Sums and products of algebraic numbers

Solution 1:

Lemma. If $F$ is a finite extension of $E$, and $a\in F$, then $a$ is algebraic over $E$.

Proof. Let $[F:E]=n$. Then $1,a,a^2,\ldots,a^n$ are linearly dependent over $F$, so there exists $c_0,c_1,\ldots,c_n\in E$, not all zero, such that $$c_01 + c_1a + \cdots + c_na^n = 0.$$ Set $f(x) = c_0 + c_1x+\cdots+c_nx^n\in F[x]$. Then $f(x)\neq 0$, and $f(a) = 0$, so $a$ is algebraic over $E$. $\Box$

Lemma. Let $K$ be an extension of $E$, and let $a,b\in K$ be algebraic over $F$. Then $[F(a,b):F(a)]\leq [F(b):F]$.

Proof. $[F(a,b):F(a)]$ is the degree of the minimal polynomial of $b$ over $F(a)$. Since the minimal polynomial of $b$ over $F$ is a multiple of the minimal polynomial of $b$ over $F(a)$, the latter has degree less than or equal to the former; the degree of the former equals $[F(b):F]$, so we are done. $\Box$

Corollary. If $a$ and $b$ are algebraic over $F$, then so are $a+b$ and $ab$.

Proof. Fix an algebraic closure of $F$; $[F(a,b):F] = [F(a,b):F(a)][F(a):F]\leq [F(b):F][F(a):F] \lt\infty$. Therefore, $F(a,b)$ is a finite extension of $F$, so each element of $F(a,b)$ is algebraic over $F$. In particular, $a+b,ab\in F(a,b)$, so they are algebraic over $F$. $\Box$