Why Are the Roots of xf(x) and xg(x) Distinct in Finite Fields?

[Math Amateur]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Proposition 6.5.5.

I need help with the proof of the proposition.

Proposition 6.5.5 and its proof read as follows:View attachment 2848
View attachment 2849In the proof of Proposition 6.5.5 Beachy and Blair write:

" ... ... Since $$F$$ is the splitting field of $$xf(x)$$ over $$K$$ with distinct roots, it must contain all $$p^n$$ distinct roots of $$xg(x)$$ ... "

Although the logic of this statement seems plausible given that $$g(x)$$ is a divisor of $$f(x)$$, I am not sure of the exact logic here ... can someone please give a rigorous explanation of exactly why this follows ...

A second question is this: how do we know that the roots of $$xf(x)$$ and $$xg(x)$$ are distinct?

Any help will be appreciated.

Peter

 

[Euge]

I am reading Beachy and Blair's book: Abstract Algebra (3rd Edition) and am currently studying Proposition 6.5.5.

I need help with the proof of the proposition.

Proposition 6.5.5 and its proof read as follows:View attachment 2848
View attachment 2849In the proof of Proposition 6.5.5 Beachy and Blair write:

" ... ... Since $$F$$ is the splitting field of $$xf(x)$$ over $$K$$ with distinct roots, it must contain all $$p^n$$ distinct roots of $$xg(x)$$ ... "

Although the logic of this statement seems plausible given that $$g(x)$$ is a divisor of $$f(x)$$, I am not sure of the exact logic here ... can someone please give a rigorous explanation of exactly why this follows ...

A second question is this: how do we know that the roots of $$xf(x)$$ and $$xg(x)$$ are distinct?

Any help will be appreciated.

Peter

Since $g(x)$ is a divisor of $f(x)$, there is a polynomial $k(x)$ such that $f(x) = g(x)k(x)$. For any root $r$ of $g$, $g(r) = 0$, which implies $f(r) = g(r)k(r) = 0k(r) = 0$. Thus $r$ is a root of $f$. It follows that all the roots of $g$ are roots of $f$. Therefore, if we can show that the roots of $f$ are distinct, then we can claim that the roots of $g$ are distinct. Since 0 is not a root of $f$ (or $g$), this will show that the roots of $xf(x)$ (and hence the roots of $xg(x)$) are distinct.

In Beachy and Blair's book (second edition), Lemma 6.5.6 implies that the roots of $f$ are distinct. The strange thing is, this lemma comes after, not before Proposition 6.5.5. I'll state the lemma here:

Let $F$ be a field of characteristic $p$. If $n$ is a positive integer not divisible by $p$, then the polynomial $x^n - 1$ has no repeated roots in any extension field of $F$.

 

[Math Amateur]

Since $g(x)$ is a divisor of $f(x)$, there is a polynomial $k(x)$ such that $f(x) = g(x)k(x)$. For any root $r$ of $g$, $g(r) = 0$, which implies $f(r) = g(r)k(r) = 0k(r) = 0$. Thus $r$ is a root of $f$. It follows that all the roots of $g$ are roots of $f$. Therefore, if we can show that the roots of $f$ are distinct, then we can claim that the roots of $g$ are distinct. Since 0 is not a root of $f$ (or $g$), this will show that the roots of $xf(x)$ (and hence the roots of $xg(x)$) are distinct.

In Beachy and Blair's book (second edition), Lemma 6.5.6 implies that the roots of $f$ are distinct. The strange thing is, this lemma comes after, not before Proposition 6.5.5. I'll state the lemma here:

Let $F$ be a field of characteristic $p$. If $n$ is a positive integer not divisible by $p$, then the polynomial $x^n - 1$ has no repeated roots in any extension field of $F$.

Thanks Euge ... ... Very much appreciate your help,

Peter

 

[Deveno]

For a field $F$, it is a theorem that any polynomial of degree $m$ in $F[x]$ has at MOST $m$ distinct roots. This is easily shown using induction on $m$ and the division algorithm (this is where we require $F$ to be a field, so that $F[x]$ is Euclidean).

