18
$\begingroup$

YACP mentions in a comment that:

There are examples of non-isomorphic fields $K$ and $L$ with $(K,+)\cong (L,+)$ and $(K^{\times} ,\cdot)\cong (L^{\times},\cdot)$

Can someone provide an example?

I have found the following thread. But there, the underlying structure is a ring. Hence, an answer to the current question is automatically an answer to the old question as well. EDIT: Sorry, the other question asks for finite commutative ring, and this won't be possible in the case of fields…

$\endgroup$
2
  • 7
    $\begingroup$ $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt{3})$? $\endgroup$
    – user27126
    Jan 1, 2014 at 21:06
  • 7
    $\begingroup$ @Sanchez This can be a good example as one knows that $\mathbb Z[\sqrt 2]$ and $\mathbb Z[\sqrt 3]$ are the rings of algebraic integers of $\mathbb Q(\sqrt 2)$ and $\mathbb Q(\sqrt 3)$, respectively and both are euclidean. But their groups of units are not so easy to handle. (Anyway, they are both isomorphic to $\{\pm 1\}\times\mathbb Z$ if I'm not wrong.) That's why I've preferred imaginary quadratic number fields. $\endgroup$
    – user26857
    Jan 1, 2014 at 22:00

1 Answer 1

Reset to default
19
$\begingroup$

An example can be the following: $\mathbb Q(i\sqrt 2)$ and $\mathbb Q(i\sqrt 7)$.

It's well known that these fields are not isomorphic: quadratic fields are isomorphic if and only if they are equal.

It is obvious that their additive groups are isomorphic.

With respect to the multiplicative groups, note that both are isomorphic to $\{\pm1\}\times\mathbb Z^{(\mathbb N)}$ (the last denotes a countable direct sum of copies of the group of integers).

$\endgroup$
1
  • 1
    $\begingroup$ You could elaborate: why 2 and 7? What is wrong with 3 or 5? - (It's because we know the units and have UFDs, but mentioning that might be fine) $\endgroup$
    – Hagen von Eitzen
    Apr 24, 2016 at 7:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .