Fields isomorphic as vector spaces but not as fields

Question

Regarding this post: Why two extension fields are isomorphic as vector spaces but not fields?

It's clear (more or less) why it's an isomorphism between vector spaces, but am having difficulty understanding why there is no isomorphism between fields. The answers say that, to prove the fields are isomorphic, you can rename the elements of one field, and get the same field.

1. So, we can simply rename "2" to "3", and then will get that $\mathbb{Q}(\sqrt{3}) \cong \mathbb{Q}(\sqrt{2})$. Where's the flaw in that reasoning?

2. Why is it that, to have an isomorphism, it's necessary that a square of an element on one field has to also be in the other field?

3. This statement seems to be false: $\mathbb{Q}(\sqrt{3}) \cong \mathbb{Q}(i)$. But $i^2=-1\in \mathbb{Q}(\sqrt{3})$. Isn't that enough to prove that those two fields are isomorphic? Why not?

Note: If you could somehow use category theory, a la Paolo Aluffi, to explain, that would be great.

Answer 1

When we say "a field $K,$" that is really a shorthand for "a field, $\mathcal K=(K,0,1,+,\times)$ where $K$ is a set, $0,1$ are elements of the set, and $+,\times$ are binary operations on the set. We are actively confusing the difference between the set $K$ and the field $\mathcal K,$ because, usually, the standard operations are implied, and it is a lot of work to keep rewriting the long form.

Even above, we really shouldn't use the symbols $0,1,+,\times$ but rather symbols $0_{\mathcal K},1_{\mathcal K},\dots,$ because there is no reason to expect the $0$ of one field to be the $0$ of another field. But that gets noisy. So we usually just write $0,1,+,\times,$ unless really needed.

You can swap the meanings of the symbols $2$ and $3$ in the set $\mathbb Q,$ but then you will need to change the operations $+$ and $\times$ to make two isomorphic fields. So you would have $1\oplus 1=3, 2\otimes x=x\oplus x\oplus x.$ This is a different field from $\mathbb Q$ with the standard operations, although it is isomorphic.

In $\mathcal K_D=(\mathbb Q(\sqrt D),0,1,+,\times)$ the operations are supposed to agree on the rationals to the usual operations. So we have the property that for any $a\in\mathcal K_2,$ there is a $b\in\mathcal K_2$ such that $a\times a+a\times a=b\times b.$

Note that property doesn't anywhere use the symbol, $2.$

Now, this would have to be true in $\mathcal K_3$ if they were isomorphic. But it is not.

Answer 2

The stuff about renaming elements is just an informal intuition for how you should think about isomorphisms (which personally I don't find particularly helpful, but your mileage may vary). To say that $\mathbb Q(\sqrt{2})$ and $\mathbb Q(\sqrt{3})$ are isomorphic as fields means that there exist field homorphisms $\varphi:\mathbb Q(\sqrt{2})\to\mathbb Q(\sqrt{3})$ and $\psi:\mathbb Q(\sqrt{3})\to\mathbb Q(\sqrt{2})$ such that $\psi\circ \varphi$ is the identity on $\mathbb Q(\sqrt{2})$, and $\varphi\circ \psi$ is the identity on $\mathbb Q(\sqrt{3})$. To prove that these fields are or are not isomorphic, you need to use this definition (or use a theorem that is a consequence of this definition); there is no other way. Intuition might help you construct the proof, but it cannot act as a substitute for it.

Here is how I would reason this out. Suppose $\varphi:\mathbb Q(\sqrt{2})\to\mathbb Q(\sqrt{3})$ is a field homomorphism. Every element of $\mathbb Q(\sqrt{2})$ can be written as $a+b\sqrt2$ for some unique $a,b\in\mathbb Q$. From the definition of a field homomorphism, it follows that for all $a,b\in\mathbb Q$, $$ \varphi(a+b\sqrt2)=\varphi(a)+\varphi(b\sqrt2)=\varphi(a)+\varphi(b)\varphi(\sqrt2) \, . $$ If $A$ and $B$ are rings which both include $\mathbb Q$ as a subring, then a ring homomorphism $\varphi:A\to B$ must satisfy $\varphi(r)=r$ for all $r\in\mathbb Q$. (This is a general theorem from ring theory which hopefully you have seen before. If not, then feel free to ask about it.) It follows that $$ \varphi(a+b\sqrt{2})=a+b\varphi(\sqrt2) $$ for all $a,b\in\mathbb Q$. Thus $\varphi$ is determined by its value at $\sqrt2$. (That is, if $f,g:\mathbb Q(\sqrt 2)\to\mathbb Q(\sqrt 3)$ are ring homomorphisms such that $f(\sqrt2)=g(\sqrt2)$, then $f=g$.)

Write $\alpha=\varphi(\sqrt 2$). Then, $$ \alpha^2-2=\varphi(\sqrt2)^2-\varphi(2)=\varphi(0)=0 \, . $$ Suppose that $\alpha=c+d\sqrt3$, where $c,d\in\mathbb Q$. Then, $$ \alpha^2-2=(c^2+3d^2-2)+2cd\sqrt3=0 \, , $$ hence $c^2+3d^2=2$ and $cd=0$. But this pair of equations does not have rational solutions. Thus, there does not exist a field homomorphism $\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3)$; it follows a fortiori that $\mathbb Q(\sqrt2)$ and $\mathbb Q(\sqrt3)$ are not isomorphic.

Said differently, $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt3)$ are not isomorphic because $\mathbb Q(\sqrt 2)$ contains an element $x$ such that $x^2-2=0$. If there were an isomorphism $\varphi:\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3)$, then $\varphi(x)$ would satisfy $\varphi(x)^2-2=0$. But there is no element $y\in\mathbb Q(\sqrt3)$ such that $y^2-2=0$. Hence, there does not exist an isomorphism $\mathbb Q(\sqrt2)\to\mathbb Q(\sqrt3)$.