when you reach Galois theory even!

As for your method, I think you are right that if a polynomial is irreducible in one field and not in the other field, then they are not isomorphic. I think you need to explicitly prove that indeed it is the case, because I think you may have "begged the question": the statement that they are algebraically indistinguishable is precisely the question.

edited Apr 01 '16 at 04:05

answered Apr 01 '16 at 03:43

Evangeline A. K. McDowell

614

it makes me think to a question : is there a name for such fields (as $\mathbb{Q}$) whose automorphism group reduces to ${Id}$ ? – reuns Apr 01 '16 at 03:47
@user1952009 It should be just the Galois group of the field itself: $\text{Gal}(\mathbb{Q}/\mathbb{Q})={\text{Id}}$. – Evangeline A. K. McDowell Apr 01 '16 at 03:49
@eri, your answer makes perfect sense, but I am stuck trying to understand why "for this isomorphism each rational number must be mapped to itself." – wjmolina Apr 01 '16 at 03:59
@JosuéMolina I have added the reasoning above in the answer, hope that answers your question. – Evangeline A. K. McDowell Apr 01 '16 at 04:08
@eri, you are right; I forgot that $\phi(m)=\phi(1+\cdots+1)=\phi(1)+\cdots+\phi(1)=1+\cdots+1=m$. Thank you! – wjmolina Apr 01 '16 at 04:21
2

I don't see how you conclude that $\phi(\sqrt{2}) = \pm i\sqrt{2}$. – Andrew Dudzik Apr 01 '16 at 05:05
@Slade I think if you use $\phi(a+b\sqrt{2})=c+d(i\sqrt{2})$, it should imply $a=c, b=d$ by comparing coefficients and using field homomorphism property so it all boils down to that equation. – Evangeline A. K. McDowell Apr 01 '16 at 05:34

score 2 · Answer 3 · answered Apr 01 '16 at 19:40

This might be seen as just an elaboration of Slade's answer, but the idea is very useful.

Let $L,K$ be to field extensions of $\mathbb Q$. As eri argues, any homomorphism between $L$ and $K$ fixes $\mathbb Q$. Now, let $\varphi\colon L\to K$ be a homomorphism and $p\in\mathbb Q[x]$ a polynomial. Note that for any $\alpha\in L$ we have $\varphi(p(\alpha)) = p(\varphi(\alpha))$ precisely because $\varphi$ is ring homomorphism and fixes $\mathbb Q$. In particular, it means that if $\alpha$ is a root of polynomial $p$, $\varphi(\alpha)$ is a root of the same polynomial as well.

Now, if there were a homomorphism $\varphi\colon\mathbb Q[i\sqrt 2]\to\mathbb Q[\sqrt 2]$, it would map $i\sqrt 2$ to a root of $x^2 + 2$ (i.e. $\pm i\sqrt 2$), but those are not elements of $\mathbb Q[\sqrt 2]$. There is no $\varphi\colon\mathbb Q[\sqrt 2]\to\mathbb Q[i\sqrt 2]$ for the same reason.

score 0 · Answer 4 · edited Apr 13 '17 at 12:20

This can be proved in the same way as here. For this reason, one might add some different arguments here: for square-free $d$, the discriminant $D$ of $\mathbb{Q}(\sqrt{d})$ is an invariant of the field. We have $d=8$ for $\mathbb{Q}(\sqrt{2})$ and $D=-8$ for $\mathbb{Q}(\sqrt{-2})$.

Also the unit group of the ring of integers is different. In case of $K=\mathbb{Q}(\sqrt{2})$ we have $\mathcal{O}_K^{\times}\cong \pm 1\times \mathbb{Z}$, whereas for $K=\mathbb{Q}(\sqrt{-2})$ we have $\mathcal{O}_K^{\times}\cong \pm 1$ according to Dirchlet's unit theorem.

Are $\mathbb Q\left(\sqrt2\right)$ and $\mathbb Q\left(i\sqrt2\right)$ isomorphic?

4 Answers4