My question is more conceptual, so I will come straight to the exercise:
Exercise: Let $M= \lbrace g,u \rbrace $ be a Set. On $M$ the Addition and the Multiplication is given by: \begin{align} g+g = u+u &= g \tag{$\alpha$} \\g+u = u+g& = u \tag{ $\beta$} \\ \\ g \cdot g = u \cdot g = g \cdot u & =g \\ u \cdot u & =u \end{align}
I know what it takes for $M$ to be a field, axiom wise. I will just focus on the additive properties, if I got it right there I can also do the multiplicative ones.
Commutativity: Follows straight from $\beta$. $\beta \implies g+u = u + g, \ \forall u,g \in M$
Existence of additive identity: $\alpha \implies g+g=g \implies g \equiv 0$
Existence of additive inverses: $\alpha \implies u + u = g =0 \implies u$ is inverse element of $u$ and I believe it is correct to say that $ \alpha \implies g+g=g =0 \implies g$ is inverse element of $g$
Associativity: Here I struggle. The set $M$ only contains two elements, to show associativity I would have (intuitively) said that I need three elements. However, there are Galois Fields which only contain two elements (we didn't address that in Class, I did exhaust google for that)
So I thought about the following 'workaround': Let $c \in M= \lbrace u,g \rbrace \implies c=u \vee c = g$. If $c=u$ we have: $$g+(u+c)=g+(u+u)\overset{\alpha}=g+g=(u+u)+g =(u+c)+g$$ and then the same for $c=g$. Is this (and the above) correct? Or do I need to formulate my thoughts differently?
Update: After the answer given below my insight into this problem is better. I know now that for associativity I need to analyze 8 cases. I have written down all those cases, but I am lost for the one if $(x,y,z) \in M^3$ with $(x,y,z)=(u,g,g) $ then I have $$ u+(g+g)\overset{\alpha}=u+(g)\overset{com.}=(g)+u=(g+g)+u=... ?$$
Associativity involves three elements, but they do not have to be distinct. So with a two element field $a+(b+c)=(a+b)+c$ means that whatever values $a,b,c$ take in the field, the equation is true. There are eight possibilities to check, which is not too bad.
Associativity can be a pain to prove if you don't have a short cut. The associativity of the two element field can be established by identifying it with $\mathbb Z/2\mathbb Z$ and using the associativity of $\mathbb Z$.
Likewise, it is easy to prove that the composition of functions is associative. We can prove that matrix multiplication is associative by associating each matrix with a function (a linear map relative to a basis), rather than computing the matrix entries element by element.