Proof for $a\cdot b \neq a$ in a field with four elements $\{0,1,a,b\}$

Question

Is there a formal (but not advanced-level, I'm still a beginner) proof for $a\cdot b \neq a$ in a Field? (Körper in German).

I looked at lots of resources, but could only find for $0a = a0 = 0$, the commutative, associative laws, and a couple of others.

The field given is with 4 distinct elements $0, 1, a, b$. The hint on my question paper says to prove by contradiction (Widerspruch).

Any pointers on how to proceed?

Answer 1

First of all, this is only true if $b \neq 1$ and $a \neq 0$. If $b = 1$, certainly $a\cdot b = a\cdot 1 = a$. If $a \neq 0$, then $a\cdot b = 0 = a$.

Now to prove it, Assume $a\cdot b = a$, $b \neq 1$ and $a \neq 0$. But $1$ is the distinct element of the field such that $x \cdot 1 = x$ for any $x$ in the field. Then $ b = 1$. Contradiction.

Answer 2

Part of the definition of a field is that for every $a \ne 0$ there exist a $a^{-1} $ so that $a^{-1}a=1$.

So if $ab=a $ then either

i) $a =0$ and $ab=0b=0=a $

or

ii) $a\ne 0$ and therefore:

$ab=a $

$a^{-1}ab=a^{-1}a $

$1*b=1$

$b=1$.

So if you are given $ab=a $ in a field you can always conclude either $b=1$ or $a=0$.