# GF(2) is a field, binary field

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 $\Bbb Z_2$. – Omnomnomnom 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? – Knight Sep 11 '16 at 21:16
• that's exactly the right idea. – Omnomnomnom Sep 11 '16 at 21:17
• Are there another way to prove that $GF(2)$ is a field? – Knight Sep 11 '16 at 21:18
• You could also apply exactly whatever logic was necessary to prove that $\Bbb Z_2$ was a field. – Omnomnomnom Sep 11 '16 at 21:19

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.