# Is there any field of characteristic 4 or any other composite number? [duplicate]

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Characteristic of a field is $0$ or prime

Is there any field of characteristic 4? Or any other composite number?

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## marked as duplicate by JavaMan, Zhen Lin, Srivatsan, Thomas Andrews, Jyrki Lahtonen♦Dec 19 '11 at 17:59

No, the characteristic must be a prime. See here for a discussion of this point. – Dilip Sarwate Dec 19 '11 at 15:38
– JavaMan Dec 19 '11 at 15:53

No, a field $F$ can only have prime characteristic:

Let $a \neq 0 \in F$. If $(mn)a = 0$ then also $m(na)=0$. If $(na) \neq 0$ then $F$ has characteristic $\leq m$. If not, then $F$ has characteristic $\leq n$. Either way, the characteristic is $< mn$. This implies that a field must have characteristic prime or zero.

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Your post should be edited to include the case that the characteristic is zero. – JavaMan Dec 19 '11 at 16:05
@AndréNicolas, Dilip: Thanks for the correction. I even confused myself. The answer should make more sense now. – Fredrik Meyer Dec 19 '11 at 16:16

This is a small variation on Fredrik Meyer's answer.

$0$ and $1 \neq 0$ are the additive and multiplicative identities in a field $\mathbb F$.

• If $1$, $1+1$, $1+1+1, \cdots$ are all distinct elements of $\mathbb F$, then the characteristic of $\mathbb F$ is said to be $0$. Note that $\mathbb F$ contains (at least) a countably infinite number of elements.

• If $1$, $1+1$, $1+1+1, \cdots$ are not all distinct, then if we have for some $i$ and $j$, $i < j$, that $$\underbrace{1+ \cdots + 1}_{i~\text{ones}} = \underbrace{1+ \cdots + 1}_{j~\text{ones}} = \underbrace{1+ \cdots + 1}_{i~\text{ones}} + \underbrace{1+ \cdots + 1}_{j-i~\text{ones}},$$ we can conclude that $\underbrace{1+ \cdots + 1}_{j-i~\text{ones}} = 0$. The smallest integer $N > 1$ such that $\underbrace{1+ \cdots + 1}_{N~\text{ones}} = 0$ is called the characteristic of the field. (The possibility that $N$ could be $1$ is ruled out by the fact $1 \neq 0$ in a field). $N$ must be a prime number because if $N$ were composite, say $N = mn$ with $m,n > 1$, then from $$0 = \underbrace{1+ \cdots + 1}_{mn~\text{ones}} = \left(\underbrace{1+ \cdots + 1}_{m~\text{ones}}\right) \times \left(\underbrace{1+ \cdots + 1}_{n~\text{ones}}\right),$$ we get that at least one of $\underbrace{1+ \cdots + 1}_{m~\text{ones}}$ and $\underbrace{1+ \cdots + 1}_{n~\text{ones}}$ is $0$, that is, fewer than $N$ $1$'s sum to $0$ in contradiction of the definition of $N$.

Thus the characteristic of field is either $0$ or a prime number.

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