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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 $e=0$?

Michael Hardy
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palio
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  • What is your definition of characteristic? – Chris Eagle Oct 31 '11 at 20:49
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    if $e+e=e$ does'nt this imply that $e=0$? – palio Oct 31 '11 at 20:55
  • 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. – Arturo Magidin Oct 31 '11 at 20:55
  • @palio: certainly not! It implies that $(1+1)e=0$, so either $e=0$ or $1+1=0$; i.e., the characteristic could be $2$. – Arturo Magidin Oct 31 '11 at 20:56
  • 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 $ne$?). – Chris Leary Oct 31 '11 at 20:59
  • 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 $e=0$ ? – palio Oct 31 '11 at 21:05
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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 $e+e=0$). – Jyrki Lahtonen Oct 31 '11 at 21:11
  • @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 me – palio Oct 31 '11 at 21:12
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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. – Arturo Magidin Oct 31 '11 at 21:13
  • @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. – Arturo Magidin Oct 31 '11 at 21:15
  • @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 !! – palio Oct 31 '11 at 21:26
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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. – Chris Eagle Oct 31 '11 at 21:59
  • @ChrisEagle: There are many results, papers, conferences about the field with one element introduced first by Tits. – markvs Feb 20 '22 at 18:26

2 Answers2

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Let $1$ denote the multiplicative identity of the field. Then, $1 \neq 0$ where $0$ is the additive identity. So every field must have at least two elements: your notion of a field with one element $0$ is not correct.

So, since the field contains $1$, it contains $1+1$, $1+1+1$, $1 + 1 + 1 + 1, \ldots, 1 + 1 + \cdots + 1$ (where the last sum has $n$ $1$s in it). All these $n$ sums cannot be distinct since the field has only $n-1$ nonzero elements. So, for some distinct $i$ and $j$, the sum of $i$ $1$s must equal the sum of $j$ $1$s, and by subtraction, the sum of $|i-j|$ $1$s must equal $0$. The smallest number of $1$s that sum to $0$ is called the characteristic of the finite field, and the characteristic must be a prime number. This is because if the characteristic were a composite number $ab$, then the sum of $a$ $1$s multiplied by the sum of $b$ $1$s, which equals the sum of $ab$ $1$s by the distributive law, would equal $0$, that is, the product of two nonzero elements would equal $0$, which cannot happen in a field.

Dilip Sarwate
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I feel as though the answer you are stating is making things more difficult. As you pointed out all integral domains are either characterstic zero or of prime characteristic, and so in particular all fields are of prime characteristic or zero characteristic. But, if a field has characteristic zero you know that it's multiplicative identity has infinite additive order and all finite groups (and thus all finite fields) have no elements of infinite order (for the reason FIRST reason you stated). Another route is, if you know group theory (but this is really overkill compared to the other two arguments) to note that $|\mathbb{F}_q|\mathbb{F}_q=0$ by Lagrange's theorem, and so all elements have order dividing $|\mathbb{F}_q|$.

To answer your last question, no, there are no non-trivial examples where $1=0$ since the ONLY ring with such an identity is the zero ring.

Alex Youcis
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