If $F$ is a finite field with $|F| = p^n$, then since $F^{\ast}$ is a finite group of order $p^n - 1$, it follows (from Lagrange) that for EVERY $\alpha \in F^{\ast}$:

$\alpha^{|F^{\ast}|} = 1$, that is: $\alpha^{p^n - 1} = 1$.

It follows, then, that every element of $F^{\ast}$ is thus a root of $x^{p^n - 1} - 1$. As we have $p^n - 1$ elements of $F^{\ast}$, and these elements ARE distinct, these must be ALL the roots of $x^{p^n - 1} - 1$, since we could have at most $p^n - 1$ such roots.

Perhaps this is why B&B do not invoke Lemma 6.5.6.

**********************

It is mildly instructive to investigate this for small values of $p$, and $n$. For example, suppose $p = 2, n = 2$. Then $p^n - 1 = 3$, and the roots we are seeking are roots of:

$x^3 - 1$.

Now we trivially have $1$ as a root, and so if $F = \{0,1,u,1+u\}$, so that $F^{\ast} = \{1,u,1+u\}$ it follows that $u,1+u$ must be roots of:

$\dfrac{x^3 - 1}{x - 1} = x^2 + x + 1$.

Note that here we have approached $\Bbb F_4$ from "the other side" than we did when we created $\Bbb F_4$ as:

$\Bbb Z_2(u) = \Bbb Z_2[x]/(x^2 + x + 1)$,

Instead of extending $\Bbb Z_2$ to a field that includes a root of an irreducible polynomial of degree 2 in $\Bbb Z_2[x]$, we started with a field of a given order, and recovered the irreducible polynomial that gives rise to the extension.

You might play around with this, and look at what you can discover with $p = 3$ and $n = 2,3,4$, for example.

 

[Math Amateur]

For a field $F$, it is a theorem that any polynomial of degree $m$ in $F[x]$ has at MOST $m$ distinct roots. This is easily shown using induction on $m$ and the division algorithm (this is where we require $F$ to be a field, so that $F[x]$ is Euclidean).

If $F$ is a finite field with $|F| = p^n$, then since $F^{\ast}$ is a finite group of order $p^n - 1$, it follows (from Lagrange) that for EVERY $\alpha \in F^{\ast}$:

$\alpha^{|F^{\ast}|} = 1$, that is: $\alpha^{p^n - 1} = 1$.

It follows, then, that every element of $F^{\ast}$ is thus a root of $x^{p^n - 1} - 1$. As we have $p^n - 1$ elements of $F^{\ast}$, and these elements ARE distinct, these must be ALL the roots of $x^{p^n - 1} - 1$, since we could have at most $p^n - 1$ such roots.

Perhaps this is why B&B do not invoke Lemma 6.5.6.

**********************

It is mildly instructive to investigate this for small values of $p$, and $n$. For example, suppose $p = 2, n = 2$. Then $p^n - 1 = 3$, and the roots we are seeking are roots of:

$x^3 - 1$.

Now we trivially have $1$ as a root, and so if $F = \{0,1,u,1+u\}$, so that $F^{\ast} = \{1,u,1+u\}$ it follows that $u,1+u$ must be roots of:

$\dfrac{x^3 - 1}{x - 1} = x^2 + x + 1$.

Note that here we have approached $\Bbb F_4$ from "the other side" than we did when we created $\Bbb F_4$ as:

$\Bbb Z_2(u) = \Bbb Z_2[x]/(x^2 + x + 1)$,

Instead of extending $\Bbb Z_2$ to a field that includes a root of an irreducible polynomial of degree 2 in $\Bbb Z_2[x]$, we started with a field of a given order, and recovered the irreducible polynomial that gives rise to the extension.

You might play around with this, and look at what you can discover with $p = 3$ and $n = 2,3,4$, for example.

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

Peter