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(While reading about classification of simple groups, I was wondering if there could be something like simple fields or something... And this thought came in mind)

When we talk about field in abstract algebra of linear algebra,

It really feels like the second binary operation-multiplication is very much dependent the first binary operation-addition.

If there is no counter example to it then it implies that study of field is totally dependent on the group theory.

Also every finite dimensional vector space over $F$ is isomorphic to $F^k$.

niraj panakhaniya
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  • Both $(F,+)$ and $(F^{\times},\cdot)$ are groups, but fields are more than that. They have a ring structure. So they are not totally dependent on group theory, but of course related. So fields are special cases of (commutative) rings, among other things. – Dietrich Burde Dec 16 '20 at 11:57
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    Do you want both fields to have the same unit and zero? If not, you could define for real numbers $A\times B = 2AB$ with $A\cdot B$ being the usual multiplication $AB$. They'll both lead to fields, but the unit in the former case is $\frac{1}{2}$ instead of $1$. – Cameron Williams Dec 16 '20 at 12:01
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    See also this post for counterexamples with $F=\Bbb R^2$, having the usual addition but very different multiplications. – Dietrich Burde Dec 16 '20 at 12:04
  • Nice catch @DietrichBurde . I actually at one point wrote a blog post about the different multiplications one can have on $\mathbb{R}^2$ via solving some terrible quadratic equations. – Cameron Williams Dec 16 '20 at 12:05
  • Found it: https://mathematics-abound.blogspot.com/2015/07/an-axiomatic-approach-to-complex-numbers.html . Fair warning: I wrote this years ago and haven't checked it for accuracy, so don't kill me if there are errors. – Cameron Williams Dec 16 '20 at 12:12

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Since $(\mathbb R,+)$ and $(\mathbb C,+)$ are isomorphic groups (assuming the axiom of choice), there is a binary operation $\times$ on $\mathbb R$ such that $(\mathbb R ,+,\times)$ is isomorphic to the field of complex numbers. Therefore $\times$ is not the same binary operation as real multiplication.

The real numbers and the complex numbers have isomorphic additive groups because they are both vector soaces of (algebraic, i.e., Hamel) dimension $2^{\aleph_0}$ over the field of rational numbers.

bof
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