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How does one go about proving that the sums and products of two algebraic numbers over a field $F$ (say $a,b\in K$, where $K/F$ is a field extension) is also algebraic?

If we call $f_a$ and $f_b$ the min. poly's of $a$ and $b$, then I'm assuming the answer involves such polynomials. Perhaps looking at their roots in splitting fields for both of them? And finding a "big" splitting field, constructed from those two other ones?

In particular, I'd like a way of explicitly constructing the minimal polynomials $\ f_{ab}$ of $ab$ and $f_{a+b}$ of $a+b$.

I read somewhere that $g(x)=\Pi_j\Pi_i (x-\alpha_i\beta_j)$ works for $ab$, where the $\alpha_i$ and $\beta_j$ are the roots of $f_a$ and $f_b$, respectively, but I do not know why $g(x)\in F[x]$. Similar remarks for $a+b$

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Why not begin by finding the integer polynomial satisfied by $3^{1/2}+5^{2/3}$. – GEdgar May 5 '12 at 16:57
There's a helpful Keith Conrad handout on this matter, but I can't find it at the moment. – Dylan Moreland May 5 '12 at 17:41
Dylan probably means Theorem 2.3 (and the example after it) in, but the reality is that the simplest answer to your initial question does not involve the minimal polynomials of $a$ and $b$. Instead it involves linear algebra and the relation between algebraicity and finite-dimensionality of a field extension. Your initial question should be answered in almost any book that discusses field theory. Read some books! Trying to prove this in general with explicit polynomials is not the right first approach, I think. – KCd May 5 '12 at 18:08
By the way, your guess about what works for $ab$ will give you some nonzero polynomial in $F[x]$ with $ab$ as a root, but it is not guaranteed to be the minimal polynomial (even though, in practice, it often will be). There's a big difference between asking for some nonzero polynomial in $F[x]$ with a particular number as a root and asking for the minimal polynomial in $F[x]$ with a particular number as a root, as the latter task involves issues of irreducibility, which is quite sensitive to the nature of the base field $F$ (can't be captured by a universal algebraic formula). – KCd May 5 '12 at 18:12
Arturo's answer is essentially the one I had in mind involving ideas from linear algebra and dimensions of field extensions, rather than explicit polynomials. – KCd May 5 '12 at 22:09
up vote 7 down vote accepted

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 $F$. $\Box$

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