Proving that $\{0,1\}$ is a field with $1+1:=0$ EDIT: Hopefully question made clearer. Unfortunately this is a question found in analysis book and I do not actually have background on abstract algebra. Sorry for the confusion arisen.
As the title says, I am trying to show that a set of $\{0,1\}$, equipped with the obvious multiplication and with $1+1:=0$, is a field. 
I encounter this question after the axioms of addition and multiplication of field in $\mathbb{R}$. I am not sure that how would I approach this question. Should I just check by brute force, e.g. check $x+y = y+x$ for all elements in $\{0,1\}$? How about associative of addition, i.e. $(x+y)+z= x+(y+z)$, how do I find myself such a third element $z$ from $\{0,1\}$?
Thanks
 A: If you have no knowledge of quotient rings such as $\:\mathbb Z/2\mathbb Z\:$ then brute-force verfification of all field axioms may be your only choice. This can be done intelligently, e.g. to verify the distributive law $\rm\:x(y\!+\!z) = xy\!+\!xz\:$ doesn't require checking all $8$ possibilities arising from the two possible values $(0$ or $1)$ for each variable. Instead, note that if $\rm\:x=1\:$ it becomes $\rm\:y\!+\!z = y\!+\!z,\:$ so is true for all $\rm\:y,z.\:$ Else $\rm\:x\ne 1\:\Rightarrow\:x=0\:$ so it becomes $\:0 = 0,\:$ so is true for all $\rm\:y,z.\:$ Therefore the distributive law is true for all $\rm\:x,y,z.\:$ Analogous optimizations exist for the other axioms.
A: This is a two element set so showing it is a field by brute force should be easy enough. 
Alternative ways to check this is to note that that the set above is $\mathbb{Z} / 2\mathbb{Z}$. Note that $\mathbb{Z}$ is a ring and $2\mathbb{Z}$ is a maximal ideal in $\mathbb{Z}$. There is a theorem that if $M$ is a maximal ideal in a commutative ring $R$, then $R / M$ is a field. 
A: You can either go with brute force, or notice that the field $\mathbb{Z}/2\mathbb{Z}$ is isomorphic to that field.
As for checking the associativity, notice that the definition doesn't require $x,y,z$ to be distinct.
