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I'm new to linear algebra and I'm reading K. Hoffman and R. Kunze book. My question is how to prove this:

Consider field $F$ that has $p^k$ elements which $p$ is a prime number and $k$ is a natural number. Show that characteristic of $F$ is $p$.

I know it for $k=1$. Actually $\mathbb{Z}_p = \{0, 1, 2, ..., p-1\}$ with addition and multiplication in hang of $p$. Then $char(F)=p$. But I don't know how to prove it for $p^k$-elements fields.

J. W. Tanner
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Harry Maguire
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  • If the characteristic was $q$ then $1$ will generate an additive group of order $q$ which is subgroup of a group of order $p^k$ – Tony Pizza Oct 06 '23 at 15:37
  • @TonyPizza How will 1 create an additive group? And what actually group or additive group mean here? –  Oct 06 '23 at 15:57
  • @4bolfazl because every element inside a group generates a subgroup. Only that, nothing else. – Tony Pizza Oct 06 '23 at 15:59
  • @TonyPizza I'm sorry I'm new to these. What does actually a group mean in this topic? And what does it mean when you say a group of order q? –  Oct 06 '23 at 16:01
  • a group of order $q$ is a set of $q$ elements with a binary operation that is associative, and there is an identity element, and each element has an inverse – J. W. Tanner Oct 06 '23 at 16:52
  • @4bolfazl group is certain structure in abstract algebra much like a field. You can read the definition online. In particular you will find that a field is both a group under addition as well as multiplication. Moreover it is commutative in both. That's why it is probably the simplest structure in abstract algebra while a group is much more general and hence possibly complicated. – Tony Pizza Oct 06 '23 at 16:54
  • @4bolfazl A field is not a group under multiplication, you have to exclude $0$. By the way, the distributive property that also holds in fields and relates $+$ and $\cdot$ is very important (more important than commutativity). – Amateur_Algebraist Oct 06 '23 at 19:50

2 Answers2

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First to clear up some points in the comments above, a field must be an abelian group and we often use $0$ not $e$ and certainly not $1$ for the additive identity. Additionally there must a multiplicative operation and mult. identity often denoted $1$ such that each non-zero element has a multiplicative inverse. Before considering finite examples, main examples include the rationals and reals with usual operations. Note that the non-zero elements form an (abelian) group under the multiplicative operation so basically it's a ring with division.

Sketch of (one possible) proof of original problem:

Write $\mathbb{Z_n}$ to mean $\mathbb{Z}/n\mathbb{Z}$ as an (additive) abelian group for any $n$.

Now suppose $F$ is a field with $p^k$ elements with $p$ prime. That is a given here but note that the order of a finite field MUST be a power of a prime, else as a ring it would have zero-divisors and not be a field.

As an abelian group, $F$ has order $p^k$ so what can it look like? It can only be the direct sum (or product) of $k$ copies of $\mathbb{Z}_p$. Suppose for example it was cyclic of order $p^k$ where $k>1$. Since all elements must have a multiplicative inverse, this would imply the group of units of $\mathbb{Z_{p^k}}$ has order $p^k-1$ however it is well known that this is only true when $k=1$. Other cases can be handled by the structure theorem for f.g. abelian groups. For example if the order was $27$ this rules out $\mathbb{Z_{27}}$ and $\mathbb{Z_9}\oplus\mathbb{Z_3}$.

Once you have that it is a direct sum of cyclic groups of prime order $p$ it follows that the characteristic is $p$.

Note that this does not address the (multiplicative) group structure of the non-zero elements at all. For example, in the unique field $F$ with $4$ elements, $F$ as solely an abelian group is isomorphic to the sum of two copies of $\mathbb{Z_2}$, and the multiplicative group of nonzero elements is cyclic of order $3$. If you list the first as $(0,0), (0,1), (1,0), (1,1)$ and the second as $a,a^2,1$ then the mapping between these is not apparent. That is covered here: What are the fields with 4 elements?

AlgTop1854
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$F$ is an extension of the prime field $\mathbb F_p$ whose characteristic is $p$ and this $p$ is not zero in $F$ (while $p=0$ in $\mathbb F_p$). It follows that if $x\in F$ then $p\cdot x=(p\cdot 1)x=0\cdot x=0$ for all $x$. This shows that the characteristic of $F$ is $p$.

Ataulfo
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  • What do you mean actually by $\mathbb{F}_p$? And how $F$ is an extension of it? How it can be shown? –  Oct 09 '23 at 14:32
  • @4bolfazl For all prime $p$ and all integer $n\ge1$ there is a unique field $F$ having $p^n$ elements. $\mathbb F_p$ is a standard notation for the field having $p$ elements (it is called "prime" because it has not a subfield properly speaking) – Ataulfo Oct 10 '23 at 12:53