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Suppose $E/F$ is a field extension and $\alpha, \beta \in E$ are algebraic over $F$. Then it is not too hard to see that when $\alpha$ is nonzero, $1/\alpha$ is also algebraic. If $a_0 + a_1\alpha + \cdots + a_n \alpha^n = 0$, then dividing by $\alpha^{n}$ gives $$a_0\frac{1}{\alpha^n} + a_1\frac{1}{\alpha^{n-1}} + \cdots + a_n = 0.$$

Is there a similar elementary way to show that $\alpha + \beta$ and $\alpha \beta$ are also algebraic (i.e. finding an explicit formula for a polynomial that has $\alpha + \beta$ or $\alpha\beta$ as its root)?

The only proof I know for this fact is the one where you show that $F(\alpha, \beta) / F$ is a finite field extension and thus an algebraic extension.

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In Herstein's Topics in Algebra this fact is proved, and although I understood the proof, I realized at some point that I didn't know how to find the minimal polynomial of $\alpha+\beta$ if I knew those of $\alpha$ and $\beta$. So I went back to the proof in Herstein's book and saw that if you read it with that question in mind, you actually get an algorithm for that. – Michael Hardy Jun 7 '12 at 14:15
up vote 19 down vote accepted

The relevant construction is the Resultant of two polynomials. If $x$ and $y$ are algebraic and $P(x) = Q(y) = 0$ and $\deg Q=n$ then $z=x+y$ is a root of the resultant of $P(x)$ and $Q(z-x)$ (where we take this resultant by regarding $Q$ as a polynomial in only $x$) and $t=xy$ is a root of the resultant of $P(x)$ and $x^n Q(t/x).$

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I think you mean "where we regard this resultant as a polynomial in $z$". – Patrick Da Silva Jun 7 '12 at 14:21
@Ragib: Nice. But we need to check that the resultant is not the zero polynomial. – falang Dec 22 '13 at 1:44
@falang If the resultant were the zero polynomial, then in particular $P(x)$ and $Q(z - x)$ would have common divisors (polynomials in $x$ of degree $\ge 1$) everywhere. – Cloudscape Feb 3 at 13:22

Let $\alpha$ have minimal polynomial $p(x)$ and let $\beta$ have minimal polynomial $q(x)$. Then $V = F[x, y]/(p(x), q(y))$ is a finite-dimensional vector space over $F$ of dimension $\deg p \deg q$ (it is not necessarily the same dimension as $F(\alpha, \beta)$, for example when $\alpha = \beta$); moreover, it has an explicit basis $$x^i y^j : 0 \le i < \deg p, 0 \le j < \deg q.$$

$xy$ and $x + y$ act by left multiplication on $V$ and one can write down explicit matrices for this action in the basis above in terms of the coefficients of $p$ and $q$. Now apply the Cayley-Hamilton theorem.

This argument proves the stronger result that if $F$ is the fraction field of some domain $D$ and $\alpha, \beta$ are integral over $D$ (hence $p, q$ are monic with coefficients in $D$) then so are $\alpha \beta, \alpha + \beta$.

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Is your argument similar in flavor to my second answer? I.e. is your construction with $F[x,y]/(p(x),q(y))$ isomorphic to the construction with the tensor product? +1 by the way. – Patrick Da Silva Jun 7 '12 at 14:35
@Patrick: yes, it's essentially identical. The matrices you get for $xy$ and $x + y$ are the Kronecker product and Kronecker sum ( of the companion matrices ( of $p$ and $q$. – Qiaochu Yuan Jun 7 '12 at 14:41
I didn't know those sum/products had the name of Kronecker. Thanks for that info! – Patrick Da Silva Jun 7 '12 at 14:58

Okay, I'm giving a second answer because this one is clearly distinct from the first one. Recall that finding a polynomial over which $\alpha+\beta$ or $\alpha \beta$ is a root of $p(x) \in F[x]$ is equivalent to finding the eigenvalue of a square matrix over $\mathbb Q$, since you can link the polynomial $p(x)$ to the companion matrix $C(p(x))$ which has precisely characteristic polynomial $p(x)$, hence the eigenvalues of the companion matrix are the roots of $p(x)$.

If $\alpha$ is an eigenvalue of $A$ with eigenvector $x \in V$ and $\beta$ is an eigenvalue of $B$ with eigenvector $y \in W$, then using the tensor product of $V$ and $W$, namely $V \otimes W$, we can compute $$ (A \otimes I + I \otimes B)(x \otimes y) = (Ax \otimes y) + (x \otimes By) = (\alpha x \otimes y) + (x \otimes \beta y) = (\alpha + \beta) (x \otimes y) $$ so that $\alpha + \beta$ is the eigenvalue of $A \otimes I + I \otimes B$. Also, $$ (A \otimes B)(x \otimes y) = (Ax \otimes By) = (\alpha x \otimes \beta y) = \alpha \beta (x \otimes y) $$ hence $\alpha \beta$ is the eigenvalue of the matrix $A \otimes B$. If you want explicit expressions for the polynomials you are looking for, you can just compute the characteristic polynomial of the tensor products.

Hope that helps,

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Technically, you could find the automorphisms of the Galois closure of $F(\alpha,\beta)$ over $F$ (assuming this extension is separable) and compute the polynomial $$ \prod_{\sigma \in \mathrm{Gal}}(x- \sigma(\alpha+\beta)) $$ or the same with $\alpha \beta$, but I don't believe this is what you are looking for. Since you can define Galois closures without knowing that $\alpha + \beta$ and $\alpha \beta$ are also algebraic, it is a legitimate way of proving it, but not a practical nor pedagogical one.

Hope that helps,

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Hm. I realize that I need the fact that $|\mathrm{Gal}(F(\alpha,\beta)/F)| (= [F(\alpha,\beta) : F]) < \infty$ for this construction to make sense, hence it's not really that much worth it, but at least it gives intuition. – Patrick Da Silva Jun 7 '12 at 14:09

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