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I want to state that a field $F$ is of characteristic zero in logical notation to an audience without referring them to the meaning of the characteristic of a field.

My first thought was the proposition $\forall x \in F \setminus \{-1\} : x + 1 \ne 0,$ but this holds for any field.

How can “of characteristic zero” be stated using quantifiers?

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    $\begingroup$ An equivalent statement is $F$ contains a subfield isomorphic to $\mathbb Q$ $\endgroup$ – Bang Pham Khoa Jun 4 at 2:09
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No. In any field the equation $$ x + 1 = 0 $$ has the unique solution $x = -1$, independent of the characteristic.

If your audience is unfamiliar with the characteristic of a field and you need to explain it, do that with an example. In $\mathbb{Z}_5$ you have $1 + 1 + 1 +1 + 1 = 0$. In a field with characteristic $0$ no sum of $1$s can be $0$.

Don't use logical symbols, use words.

Edit in response to the edited question.

If you must use a formal statement you might say

For every positive integer $n$ the sum of $n$ $1$s is not $0$.

or

The smallest subfield is infinite.

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  • $\begingroup$ It's going in a list of axioms all written in the form $\forall a, b, c, \ldots \in F : P(a, b, c, \ldots)$. Would it suffice to write $\forall n \in \mathbb{N} : \sum_{i=1}^n 1 \ne 0$? $\endgroup$ – holomenicus Jun 4 at 2:42
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    $\begingroup$ @holomenicus That should do, with the implicit assumption that $0 \not \in \mathbb{N}$. I'm not a logician - if first order logic is what you need I trust that sum as a function of $n$ can be written in first order logic. $\endgroup$ – Ethan Bolker Jun 4 at 3:01
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No.

The statement is equivalent to

for all $x\in F$, if $x+1=0$ then $x=-1$.

This is true in every field.

To explain characteristic zero, how about just "adding $1$ and $1$ and $1$ and so on [finitely many times] will never give you zero". Omit the "finitely many times" if you think it will confuse your audience. And I would certainly suggest briefly giving an example of a field with non-zero characteristic, otherwise they will just say "that's obvious, what's the point?"

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    $\begingroup$ +1 I was just finishing the same answer when yours appeared. $\endgroup$ – Ethan Bolker Jun 4 at 2:12

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