Let $E/F$ be a field extension.Let $A$ be the set of algebraic elements over $F$: elements $a$ so that there exists an $f\in F[x]$ so that $f(x)$. The lecturer has mentioned that $A$ is a subfield of $F$ and I can't understand why.
I've searched for an answer here and got the main idea: if $[F(a): F], [F(b), F]$ are finite, then so must be $[F(a, b):F]$. But I can't see why it's so.
My attempt: we know that there are degrees $m, n$ so that $1, a, ... , a^m$ is a basis for $F(a)$ and $1, b, ... , b^n$ is a basis for $F(b)$. It seems intuitive to claim that $\left \{ a^ib^j \right \}$ is a basis for $F(a, b).$ It is certainly true for the ring $F[a, b]$, but, for example, why would $\frac{1}{a^2b+ab+1}$ be a linear combination of such?
And if I'm already asking: the intuitive way to attack this problem is, given a polynomial with $a$ as a root and one with $b$ as a root, constructing a polynomial with $a+b$ or $ab$ as a root. It seems like there is no good way to do so. Am I right?
Thank you!
There are two solutions : one is going in with your hands and finding polynomials that work for $a+b, ab$.
Another way is a more linear kind of thing.
Consider $a,b$ algebraic over $F$. Then $ab, a+b \in F(a,b)$.
But $b$ is a root of a certain polynomial with coefficients in $F$, so we can say the same about a polynomial with coefficients in $F(a)$, therefore $[F(a)(b):F(a)]$ is finite.
But also $[F(a) : F]$ is finite.
Now the telescopic basis theorem states that $[F(a,b): F(a)][F(a): F] = [F(a,b) : F]$. But $F(a,b) = F(a)(b)$ and so $[F(a,b):F(a)]$ is finite and thus $[F(a,b):F]$ is also finite : $a+b, ab$ are algebraic over $F$.
The argument you didn't see is the telescopic basis theorem. It's proved by noting that if $(u_i)$ is a basis of $F(a,b)$ over $F(a)$ and $(v_j)$ of $F(a)$ over $F$, then $(v_j u_i)$ is a basis of $F(a,b)$ over $F$ (which is simple to show, once you know what you have to show)