# Show That There Is One and (Essentially) Only One Field With 3 Elements

My thought process is the following: By default, 0 and 1 have to be in this set of 3 elements since they are the neutral additive and neutral multiplicative elements, respectfully. So the set of the 3 elements that make up this field must be {0,1,x}. But for any element "a" in the set, there must be some "-a" element such that a + (-a) = 0. Therefore there is only one option for x; x=-1. My set must then be {-1,0,1}. Is this correct? It seems too simple, so I wanted a second opinion.

• It is right ... but what have it to do with real analysis? Commented Aug 23, 2015 at 19:32
• What does this have to do with real analysis? It looks like algebra to me. Commented Aug 23, 2015 at 19:33
• No, it is not right. What if $1+1=0$ and $x$ is another element? You have to handle the case where $(-1)=1$. @user251257 Commented Aug 23, 2015 at 19:35
• @ThomasAndrews Well we need to have that $x+1$ is one of $0,1,x$. If $x+1=0$ then $x=-1$ since inverses are unique. If $x+1=1$ then $x=0$, which means we don't have three elements. If $x+1=x$ then $1=0$, which means the field is trivial. So we need to have $x=-1$ necessarily. In your example, we'd then have $x=1$ and the field has only two elements. So it's not a counterexample. Commented Aug 23, 2015 at 19:40
• No, you are wrong. You are actually missing the point of the definitions - the above field you are creating with three elements is exactly the field of $\mathbb Z$ modulo $3$. So if you are counting the field of two elements as "not a regular set," whatever that means, then you are also calling the field of $3$ elements "not a regular set." The key to getting math is reading and understanding the definition. @FrenchToastCrunch Commented Aug 23, 2015 at 19:48

Your answer is correct, your reasoning is not. I realize you're new to abstract algebra, so I'll just say this: when we write down things like $1$ and $0$, we mean only that these things are the multiplicative and additive identities, respectively. We do NOT think of them as subsets of the real line. For example, we can define a perfectly good field with only two elements $\mathbb{Z}_2=\{0,1\}$ where multiplication is defined as usual, and addition is defined as $1+1=0$.
Now, we want to show that there's only one field with three elements. We know $0$ and $1$ must be in this field, and that $0\neq 1$. So if we call the remaining element $x$, we have to find if there is more than one possible thing for $x$ to be.
We know, by closure, that $x+1$ has to be in our field, so either $x+1=0$, $x+1=1$, or $x+1=x$. In the first case, we get $x=-1$, where by $-1$ I do NOT mean "negative one", but rather simply "the additive inverse of one". In the second case we get $x=0$, which is wrong because we want $x$ to be a NEW element of our set. In the third case, we get $1=0$, which is also bad. So we know $x=-1$!
Now, we still haven't fully determined our field. Why? Well, we know there's only one possible third element, but we still have to fully define what addition does on the elements. In particular, we want to show that there's only ONE way to define addition on the elements consistent with the field axioms. We already know what happens if you take $1+0$, $1+-1$, and $-1+0$. We just need to figure out what happens when you take $1+1$ and $-1+-1$. We know $1+1$ has to be in our set, so $1+1=1,0,-1$. If $1+1=1$ then $1=0$, if $1+1=0$ then $-1=1$ and there's only two elements in your field. So $1+1=-1$. Similarly, $-1+-1=1$. Then you just have to check there's only one way to define multiplication (there is), and you're done.