Finite Field of Characterisitic p .... ....

[Math Amateur]

Homework Statement

I need help with Exercise 1 of Dummit and Foote, Section 13.2 : Algebraic Extensions ..

I have been unable to make a meaningful start on the problem ... ...

Exercise 1 of Dummit and Foote, Section 13.2 reads as follows:

Homework Equations

A relevant definition is the definition of the characteristic of a field ...

This definition along with some remarks by D&F reads as follows:

A proposition which may be relevant is D&F Proposition 1 of Section 13.1 ... ...

This proposition reads as follows:

The Attempt at a Solution

I have been unable to make a meaningful start on this problem ...

BUT ... further I hope to understand why D&F put this exercise at the end of a section on algebraic extensions ... the exercise seems a bit remote from the subject matter ... ...

Hope someone can help ...

Peter

 

[andrewkirk]

The problem is solved by considering the field as a vector space over the field $\langle 1_F\rangle \cong \mathbb Z_p$. If the vector space has dimension $n$ then it has a basis of size $n$ and we can show that the order of the vector space is $p^n$ by considering the number of different $n$-tuples whose components are in $\mathbb Z_p$.

I think the reason it comes at the end of the foregoing chapter is that that chapter (IIRC) has shown how a field can be regarded as a vector space.

The vector space will be one-dimensional if the field is isomorphic to $\mathbb Z_p$, and so will have order $p$ and $\{1_F\}$ is a basis. If it is not isomorphic, the field generated by $1_F$ will be a subfield (call it $S$) of order $p$. The other elements of the basis can be found by considering what elements are needed to extend $S$ to $F$.

 

[Math Amateur]

The problem is solved by considering the field as a vector space over the field $\langle 1_F\rangle \cong \mathbb Z_p$. If the vector space has dimension $n$ then it has a basis of size $n$ and we can show that the order of the vector space is $p^n$ by considering the number of different $n$-tuples whose components are in $\mathbb Z_p$.

I think the reason it comes at the end of the foregoing chapter is that that chapter (IIRC) has shown how a field can be regarded as a vector space.

The vector space will be one-dimensional if the field is isomorphic to $\mathbb Z_p$, and so will have order $p$ and $\{1_F\}$ is a basis. If it is not isomorphic, the field generated by $1_F$ will be a subfield (call it $S$) of order $p$. The other elements of the basis can be found by considering what elements are needed to extend $S$ to $F$.

Thanks Andrew ... I vaguely follow ...

Are you able to formalise the proof a bit ...

Peter

 

[andrewkirk]

Use the distributive law to prove that the vector space over $\mathbb Z_p$ generated by $1_F$ is a subfield of $F$ of order $p$, which we can label as $S$.
Hence $F$ is a field extension of $S$.
In the previous work (somewhere in D&F, else see the wiki article on Field Extensions) we've seen that a field extension is a vector space over the subfield (ie we mean the subfield $S$ is the scalar field from which scalar multiplication in the vector space $F$ is defined).
Since $F$ is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to $S^n$ for some finite $n\in\mathbb N$.
The canonical element of $S^n$ is an ordered $n$-tuple of elements of $S$. Since there are $p$ distinct elements of $S$, that means there are $p^n$ distinct elements of $S^n$.
Hence, via the vector space isomorphism, we know that $|F|=|S^n|=p^n$.

You might find it helpful to play around with a couple of examples.
(1) What are the elements of the field $\mathbb Z_3(\sqrt 2)$, and what is the characteristic and order of this field?
(2) What are the elements of the field $\mathbb Z_3(\sqrt{-1})$, and what is the characteristic and order of this field?

 

[Math Amateur]

Use the distributive law to prove that the vector space over $\mathbb Z_p$ generated by $1_F$ is a subfield of $F$ of order $p$, which we can label as $S$.
Hence $F$ is a field extension of $S$.
In the previous work (somewhere in D&F, else see the wiki article on Field Extensions) we've seen that a field extension is a vector space over the subfield (ie we mean the subfield $S$ is the scalar field from which scalar multiplication in the vector space $F$ is defined).
Since $F$ is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to $S^n$ for some finite $n\in\mathbb N$.
The canonical element of $S^n$ is an ordered $n$-tuple of elements of $S$. Since there are $p$ distinct elements of $S$, that means there are $p^n$ distinct elements of $S^n$.
Hence, via the vector space isomorphism, we know that $|F|=|S^n|=p^n$.

You might find it helpful to play around with a couple of examples.
(1) What are the elements of the field $\mathbb Z_3(\sqrt 2)$, and what is the characteristic and order of this field?
(2) What are the elements of the field $\mathbb Z_3(\sqrt{-1})$, and what is the characteristic and order of this field?

Thanks so much Andrew ... much clearer now ...

BUT ... I'm struggling a bit when I come to the following step ... ... :

" ... ... Since $F$ is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to $S^n$ for some finite $n\in\mathbb N$. ... ... "Can you give an indication of why this statement is true ... ... ?

Can you indicate where it might be proved in full ...?

Peter

 

[andrewkirk]

I'm struggling a bit when I come to the following step ... ... :

" ... ... Since $F$ is finite, it must be a finite-dimensional vector space and hence, from linear algebra, we know it must be isomorphic to $S^n$ for some finite $n\in\mathbb N$. ... ... "Can you give an indication of why this statement is true ... ... ?

Can you indicate where it might be proved in full ...?

Peter

There is an argument in this wiki section.
The general theorem is that an $n$ dimensional vector space $V$ over field $S$ is isomorphic to the vector space that is the direct sum of $n$ copies of $F$, aka $F^n$.
We can prove it by, given a basis $b_1,...,b_n$ for $V$, defining a map from $V$ to $S^n$ that is the linear extension of the map that takes $b_k$ to the element of $S^n$ whose components are all $0_S$ except for a $1_S$ in the $k$th position.

Normally one would use $F$ for the field, but I've used $S$ to be consistent with the notation used above, since $F$ is already being used for something else (the vector space $V$).

It is straightforward, but instructive, to show that the map is an isomorphism.

 

[Math Amateur]

Thanks for all your help Andrew ...

... I think I now understand this exercise, thanks to you ...

I have a solution that is a very slight variant of your solution ...

============================================================================

$F$ finite field of characteristic $p$$\Longrightarrow$ prime subfield of $F$ is isomorphic to $\mathbb{F}_p$ and $p$ must be prime ... (Lidl and Niederreiter, Introduction to Finite Fields ... ... , Theorem 1.78 ... ... )also$F$ finite means that $F$ has finite dimension over $\mathbb{F}_p$, say dimension of $F$ over $\mathbb{F}_p = n$$\Longrightarrow$ Basis for $F$ over $\mathbb{F}_p$ has $n$ elements$\Longrightarrow$ all elements of $F$ are uniquely expressible as

$c_1 v_1 + c_2 v_2 + \ ... \ ... \ \ c_n v_n$ ... ... ... (1)where $c_1, c_2, \ ... \ ... \ \ , c_n \in \mathbb{F}_p$

and

$v_1, v_2, \ ... \ ... \ \ , v_n \in F$
$\Longrightarrow$ number of elements in $F, |F| = p^n$

since each $c_i$ in expression (1) has $p$ possibilities ... ...=========================================================================Is the above correct?Could someone please critique solution/proof ... is it rigorous ...?Peter