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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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    $\begingroup$ 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. $\endgroup$ – Michael Hardy Jun 7 '12 at 14:15
  • $\begingroup$ Related: math.stackexchange.com/questions/1277753, math.stackexchange.com/questions/331017 $\endgroup$ – Watson Nov 29 '18 at 20:03
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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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    $\begingroup$ I think you mean "where we regard this resultant as a polynomial in $z$". $\endgroup$ – Patrick Da Silva Jun 7 '12 at 14:21
  • $\begingroup$ @Ragib: Nice. But we need to check that the resultant is not the zero polynomial. $\endgroup$ – falang Dec 22 '13 at 1:44
  • $\begingroup$ @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. $\endgroup$ – Cloudscape Feb 3 '16 at 13:22
  • $\begingroup$ 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? $\endgroup$ – BharatRam Aug 4 '16 at 14:19
  • $\begingroup$ @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. $\endgroup$ – AmbretteOrrisey Dec 14 '18 at 16:30
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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,

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    $\begingroup$ 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. $\endgroup$ – AmbretteOrrisey Dec 14 '18 at 16:25
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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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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jun 7 '12 at 14:35
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    $\begingroup$ @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 $q$. $\endgroup$ – Qiaochu Yuan Jun 7 '12 at 14:41
  • $\begingroup$ I didn't know those sum/products had the name of Kronecker. Thanks for that info! $\endgroup$ – Patrick Da Silva Jun 7 '12 at 14:58
  • $\begingroup$ 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. $\endgroup$ – Juan Carlos Ortiz Apr 22 '19 at 19:03
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    $\begingroup$ @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 $q$. $\endgroup$ – Qiaochu Yuan Apr 22 '19 at 20:36
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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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    $\begingroup$ 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. $\endgroup$ – Patrick Da Silva Jun 7 '12 at 14:09
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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).

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