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.

  • 1
    $\begingroup$ If you know that $\Bbb Z_2$ is a field, then yes: you should build an isomorphism between $GF(2)$ and $\Bbb Z_2$. $\endgroup$ – Omnomnomnom Sep 11 '16 at 21:13
  • $\begingroup$ $Z_2$ 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? $\endgroup$ – Knight Sep 11 '16 at 21:16
  • $\begingroup$ that's exactly the right idea. $\endgroup$ – Omnomnomnom Sep 11 '16 at 21:17
  • $\begingroup$ Are there another way to prove that $GF(2)$ is a field? $\endgroup$ – Knight Sep 11 '16 at 21:18
  • $\begingroup$ You could also apply exactly whatever logic was necessary to prove that $\Bbb Z_2$ was a field. $\endgroup$ – 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.


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