Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 18 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 $x^n Q(t/x).$

share|cite|improve this answer
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 $\alpha \beta, \alpha + \beta$.

share|cite|improve this answer
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,

share|cite|improve this answer

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,

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.