# How to prove that the sum and product of two algebraic numbers is algebraic? [duplicate]

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.

• 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. Jun 7 '12 at 14:15
• Nov 29 '18 at 20:03

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 $F$ (living in some algebraic extension of $F$), 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,

• Definitely clearly a distinct answer, that! ¶ So it looks like there is a choice then between two distnct methods: that of finding the determinant of a dense (m+n)×(m+n) matrix, & that of finding the determinant of a sparse mn×mn matrix. Casting in the mind the forms of the two matrices, it seems quite remarkable that they must yield the same result. I wouldn't be surprised either if it should transpire that using fairly ordinary algorithms for calculation of determinant the sheer amount of calculation were prettymuch exactly the the same. Dec 14 '18 at 16:25

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 $x^n Q(t/x).$

• I think you mean "where we regard this resultant as a polynomial in $z$". Jun 7 '12 at 14:21
• @Ragib: Nice. But we need to check that the resultant is not the zero polynomial. 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. Feb 3 '16 at 13:22
• Will this work for the sum and product of integral elements over a commutative ring? Will the resultants obtained this way be monic if $P$ and $Q$ are monic? Aug 4 '16 at 14:19
• @Patrick da Silva -- and you mean, leaving t or z as a symbol rather than substituting the value of t or z? I'm sure it must mean that ... but I do agree that it could have been & ought to have been made clearer in the exposition. Dec 14 '18 at 16:30

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 $\alpha \beta, \alpha + \beta$.

• 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. 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 (en.wikipedia.org/wiki/Kronecker_product) of the companion matrices (en.wikipedia.org/wiki/Companion_matrix) of $p$ and $q$. Jun 7 '12 at 14:41
• I didn't know those sum/products had the name of Kronecker. Thanks for that info! Jun 7 '12 at 14:58
• This is a great proof! Just curious... wouldn't you have to prove that the map $mult: F(u,v)\to GL(V)$ is injective? In other words, say $f$ is the char poly of $xy$, we know that $f(uv)$ acts as 0 on $V$, but to know that $f(uv)=0$ we need injectivity of the aforementioned "multiplication" map, no? Edit: Nevermind, injectivity is just a consequence of the fact that $F(u,v)$ is a domain. Apr 22 '19 at 19:03
• @Juan: yes, and by making it slightly more explicit you can even show that the characteristic polynomials have coefficients which are polynomials in the coefficients of $p$ and $q$. Apr 22 '19 at 20:36

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,

• 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. Jun 7 '12 at 14:09

Consider fields $$E \supseteq F$$, and elements $$\alpha, \beta \in E$$ algebraic over $$F$$. We want to show $$\alpha + \beta$$, $$\alpha \beta$$ are algebraic over $$F$$ too. If even one of $$\alpha, \beta$$ are $$0$$, the result is trivial, so let's take both $$\alpha, \beta$$ to be non-zero.

We have $$\alpha ^m + a_{m-1} \alpha ^{m-1} + \ldots + a_0 = 0$$ ( each $$a_i \in F$$ ), and $$\beta ^n + b_{n-1} \beta ^{n-1} + \ldots + b_0 = 0$$ ( each $$b_j \in F$$ ).

(The first equation lets us express all powers of $$\alpha$$ as $$F$$-combinations of $$1, \alpha, \ldots, \alpha ^{m-1}$$. Similarly for $$\beta$$)

Let $$Z := \, [ \, \alpha ^0 \beta ^0, \alpha ^0 \beta ^1, \ldots, \alpha ^0 \beta ^{n-1} ; \alpha ^1 \beta ^0, \ldots, \alpha ^1 \beta ^{n-1} ; \ldots ; \alpha ^{m-1} \beta ^{0}, \ldots, \alpha ^{m-1} \beta ^{n-1} \, ]^{T} \in E^{mn}$$

Now notice we can express $$(\alpha + \beta)Z$$ as $$M_1 Z$$ with $$M_1 \in F^{mn \times mn}$$. So $$( ( \alpha + \beta ) I - M_1 ) Z = 0$$, and as $$Z \neq 0$$ we have $$\det( (\alpha + \beta)I - M_1 ) = 0$$. Hence $$\alpha + \beta$$ is a root of the polynomial $$P(t) := \det( tI - M_1 ) \in F[t]$$, and is therefore algebraic over $$F$$. Similarly we can show $$\alpha \beta$$ is algebraic over $$F$$ (Write $$\alpha \beta Z = M_2 Z$$ and proceed as above).