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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Arthur Nov 11 '19 at 12:05
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Nov 11 '19 at 12:08
  • $\begingroup$ yes. Can it have anything other than 1? $\endgroup$ – user6394019 Nov 11 '19 at 12:08

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.

  • $\begingroup$ I understand. Can you also explain why char(Q)=char(R)=char(C)=0? $\endgroup$ – user6394019 Nov 11 '19 at 13:45
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    $\begingroup$ @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 $\operatorname{char}(\Bbb Q) = 0$. $\endgroup$ – Arthur Nov 11 '19 at 13:50

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