Finding degree of minimal polynomial of $a$ over $F$

Question

So while working through Abstract Algebra by Dan Saracino I've run into some confusion. I'm on a question that states the following:

Suppose $E$ is an extension field of $\mathbb{Z_5}$ and $E$ has exactly $78125$ elements. Find $\deg(a/\mathbb{Z_5})$ for every $a \in E- \mathbb{Z_5}$.

What I'm wondering is are there $78120$ elements that can be $a$ or am I over complicating this?

Answer 1

Notice that $5^7 = 78125$, so $E$ is a degree $7$ extension over $\mathbb{Z}_5$. Before we continue, recall two important theorems that'll come in handy:

• If $f(x) \in F[x]$ is irreducible and of degree $n$, then $F[x]/\langle f(x) \rangle \cong F[\alpha]$, where $\alpha$ is a root of $f(x)$, and further $F[\alpha]$ is a degree $n$ extension over $F$.

• If we have a tower of fields $F \subset L \subset K$, then $[K:F] = [K:L] \cdot [L:F]$.

Given the above, let's say $a \in E \setminus \mathbb{Z}_5$. Then we have $\mathbb{Z}_5 \subsetneq \mathbb{Z}_5[a] \subseteq E$. The first result tells us that if $m \in \mathbb{Z}_5[x]$ is the minimal polynomial for $a$, then $\deg(m) = [\mathbb{Z}_5[a]:\mathbb{Z}_5]$, which must be strictly greater than $1$ (why?).

Can you take it from here?

Answer 2

We calculate that

$78,125 = 5^7; \tag{1}$

it follows that $E$ is isomorphic to $GF(5^7)$, and that $[E:\Bbb Z_5] = 7$. Consider, for any $a \in E$, the field $\Bbb Z_5(a) \subset E$; we have

$[E:\Bbb Z_5(a)][\Bbb Z_5(a): \Bbb Z_5] = [E:\Bbb Z_5] = 7; \tag{2}$

$7$ being prime in $\Bbb Z$, its only positive integral divisors are $1$ and $7$ itself. Thus $[\Bbb Z_5(a):\Bbb Z_5] = 1$ or $[\Bbb Z_5(a):\Bbb Z_5] = 7$. In the former case, $a$ satisfies a first degree (linear) monic polynomial $f(x) \in \Bbb Z_5[x]$; since all such $f(x)$ are of the form $x - \beta$ for $\beta \in \Bbb Z_5$, this implies $a \in \Bbb Z_5$; also, every $a \in \Bbb Z_5$ satisfies $x - a$; thus we conclude that $[\Bbb Z_5(a): \Bbb Z_5] = 1$ precisely when $a \in \{ 0, 1, 2, 3, 4 \} = \Bbb Z_5$. For all other $a$, that is $a \in E - \Bbb Z_5$, we must have the second alternative, $[\Bbb Z(a); \Bbb Z] = 7$. Thus each such $a$ satisfies an irreducible polynomial $f(x) \in \Bbb Z_5[x]$ with $\deg f = 7$; the degree of $a$ over $\Bbb Z_5$,

$\deg(a/\Bbb Z_5) = 7. \tag{3}$

It is our great good fortune to have these methods of elementary field theory at our disposal; they save us from checking each of the $78,120$ elements of $E - \Bbb Z_5$ individually!

Note: Generalization: It appears that a similar argument may be applied to any field extension $E/F$ with $[E:F] = p$, a prime. Then for $a \in E - F$, $[E:F(a)][F(a):F] = p$ so $[F(a):F] = 1$ or $[F(a):F] = p$ etc. etc. etc. End of Note.