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knowing that the characteristic of an integral domain is $0$ or a prime number, i want to prove that the characteristic of a finite field $F_n$ is a prime number, that is $\operatorname{char}(F_n)\not = 0$. the proof i'm reading uses the fact that within $e,2e,3e,\ldots$ there are two that are necessarily equal $ie=(i+k)e$ so $ke=0$ for some positive $k$. But can i say the following argument: $F_n=\{0,e,x_1,\ldots ,x_{n-2}\}$ ( $e$ is the multiplicative identity ) and since $e\not = 0$ then $e\not = 2e\not =\cdots ,\not = ne$ so we have $n$ distinct elements $ie, i=1,\ldots,n$ of $F_n$ hence one of them must equal $0$; $ie=0$ for some $i\in \{2,\ldots,n\}$, moreover the trivial field with one element $e=0$ has obviously characteristic $1$. Is there a non-trivial field where $e=0$?

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    What is your definition of characteristic?2011-10-31
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    if $e+e=e$ does'nt this imply that $e=0$?2011-10-31
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    On what grounds do you assert that $e\neq 2e\neq\cdots,\neq ne$"? It's not even true in general: in the field of 4 elements, you do **not** have any element $e$ with $e,2e,3e,4e$ pairwise distinct.2011-10-31
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    @palio: certainly not! It implies that $(1+1)e=0$, so either $e=0$ **or** $1+1=0$; i.e., the characteristic could be $2$.2011-10-31
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    You might want to try considering the prime factorization of the integer $k$ because you need to find a prime $p$ for which $pe=0$ (if $k$ factors as $k=mn$, what can you say about the elements $me$ and $ne$?).2011-10-31
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    you are using substraction in a case and not in the other: since $e+e=e$ then $e+e+e=e+e=e$ then by substracting $-e$ you get $e+e+e-e=e-e=0$ hence $e+e=0$ but why didn't you substruct $-e$ from the beginning i mean since $e+e=e$ then $e+e-e=e-e=0$ so $e=0$ ?2011-10-31
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    @palio $e+e=e$ is not possible (except in the zero ring), and it does imply that $e=0$. But what happens in characteristic two is $e+e+e=e$ (and consequently $e+e=0$).2011-10-31
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    @Arturo Magidin : here $ne$ i not a real multiplication, it is only a formal way to write $e+e+...+e$; $n$ times. so saying that $1+1=0$ has no meaning to me2011-10-31
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    @palio: I misread what you wrote; from $e+e=e$ you *can* conclude $e=0$; but what if $e\neq 2e$, $2e\neq 3e$, but $e=3e$? Then you have $e=e+e+e$, from which you get $0=e+e$. Note in particular that just because $e\neq 2e$, and $2e\neq 3e$, this does *not* tell you that $e\neq 3e$. Inequality is not transitive.2011-10-31
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    @palio: I **know** it's not real multiplication. But every field has an element which is the multiplicative identity, and that is usually called $1$. So $1+1$ *does* make sense: it means the multiplicative identity added to itself. And for every element $e$ in the field, you have $e+e = 1e+1e = (1+1)e$ by distributivity. You never said what $e$ was, I assumed it was merely some element in the field. If you meant it to be the multiplicative identity, you needed to say so.2011-10-31
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    @Arturo Magidin : i'm sorry arturo i thaught it was clear from my notation of $F_n=\{0,e,x_1,...,x_{n-2}\}$ but yes by $e$, I meant the multiplicative identity of $F_n$. But the example you gave of $e\not = 3e$ explains it all. thank you !!2011-10-31
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    There is no "trivial field with one element". The definition of a field includes the statement that the multiplicative identity is not the additive identity.2011-10-31

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