# Why two extension fields are isomorphic as vector spaces but not fields?

I understand that $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$ are isomorphic as vector spaces but not as fields. However, I do not understand why that is true. What is happening when they are vector spaces so that they are now isomorphic?

## 2 Answers

Fields have arithmetic properties which have to be preserved under isomorphism. So for example, if $\mathbb Q(\sqrt{3})$ and $\mathbb Q(\sqrt{2})$ are isomorphic, then since $3$ is a square in the first field it has to be a square also in the second, and you can easily check that this is not possible. On the other hand the $\mathbb Q$-vector space structure "sees" only the $\mathbb Q$-dimension, which is $2$ in both cases and if you know elementary linear algebra this tells you that those two fields are isomorphic as vector spaces .

Being isomorphic as fields means that you have two different fields whose sum and product tables are essentially the same (you only have different "names").

You can see that every isomorphism of your fields must fix $\mathbb{Q}$, so 3 has to appear in both of the multiplication tables as a square (which is clearly not possible).

On the other hand, when dealing with vector spaces, you just have additive groups and a scalar product. The additive tables of your structures are isomorphic (it would be the same for each $\mathbb{Q}(\sqrt{d}))$ so you actually have a vector space isomorphism.