Algebraic numbers are a field I want to prove that algebraic numbers are a field using extensions field theory. This seems to be very easy, so I feel strange for not understanding this.
The exercise says: let $E/F$ be an extension and $a,b\in E,a\neq 0$ algebraics over $F$. Then $a+b,ab,a^{-1}$ are algebraic. (Hint: $F(a,b)$ is a finite dimensional space over $F$).
I understood immediately the hint. $[F(a,b):F(b)]$ is finite because $a$ is algebraic over $F$, hence over $F(b)$. $[F(b):F]$ is finite, and therefore $[F(a,b):F]$ is finite. So, $F(a,b)$ is a finite dimensional space over $F$.
I also see that $a+b,ab,a^{-1}\in F(a,b)$.
But I don't really see why this implies that $a+b,ab,a^{-1}$ are algebraic over $F$ (I read in another page that this is trivial). Can you explain to me?
Thank you.
 A: Let $K/F$, and let $\alpha \in K$. Then $\alpha$ is algebraic over $F$ if and only if $[F(\alpha):F]<\infty$:
If $\alpha$ is algebraic over $F$, I'm sure you've seen in your textbook that $F(\alpha)$ is a vector space with basis $\{1,\alpha,\alpha^2, \dots, \alpha^{n-1}\}$ where $n$ is the degree of the minimal polynomial of $\alpha$ over $F$.
Conversely, if $[F(\alpha):F]=n<\infty$ then the set $\{1,\alpha, \dots, \alpha^n\}$ consists of $n+1$ vectors in an $n$-dimensional vector space, so it's not linearly independent. But this means exactly that $\alpha$ is a root of a nonzero polynomial of degree $n$. 
Now since $\alpha + \beta$, $\alpha \beta$, etc. are in a finite-dimensional extension of $F$, the extensions $F(\alpha+\beta)/F$, $F(\alpha \beta)/F$, etc. are all finite. 
A: You've pretty much got it. Now, in our case $F = \mathbb{Q}$ and $E = \mathbb{A}$ which I'll use as a letter for the algebraic numbers since I can't think of a better one. In particular, we have that for any $a, b \in \mathbb{A}$, we know $$\mathbb{Q} \subset \mathbb{Q}(a,b) \subset \mathbb{A}.$$ But since $a+b$, $ab$, and $a^{-1}$ (also $b^{-1}$ but whatever) are in $\mathbb{Q}(a,b)$, they must also be in $\mathbb{A}$, since $\mathbb{Q}(a, b)$ is a subfield. Since $a, b$ were arbitrary in $\mathbb{A}$, we're done!
Often we look at field extensions and lose the perspective of seeing the field we extended as a subfield of the extension but it's helpful to see it this way here.
A: If $L/F$ is any field extension and $c\in L$ has the property that $[F(c):F]$ is finite, then $c$ is algebraic over $F$. Of course, if you already know that $[L:F]$ is finite, then the fact that $[L:F]=[L:F(c)][F(c):F]$ implies that $[F(c):F]$ is finite for every $c\in L$.
In your situation, you have $L=F(a,b)$ and $c\in\{a+b,ab,a^{-1}\}$.
