MHB Finite Fields and Splitting Fields

[Math Amateur]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Theorem 6.5.2.

I need help with the proof of the Theorem.

Theorem 6.5.2 and its proof read as follows:View attachment 2842In the conclusion of the proof, Beachy and Blair write the following:

" ... ... Hence, since F is generated by these roots, it is a splitting field of f(x) over its prime subfield."

I am concerned about how exactly Beachy and Blair reach their conclusion that F is a splitting field for f(x) ... ...

Now Beachy and Blair define a splitting field as follows:View attachment 2843Now the first condition in the definition is achieved in the proof of the Theorem ... but how/why do we have the second condition for a splitting filed hold ... the implication of Beachy and Blair is that F being generated by the $$p^n$$ roots is the same as adjoining these roots to K, where K is the prime subfield of F ... but how/why does this follow?

Further, is it always the case that F being generated by n roots is the same as adjoining these roots to the subfield in question - or is Beachy and Blair's conclusion true because K is the prime subfield?

I would appreciate some help in clarifying the above point.

Peter

 

[Deveno]

Well, you could look at it this way:

Suppose $E$ is an extension of $\Bbb Z_p$, but a subfield of $F$.

Then $E$ cannot contain ALL of the roots of $f(x) = x^{p^n} - x$, because it doesn't have enough elements. So $f$ cannot split in any subfield of $F$ and it DOES split in $F$.

Now for any subset $S \subseteq F$, we certainly have $\Bbb Z_p(S) \subseteq F$ (they might be equal, they might not, it depends on $S$).

In particular, $\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$, where $\alpha_j$ is any non-zero element of $F$.

Just as clearly: $F^{\ast} = \{\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}\} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1})$

This means:

$F^{\ast} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$.

Now $F$ and $F^{\ast}$ differ by one element, $0$. Since $0 \in \Bbb Z_p$, we have:

$\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) = F$

which is condition 2.

Note that the list of roots is somewhat redundant, because $p - 1$ of them are already in $\Bbb Z_p$.

 

[Math Amateur]

Well, you could look at it this way:

Suppose $E$ is an extension of $\Bbb Z_p$, but a subfield of $F$.

Then $E$ cannot contain ALL of the roots of $f(x) = x^{p^n} - x$, because it doesn't have enough elements. So $f$ cannot split in any subfield of $F$ and it DOES split in $F$.

Now for any subset $S \subseteq F$, we certainly have $\Bbb Z_p(S) \subseteq F$ (they might be equal, they might not, it depends on $S$).

In particular, $\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$, where $\alpha_j$ is any non-zero element of $F$.

Just as clearly: $F^{\ast} = \{\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}\} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1})$

This means:

$F^{\ast} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$.

Now $F$ and $F^{\ast}$ differ by one element, $0$. Since $0 \in \Bbb Z_p$, we have:

$\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) = F$

which is condition 2.

Note that the list of roots is somewhat redundant, because $p - 1$ of them are already in $\Bbb Z_p$.

Thanks Deveno ... Just working through your post carefully now ...

Peter

 

[Math Amateur]

Well, you could look at it this way:

Suppose $E$ is an extension of $\Bbb Z_p$, but a subfield of $F$.

Then $E$ cannot contain ALL of the roots of $f(x) = x^{p^n} - x$, because it doesn't have enough elements. So $f$ cannot split in any subfield of $F$ and it DOES split in $F$.

Now for any subset $S \subseteq F$, we certainly have $\Bbb Z_p(S) \subseteq F$ (they might be equal, they might not, it depends on $S$).

In particular, $\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$, where $\alpha_j$ is any non-zero element of $F$.

Just as clearly: $F^{\ast} = \{\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}\} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1})$

This means:

$F^{\ast} \subseteq \Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) \subseteq F$.

Now $F$ and $F^{\ast}$ differ by one element, $0$. Since $0 \in \Bbb Z_p$, we have:

$\Bbb Z_p(\alpha_1,\alpha_2,\dots,\alpha_{p^n-1}) = F$

which is condition 2.

Note that the list of roots is somewhat redundant, because $p - 1$ of them are already in $\Bbb Z_p$.

Thanks Deveno ... BUT ... I am slightly uneasy that I am fully understanding your analysis ... so forgive me if I check a couple of points with you to see if I understand fully what is going on ... ...

We have $$ \mathbb{Z}_p = \{ 0, 1, 2, 3 ... ... p-2, p-1 \} $$

Then we add elements $$ \alpha_j $$ to create the field $$ \mathbb{Z}_p ( \alpha_1, \alpha_2, \ ... \ ... \ \alpha_{{p^n} - 1} )$$.

So $$ \mathbb{Z}_p ( \alpha_1, \alpha_2, \ ... \ ... \ \alpha_{{p^n} - 1}$$ is the minimal field containing the elements of $$ \mathbb{Z}_p = \{ 0, 1, 2, 3 ... ... p-2, p-1 \} $$ and the elements $$ \alpha_1, \alpha_2, \ ... \ ... \ , \alpha_{{p^n} - 1} $$

Is that correct?

If so, then some of the elements $$ \alpha_j $$ are equal to the elements of $$ \mathbb{Z}_p $$.

If the above is correct, then, further, of course, the elements of F* also contain the elements $$ \mathbb{Z}_p $$.

Is my analysis of the situation correct?

Would appreciate some clarification and help.

Peter
***EDIT*** I just noticed that you write:

" ... ... Note that the list of roots is somewhat redundant, because $p - 1$ of them are already in $\Bbb Z_p$"

I suspect that this answers my question ... ...

 

[Deveno]

Thanks Deveno ... BUT ... I am slightly uneasy that I am fully understanding your analysis ... so forgive me if I check a couple of points with you to see if I understand fully what is going on ... ...

We have $$ \mathbb{Z}_p = \{ 0, 1, 2, 3 ... ... p-2, p-1 \} $$

Then we add elements $$ \alpha_j $$ to create the field $$ \mathbb{Z}_p ( \alpha_1, \alpha_2, \ ... \ ... \ \alpha_{{p^n} - 1} )$$.

So $$ \mathbb{Z}_p ( \alpha_1, \alpha_2, \ ... \ ... \ \alpha_{{p^n} - 1}$$ is the minimal field containing the elements of $$ \mathbb{Z}_p = \{ 0, 1, 2, 3 ... ... p-2, p-1 \} $$ and the elements $$ \alpha_1, \alpha_2, \ ... \ ... \ , \alpha_{{p^n} - 1} $$

Is that correct?

If so, then some of the elements $$ \alpha_j $$ are equal to the elements of $$ \mathbb{Z}_p $$.

If the above is correct, then, further, of course, the elements of F* also contain the elements $$ \mathbb{Z}_p $$.

Is my analysis of the situation correct?

Would appreciate some clarification and help.

Peter
***EDIT*** I just noticed that you write:

" ... ... Note that the list of roots is somewhat redundant, because $p - 1$ of them are already in $\Bbb Z_p$"

I suspect that this answers my question ... ...

Yep.