# Show that $a^2 = b$ in a field $\mathbb{F} = \{0,1,a,b\}$

I need some help proving the following:

Let $\mathbb{F} = \{0, 1, a, b\}$ be a field with four elements. Prove that $a^2 = b$.
You can use $a \cdot 0 = 0$ without proving it.

Attempted solution:

$$a^2 = b \\ aa = b \\ aa + 0 = b + 0$$ We know $a \cdot 0 = 0$ so we can substitute for zero $$aa + a \cdot 0 = b$$ Using the additive inverse of $aa$ we get: $$(aa) + (-aa) + a \cdot 0 = b + (-aa) \\ a \cdot 0 = b + (-aa)$$ I’m 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.

-
Can the multiplicative group of F in this case be regarded as the Klein four group? Thanks in any case. – awllower Feb 21 '11 at 12:27

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 $a^2 = 0$?

-
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).

-