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Apologies for the super-basic question. Looking at Wikipedia's definition of a finite field $F$, it is a set with $n$ elements with two binary operations that produce from an ordered pair of $F$ an uniquely determined element of $F$, called addition and multiplication. Addition and multiplication must be associative, commutative and distributive. Every element of $F$ must have an additive inverse, and every non-zero element of $F$ must have a multiplicative inverse. Division by zero is prohibited. Any element add zero is itself, and any element times one is itself, and no other elements of $F$ satisfy these identities in the place of zero, one.

Everyone also seems to say that if $F$ has a finite $n$ number of elements, $\underbrace{1+1+1+\dots+1}_{\text{$n$ times}}=n=0$. I do not see how this follows from the field axioms!

Wikipedia also provides an example of $F_4$, which has four elements $O, I, A, B$ where $O,I$ are the multiplicative and additive identities respectively. They state as if it's completely obvious that $A+B=I,\,A\cdot B=I$ and $A\cdot A=B$. My thoughts on these three equations:

In the first equation, as $A,B$ are not the additive identity, $A+B\neq A, A+B\neq B$, but since $B$ could well be $A$'s additive inverse, as such a thing is defined to exist, $A+B$ could be, to my mind, $O$, as well as $I$. Likewise $A\cdot B$ must be neither $A$ nor $B$, but again it could be $O$ or $I$, as I saw nowhere in the field axioms any statement declaring that any element multiplied with zero must be zero, and zero is the only such number for which that is true. Indeed, nowhere in the axioms do I see anything that necessitates that multiplication must be defined as repeated addition. Clearly it has to be, but I can't see how to prove it, and therefore I can't see how to prove that the zero of a field is the only number for which multiplication by it always takes the result to zero.

My confusion is clear; since the axioms define properties of operations, but not rules for what these operations are (as it is a general, abstract algebraic idea), in any finite field, how on earth can you determine what $x\cdot y=?$ and $x+y=?$, as the normal ideas of $1+1=2$ that you might build upon do not hold for all fields! What is the fundamental logic basis for the arithmetic here?

Many thanks!

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  • $\begingroup$ (1) $\underbrace{1+1+\dots+1}_{n}=n=0$ follows immediately from $(\mathbb{F},+)$ is a group and Lagrange theorem in group theory. (2) You have the additive and multiplicative identity the other way round for $O,I$. (3) $B$ is not the additive inverse of $A$ because field of 4 elements must have characteristic 2. $\endgroup$ Jun 29, 2021 at 15:26
  • $\begingroup$ @user10354138 But I am seeking to understand the assertion that a field must have characteristic! Therefore answers relying on the characteristic property are circular to me... $\endgroup$
    – FShrike
    Jun 29, 2021 at 15:30
  • $\begingroup$ You should check that by definition a ring of characteristic 0 is necessarily infinite. This answers your question. @IdioticShrike $\endgroup$
    – Randall
    Jun 29, 2021 at 16:47

5 Answers 5

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All of these follow from the field axioms. $0$ times any element is $0$ because:

$0x=(0+0)x=0x+0x$

Now add the additive inverse of $0x$ to both sides to get $0x=0$.

Next, if $xy=0$ and $x\ne 0$ then you can multiply the equation by $x^{-1}$ to get $y=0$. So $xy=0$ in a field implies that at least one of $x,y$ is $0$.

Finally, you ask why a finite field has nonzero characteristic. So suppose $F$ is a finite field. Let me denote the sum $1+...+1$ $n$ times simply by $n^F$. Since the field is finite, there must be some natural $n<m$ such that $n^F=m^F$. And then:

$(m-n)^F=\underbrace{1+1+1+\dots+1}_{\text{$m-n$ times}}=0$

So indeed $F$ has finite characteristic.

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  • $\begingroup$ To clarify: $n\lt m:n=m$ must be the case as a finite field's operations must ""wrap round"" like in modular arithmetic, where two numbers that are not the same end up being equal under the field? $\endgroup$
    – FShrike
    Jun 29, 2021 at 15:34
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    $\begingroup$ Yes, this is the same idea. Thing is, if $m\ne n$ would always imply $m^F=n^F$ then $F$ would already have infinitely many elements. (while we assumed $F$ is finite) $\endgroup$
    – Mark
    Jun 29, 2021 at 15:37
  • $\begingroup$ Thank you for putting up with the elementary question! This is the sort of thing a Wikipedia article always leaves out due to assumed experience... $\endgroup$
    – FShrike
    Jun 29, 2021 at 15:38
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Consider the $n+1$ elements $0, 1, 1+1, \ldots \sum_{i=1}^{n} 1$.

By the pigeonhole principle, two of them are equal. Say $x = y$ with $x$ before $y$ in the sequence. Then $y-x = 0$ is a sum of copies of $1$, and the field has positive characteristic.

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A field is a finite group under addition. In a finite group every element has a finite order which is a divisor of the cardinality of the the group. This is called Lagrange's theorem. The characteristic of a finite field is just the order of $1$.

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  • $\begingroup$ You don't really need Lagrange's theorem to show that every element in a finite group has finite order. $\endgroup$
    – Mark
    Jun 29, 2021 at 15:30
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Though, as you say, an element in an arbitrary field such as 1 + 1 + 1 is only defined abstractly, we can speak about it and see what its properties should be. Suppose we define the sequence $x_n = \sum_{i = 1}^n 1$. If the field is finite, then at some point $x_n = x_m$ for some $m \neq n$. Subtracting the difference then yields $ \sum_{i = 1}^{n - m} 1 = 0$.

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Question: "Why must any finite field have a non-zero characteristic?"

Answer: Any commutative unital ring $A$ has a canonical map

$$\phi: \mathbb{Z} \rightarrow A$$

defined by $\phi(1):=1_A$ and $\phi(n):=n(1_A)$ wher $1_A \in A$ is the multiplicative unit. Since $A$ is a finite field it follows $\mathbb{Z}/ker(\phi)$ is a domain, hence $ker(\phi)$ must be a prime ideal. Since $A$ is finite it follows $ker(\phi)=(p)$ must be generated by a non-zero prime $p$ since the prime ideals in $\mathbb{Z}$ are $(0)$ - the zero ideal and $(p)$ for varying $p$.

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    $\begingroup$ I won't downvote (since this is a perfectly good answer, for someone), but I really doubt using several reasonably technical theorems will illuminate the OP's very basic misconception. I could be wrong! $\endgroup$ Jun 29, 2021 at 15:34

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