# Field Proof using Axioms

Need some help proving the following:

Question:

Let F = {0, 1, a, b} be a field with four elements. Prove that a^2 = b


You can use a(0) = 0 without proving it.

Attempted solution:

a^2 = b

aa = b

aa + 0 = b + 0

We know a*0 = 0 so we can substitute for zero

aa + a*0 = b

Using the additive inverse of aa we get:

(aa) + (-aa) + a*0 = b + (-aa)

a*0 = b + (-aa)

Im still not getting the concept of how to prove things, maybe a little insight into what are possible steps to approach problems like these.

Thats as far as i got, any help is appreciated.

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the three questions you asked in the past hour are very related, and i could imagine them being on a problem set. you should add the homework tag if it is so. –  milcak Jan 31 '11 at 0:25
@milcak: Look at meta for the homework policy. Some people might flame you if you continue suggesting some questions are homework. –  Yuval Filmus Jan 31 '11 at 0:29
cool i changed it to homework :) –  1337holiday Jan 31 '11 at 0:34
@Yuval although I was right this time, I would like to read the homework policy. where can i find it? –  milcak Jan 31 '11 at 0:37
@1337holiday Thanks, this way you will get hints only, then you'll gain more. –  milcak Jan 31 '11 at 0:39

Note that $0$ and $1$ are distinguished elements, so none of the others should be equal to them. Now, $b \cdot a$ cannot be equal to $a$ or $b$ becaus then $b$ or $a$ would be $1$, respectively, but that cannot happen as they are distinct elements. Similarly, $b \cdot a$ cannot be $0$. Hence, $b \cdot a = 1$.

Use this type of reasoning to think what $a^2$ cannot be, and you'll find the anwser: what is $a$ if $a^2 = 0$?

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But how did you arrive at b⋅a = 1, i didnt understand that? –  1337holiday Jan 31 '11 at 0:41
Just because there are 4 options (as there are four elements). And I showed that $b \cdot a$ cannot be the other three, hence it must be the fourth.this is an easiest way to argue, i.e prove what $a^2$ can't be. i had the very same question on a problem set in my first year algebra course and that is how i solved it. –  milcak Jan 31 '11 at 0:44
Right i see. So essentially there are 4 possibilities to a^2 = 0. The first is a^2 = 0 cannot be true since that makes a = 0. The second is that a^2 = 1, which also cannot be true since that means a = 1. The third is a^2 = a which also is not true since that makes a = 1. The fourth then must be true which is a^2 = b. Is this correct? –  1337holiday Jan 31 '11 at 0:52
Yes that is correct. –  milcak Jan 31 '11 at 0:55
You can also check that $b = a^{-1} = a^2 = a+1$ –  milcak Jan 31 '11 at 1:01

Consider that the map $x \mapsto ax$ must be bijective (because the field is finite -- should be easy to show). We know, $$a \cdot 0 = 0$$ $$a \cdot 1 = a$$ This means $a\cdot a = b$ or $a \cdot a = 1.$ But in the latter case we get $a \cdot b = b,$ which implies $a = 1,$ which is not true (by uniqueness of 1).

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