A finite sum of $1$ equals $0$ in a field with finitely many elements.

I need to prove that for field $\mathbb{F}$ with finitely many elements $\exists n \in \mathbb{N}$, $\underbrace{1+1+1+...+1}_{n \text{ times }} = 0$.

I can see why this is true, since there is a finite number of elements. A field with elements {$0,1$}, $1+1 = 0$, so $n = 2$. For set {$0,1,2$}, $1+1+1=0$, so $n$ is equal to the cardinality of the set.

I can't however, see how to prove this Mathematically.

What I want to know: Can I prove this via the field axioms solely?

If my title is poorly labeled, please edit, or inform me to edit!

Thanks for any help!

• Just to avoid a possible misconception: while it is true that in a field of $n$ elements, $n \cdot 1 = 0$, this is not necessarily the smallest possible $n$. Unless $n$ is prime, the elements of a field of $n$ elements are not (represented by) $0$, $1$, $\dots$, $n-1$. So, where you say "so $n$ is the cardinality of the set", you have to be more careful. – Magdiragdag Mar 24 '14 at 7:56
• I've corrected "finite elements" to "finitely many elements". The former means something quite different. – Hurkyl Mar 24 '14 at 8:12
• @Display Name: Magdiragdag is right. the word "$Characteristic$" in your label implies "minimum" of such $n$'s. but your question need not minimal $n$. it wants only an $n$. – user 1 Mar 24 '14 at 8:13
• @ucf That is certainly true! Thank you for that. – Display Name Mar 24 '14 at 10:06

A field is a group with respect to $+$ and it's identity is $0$. in a group the order of any element divides the order of group (number of element).

• Just to finish the argument: say $n$ is the order of $1$, then $n \cdot 1 = 0$. – Magdiragdag Mar 24 '14 at 7:50
• I imagine I will need to prove that property, is that simple? That is actually a really simple way to attack the problem, thank you! – Display Name Mar 25 '14 at 1:16
• @Display Name: that property is proved in almost all books on Group Theory. – user 1 Mar 25 '14 at 6:18

To add to what has already been said suppose $p,q\in \mathbb N$ and that $pq\cdot 1=0$ (where this means adding $pq$ ones together), then $(p\cdot 1)(q\cdot 1)=0$ so that either $p\cdot 1=0$ or $q\cdot 1=0$.

Hence in a field the least $n$ for which $n\cdot 1=0$ is a prime (in an infinite field like $\mathbb R$ there may be no such $n$).

The usual trick with finite structures is that when you produce a sequence of things (e.g. by repeatedly adding $1$), that eventually you have to have a repeat.

If the same field element appears twice on the list, what can you say about that? (hint: convert this fact into an equation)

As an aside, $n$ can be smaller than the cardinality of the field. e.g. the finite field with 9 elements has characteristic $3$. You can construct the finite field with 9 elements by adding $\mathbf{i}$ -- i.e. a square root of $-1$. Then the set of elements $(a+b \mathbf{i})$ with $a,b \in \mathbf{F}_3$ turns out to be a field; the proof is pretty much the same as proving the complex numbers (constructed in the same manner but starting from the reals) are a field.

Let $1$ be the multiplicative identity of the finite field $\mathbb F$ where $|\mathbb F| = q$. Now, consider the following list of field elements: $$1\\ 1+1\\ 1+1+1\\ 1+1+1+1\\ \cdots\\ \underbrace{1 + 1 + \cdots + 1}_{q+1 ~\text{ones}}$$ (I refuse to call these elements $1, 2, 3, 4$ etc.) Since the field has $q$ elements, these $q+1$ field elements that we have listed above cannot all be distinct elements of the field: at least two elements in the list must be the same. Let $\underbrace{1+1+\cdots + 1}_{i~\text{ones}}$ be an element that is a repeat of an element that has been listed previously: that is, $$\underbrace{1+1+\cdots + 1}_{i~\text{ones}} = \underbrace{1+1+\cdots + 1}_{j~\text{ones}} ~\text{for some} ~ j < i.$$ Thus, we have that \begin{align} \underbrace{1+1+\cdots + 1}_{j~\text{ones}} = \underbrace{1+1+\cdots + 1}_{i~\text{ones}} = (\underbrace{1+1+\cdots + 1}_{j~\text{ones}}) + (\underbrace{1+1+\cdots + 1}_{i-j~\text{ones}}) \end{align}

It follows that $\underbrace{1+1+\cdots + 1}_{i-j~\text{ones}} = 0$. Note that $i$ might be as large as $q+1$ but since $j \geq 1$, we conclude that

there exists a positive integer $n \leq q = |\mathbb F|$ such that $$\underbrace{1+1+\cdots + 1}_{n~\text{ones}} = 0$$

Note that it is not claimed that $n$ is the smallest such integer or that $n$ is a prime, just a proof that a sum of $n$ copies of the multiplicative identity equals $0$.