GF(2) is a field, binary field

Question

I'm trying to prove that $GF(2)$ with the XOR and AND operations is a field, but I do not know how to prove this creating an isomorphism and not proving all the properties for be a field.

Is it correct think this, building an isomorphism to $Z_2$ ?, How can I prove this statement more easily?

Thanks for your time and help.

If you know that $\Bbb Z_2$ is a field, then yes: you should build an isomorphism between $GF(2)$ and $\Bbb Z_2$. — Ben Grossmann, Sep 11 '16 at 21:13
$Z_2$ is the set of classes formed by {[0],[1]} and $GF(2)$ is the set formed by {0,1}, then I can construct a function f that sends [0] to 0 and [1] to 1, then clearly f is bijective, is it correct? — Rosa Maria Gtz., Sep 11 '16 at 21:16
You could also apply exactly whatever logic was necessary to prove that $\Bbb Z_2$ was a field. — Ben Grossmann, Sep 11 '16 at 21:19
Is it possible to construct a ring and then a maximal ideal that forms all GF(2)? then GF(2) is a field — Rosa Maria Gtz., Sep 11 '16 at 21:25
That seems excessive. If you know that $GF(2)$ is a ring, it suffices to show that it is commutative and that the non-zero element has a "multiplicative inverse". — Ben Grossmann, Sep 11 '16 at 21:26

score 1 · Accepted Answer · edited Apr 13 '17 at 12:21

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You do not need to "create an isomorphism". You verify that $GF(2)$ is a finite ring (this is almost obvious), which has no zero divisors. Then you can use a well-known fact - for a proof see this MSE-question, that every such finite integral domain is a field. Or you verify the field axioms directly, of course.

edited Apr 13 '17 at 12:21

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answered Sep 11 '16 at 21:37

Dietrich Burde

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GF(2) is a field, binary field

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