# Characteristic of a field is a minimal number

I have a an issue with certain terminology of field properties.

Characteristic is defined as as the minimal number such as $$1+1+1+1...=0$$

is there a possiblity that, there exists any other number than 1 that satisfies this condition?

suppose I have : $$\mathbb{Z}/7$$ then $$1+1+1+1+1+1+1=0$$ I can't think of any other minimal number that makes it equal.

• Do you mean to ask whether a field can have more than one characteristic at the same time? Or do you mean to ask why we are adding together copies of $1$ rather than some other number? – Arthur Nov 11 '19 at 12:05
• Chiming in with Arthur. In the latter case do observe that $$a+a+\cdots+a=a(1+1+\cdots+1)$$ for all the elements $a$ of the field. – Jyrki Lahtonen Nov 11 '19 at 12:08
• yes. Can it have anything other than 1? – user6394019 Nov 11 '19 at 12:08

## 1 Answer

There is a very good reason for unsing $$1$$ when defining the characteristic: The advantages of using $$1$$ specifically is that

1. it gives the largest characteristic among elements of the ring
2. any other element is going to yield "characteristic" which is a divisor of the $$1$$-characteristic
3. $$1$$ is the one element we can count on existing in the ring

For instance, in the ring $$\Bbb Z/7\Bbb Z$$, we get that $$1+1+1+1+1+1+1 = 0$$, but no other smaller sum of $$1$$'s is equal to $$0$$. Now, it also turns out that we get the same answer for any other non-zero number. For instance, $$2+2+2+2+2+2+2 = 0$$, but no other smaller sum of $$2$$'s is going to give $$0$$ (apart from the empty sum). In fields, you could use whatever non-zero number you want. This is where point 3 above comes in, though: In $$\Bbb Z/7\Bbb Z$$, using the number $$5$$ works, but in $$\Bbb Z/5\Bbb Z$$ it doesn't. Declaring that we will use $$1$$ means we aviod this issue.

As a different example, note that characteristic is also defined for non-field rigs (at least as long as they are commutative and unital). For instance, $$\Bbb Z/6\Bbb Z$$ has characteristic $$6$$, because $$1+1+1+1+1+1 = 0$$. However, if we use $$2$$ then we get $$2+2+2 = 0$$ and if we use $$3$$ we get $$3+3 = 0$$. So different numbers give different characteristics.

However, no matter which element you take, the "characteristic" that you get will be a divisor of $$6$$, and no matter which element you take, adding $$6$$ of them together will give you $$0$$, and no smaller number than $$6$$ works for all elements simultaneously. So the number $$6$$ still very much characterises the additive structure of the ring. So that's what we use.

• I understand. Can you also explain why char(Q)=char(R)=char(C)=0? – user6394019 Nov 11 '19 at 13:45
• @user6394019 That's just a convention. There is no way to write $0$ as a finite, non-empty sum of $1$'s in those fields, so they do not have positive finite characteristic the way $\Bbb Z/7\Bbb Z$ does, for instance. It could have been called $\operatorname{char}(\Bbb Q) = \infty$, but the established convention is to call it $\operatorname{char}(\Bbb Q) = 0$. – Arthur Nov 11 '19 at 13:50