Roots Appear in Quanta

Alexander R. Perlis

We start with a special case. Consider an irreducible quintic polynomial

f(X) = X^{5} + a_{1}X^{4} + a_{2}X^{3} + a_{3}X^{2} + a_{4}X + a_{5}

with rational coefficients and with three real roots and one pair of complex conjugate roots. For example, f(X) could be X^{5} − 10X + 5.

Question. If α is a root of f, then how many roots of f lie in the field Q(α)?

The field Q(α) is obtained by adjoining the root α to Q. Thus Q(α) contains at least one root of f, and of course it can contain at most five roots of f.

Answer. The number r(f) of roots of f in Q(α) is 1. We prove that, for an arbitrary irreducible polynomial f and root α, r(f) divides the degree of f. For the quintic under discussion, adjoining one of the real roots cannot possibly produce the nonreal roots, so r(f), being a divisor of 5, must be 1.

An informal survey of books and colleagues indicates that the divisibility result “r(f) divides the degree” is not well known. In what follows, K is a field and, unless stated otherwise, all roots and field extensions are taken in a fixed algebraic closure K of K. When K = Q, we always take K inside the complex numbers so that we can speak of real roots and nonreal roots.

Theorem 1. Let f(X) in K[X] be an irreducible polynomial, and let α be a root of f. Set

r_{K}(f) := number of roots of f that lie in K(α),
s_{K}(f) := number of fields of the form K(α′), where α′ is a root of f.

Then r_{K}(f) is independent of the choice of α, and

r_{K}(f) · s_{K}(f) = cardinality of the set of roots of f.

In particular, r_{K}(f) divides the degree of f.

Concerning the last statement of the theorem: the cardinality of the set of roots of f is known as the separable degree of f, and it is well known that the separable degree divides the usual degree.

Proof. For this proof, let “root” mean “root of f” and let “stem field” signify a field of the form K(α′), where α′ is a root. Since f is irreducible, each stem field is K-isomorphic to the abstract field K[X]/(f(X)), whence any two stem fields are K-isomorphic. Isomorphisms take roots to roots, so r_{K}(f) is the same for each stem field. Each root α′ lies in precisely one stem field: it lies in K(α′), and if it also lies in K(α′′), then K(α′) ⊆ K(α′′), but because the two stem fields have the same degree over K (they are K-isomorphic), we must have K(α′) = K(α′′). In summary, the set of roots is partitioned by the stem fields into s_{K}(f) collections with r_{K}(f) roots in each collection, making r_{K}(f)·s_{K}(f) the cardinality of the set of roots.

The symbol r_{K}(f) is determined both by the polynomial f and by the base field K. When K is understood, as it was earlier when K = Q, the simpler notation r(f) can be used. There doesn’t seem to be an established name for the quantity r_{K}(f), and I propose: root quantum number of f over K. While this name initially sounds rather fancy for a simple concept, the following theorem shows that the roots of f really do come bundled in collections of size r_{K}(f).

Theorem 2. Let f(X) in K[X] be irreducible. If L/K is a field extension (not necessarily algebraic), then the number of roots of f in L is a multiple of r_{K}(f).

Proof. The proof of Theorem 1 exhibits a partition of the set of roots of f into collections of equal size r_{K}(f), where each collection has the property: in any field extension of K, the presence of one of the roots implies the presence of the remaining ones.

Remark. We can also see that the cardinality of the set of roots of f lying outside a given extension L/K (counted in an algebraically closed field containing L) is a multiple of r_{K}(f). Theorem 1 shows that r_{K}(f) divides the total number of roots, and Theorem 2 shows that r_{K}(f) divides the number of roots in L, so r_{K}(f) also divides the difference of these two numbers.

Corollary. If f(X) in Q[X] is irreducible, then the number of real roots of f is a multiple of r_{Q}(f). The same can be said about the number of nonreal roots.

Proof. Keeping the remark in mind, take L = R in Theorem 2.

Theorem 2 may be summarized as follows: roots appear in quanta. This places combinatorial restrictions on the way f can factor. For example, if f(X) in K[X] is irreducible and separable of degree 15, with α a root, then the factorization of f over K(α) cannot have the following form:

(linear)(linear)(linear)(quadratic)(quadratic)(octic).

To see this, assume for the sake of contradiction that the factorization of f over K(α) has the form indicated. Since f is separable, the three linear factors correspond to distinct roots of f in K(α), so r_{K}(f) = 3. The field L obtained from K(α) by adjoining the roots of the two quadratic factors has degree at most 4 over K(α). Thus L contains none of the roots of the octic factor, so L contains precisely seven of the roots of f. This contradicts the fact that the number of roots of f in L must be a multiple of three.

The interested reader can check that the root quantum number has the following three descriptions in terms of Galois theory. Let f be irreducible and separable over K, with Galois group G, viewed as a permutation group on the set of roots of f. Let H ⊂ G be the subgroup fixing a root α. Then:

i. r_{K}(f) is the number of roots fixed by H; ii. r_{K}(f) is the cardinality of Aut(K(α)/K); and iii. r_{K}(f) is the index [N_{G}(H) : H] of H in its normalizer.

Finally, it is instructive to think about the triples (K, n, r) that indicate the existence of an irreducible polynomial f(X) in K[X] of degree n with root quantum number r. The necessary condition discussed in this note is that r must divide n. Here are some exercises involving these triples:

1. Show that (Q, 2, 1) does not appear.

2. Find a field K for which (K, 2, 1) does appear.

3. Let r divide n. Show that there exists K for which (K, n, r) appears.

4. (Advanced) Let r divide n. Except for (Q, 2, 1), show that (Q, n, r) appears.

Solutions can be obtained from the author.

ACKNOWLEDGMENTS. David Marshall and Robert Perlis helped improve earlier versions of this note. Comments from two anonymous referees led to additional improvements. To all of you: my sincere thanks.

Department of Mathematics The University of Arizona Tucson, Arizona 85721–0089 aprl@math.arizona.edu