abstract-algebraextension-fieldfield-theoryring-theory
Let $E$ be an extension field of a finite field $F$ , where $F$ has $q$ elements. Let $a \in E$ be algebraic over $F$ of degree $n$. Prove that $F(a)$ has $q^n$ elements.
I am not sure how to do this one, but furthermore, what does $a$ being algebraic over $F$ of degree $n$ mean? Does it mean the polynomial $a$ solves in $F$ is of degree $n$?
$F(\alpha)$ means the smallest field containing both $F$ and $\alpha$.
$\gamma$ algebraic over $F$ means that there is a non-zero polynomial $p(X) \in F[X]$ (i.e., a polynomial with coefficients in $F$) with $p(\gamma) = 0$. (And transcendental means such a polynomial does not exist).
Now the problem itself. The situation is as follows.
Since $\alpha$ is algebraic over $F(\beta)$, there is a non-zero polynomial $f(X) \in F(\beta)[X]$ with $f(\alpha) = 0$. The coefficients are elements of $F(\beta)$, but clearing denominators we may as well assume they are elements of $F[\beta]$.
So, $f(\alpha)$ is a polynomial expression in both $\alpha$ and $\beta$ and we can see it as a polynomial expression $g(\beta)$ in $\beta$ with coefficients in $F[\alpha]$, i.e., $g(Y) \in F[\alpha][Y]$. (To be precise, there is a polynomial $h(X,Y) \in F[X,Y]$ such that $f(X) = h(X,\beta)$ and $g(Y) = h(\alpha,Y)$.) Now $0 = f(\alpha) = g(\beta)$.
What is still left to show is that $g(Y)$ is not the zero polynomial, i.e., that not all its coefficients are $0$. But its coefficients are of the form $c(\alpha)$ with $c(X) \in F[X]$ and because $\alpha$ is transcendental over $F$, $c(\alpha)$ is $0$ only if $c(X) = 0$. So, if $g(Y)$ were the zero polynomial, so would $f(X)$ be.
Example. Take $\alpha = T^2$ and $\beta = T^3$ in the field ${\mathbb Q}(T)$ of rational functions over ${\mathbb Q}$. Then $\alpha$ is transcendental over ${\mathbb Q}$. Also, $\beta$ is algebraic over ${\mathbb Q}(\alpha)$ as it satisfies $\beta^2 - \alpha^3 = 0$ (i.e., $\beta$ is a root of the polynomial $Y^2 - \alpha^3$ over ${\mathbb Q}(\alpha)$). Exactly the same relation shows that $\alpha$ is algebraic over ${\mathbb Q}(\beta)$ (as $\alpha$ is a root of the polynomial $\beta^2 - X^3$ over ${\mathbb Q}(\beta)$).
It's safe to add one element at a time.
If $K = F(\alpha)$ is a field extension of $F$, then $K \cong F[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ over $K$. Thus, if $R$ is any ring over $F$, then the ring homomorphisms $K \to L$ are in one-to-one correspondence with elements $\beta \in R$ satisfying $f(\beta) = 0$.
Now let $K$ and $F(\alpha)$ be field extensions of $F$. You construct the ring $R = K[x] / f(x)$, where $f(x)$ is the minimal polynomial of $\alpha$ (over $F$). It is indeed possible that $R$ is not a field; this happens if and only if $f(x)$ is not irreducible.
So to fix things, what you need is to find an irreducible polynomial $g(x)$ over $K$ such that one of the roots of $g(x)$ is a root of $f(x)$....
An alternative approach, I think, is to recognize that $R$ constructed above is a product of fields.
Best Answer
The statement "$a$ is algebraic over $F$ of degree $n$" means two things together:
In other words, the statement means that the minimal polynomial of $a$ over $F$ has degree $n$. For example, convince yourself that the following claims are true:
Here's a second characterization of degree, one which you'll find useful in solving your problem. If $a$ is algebraic over $F$ of degree $n$ then the set $\{1,a,a^2,\dots,a^{n-1}\}$ forms a basis for the vector space $F(a)$. In other words, we can think of $F(a)$ as being a vector space over $F$ and the dimension of the vector space is the degree of $a$. For vector spaces over finite fields, what is the relationship between dimension and cardinality?