# What exactly constitutes a field?

I just started learning linear algebra and I'm having trouble understanding some really basic concepts which I would like cleared before moving on.

1. Is the additive or multiplication inverse of an element x in a field necessarily in the field? I know it doesn't (or else -1 would have belong to field {0,1}) but some definitions I've read or heard seem to say otherwise.

2. Why is the characteristic of a field with four elements equal to two? How do I visualise this? What about the set {0,1,2,3}? The characteristic must be prime but I still can't wrap my mind around the fact that it's characteristic can't be anything other than four.

I know these are ridiculously basic concepts, but perhaps I've been confusing myself by checking out completely different resources at a time. Any suggestions on where to start learning anew?

• In the field $\{0,1\}$, what you think of as being $"-1"$ is actually the element $1$. We have $1+1 = 0$ -- $1$ is its own additive inverse. So yes, both the additive and multiplicative inverses of each element must be in the field (excluding multiplicative inverse for $x=0$). Dec 24, 2017 at 7:54

By definition, if $F$ is a field, then each element has an additive inverse and each nonzero element has a multiplicative inverse.
In the field $\Bbb F_2$ of two elements $-1=1$ (the arithmetic in $\Bbb F_2$ is done modulo $2$.)
The field of four elements is $\Bbb F_4=\{0,1,a,b\}$ where $\{0,1\}$ is a copy of $\Bbb F_2$, $a$ is a solution of $a^2+a+1=0$ and $b=a+1$. This entails $1+1=0$ (so the field has characteristic $2$), $a+a=0$, $a+b=1$, $a^2=a$, $ab=1$ etc.
Within your second question, you ask: “What about the set $\{0,1,2,3\}$? The characteristic must be prime”. Well, sets of numbers don't have a characteristic. The characteristic of a field is the smallest natural $n$ such that$$\overbrace{1+1+\cdots+1}^\text{n times}=0$$if such a number $n$ exists and it is $0$ otherwise. And, yes, it must be a prime number. It can't be $4$, because that would mean that $1+1+1+1=0$ and that no smaller sum of $1$'s is $0$. But$$0=1+1+1+1=(1+1)(1+1)\implies1+1=0.$$