# 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

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Are you able to use the fact that $\mathbb{Z}_p$ is a field if and only if $p$ is prime? – user33719 Jun 15 '12 at 2:26
Well, if you are trying to show it's a field, technically you need to tell us also what the multiplication is... Presumably, the "obvious" one, but it should still be said. – Arturo Magidin Jun 15 '12 at 3:21
Note that "For every $x,y,z\in F$, $(x+y)+z = x+(y+z)$" does not require that $x$, $y$, and $z$ be distinct. In fact, they can all be the same element! While we should certainly not use the same name for different things, there is no problem with using different names for the same thing. – Arturo Magidin Jun 15 '12 at 3:28
question edited. the operation of addition and multiplication is the one in $\mathbb{R}$, i.e. the axioms that you have when you first encounter analysis. – Daniel Jun 15 '12 at 3:29
@user33464: You can't have the "usual addition of $\mathbb{R}$" if you are also defining $1+1=0$.... – Arturo Magidin Jun 15 '12 at 3:29

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.

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Thanks a lot! I was really thinking about crushing through the 8 expressions by checking them one by one. – Daniel Jun 15 '12 at 3:47

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.

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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.

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