Math 55a: Honors Abstract Algebra
Homework Assignment #10 (10 November 2010):
Linear Algebra X:
Signatures and real polynomials; determinants and distances; Pontrjagin duality0
Quotes? We don’t need no stinkin’ quotes!
—adapted or misquoted from Blazing Saddles (1974), in turn adapted or mis-
quoted from The Treasure of the Sierra Madre1
Using the signature of a pairing to describe the roots of a real polynomial without repeated
factors:
1. i) (Chinese Remainder Theorem for polynomials) Let Fbe a field, P∈F[X] a nonzero
polynomial, and assume P=Qm
j=1 Pjfor some polynomials Pjsuch that there is
no polynomial Qof positive degree that divides more than one of the Pj. Let Abe
the F-algebra F[X]/(P), and for each jlet Ajbe the F-algebra F[X]/(Pj). [Recall
that (f) is the ideal in A[X] generated by f.] Prove that the algebra homomorphism
A→Qm
j=1 Ajwhose j-th coordinate is reduction mod Pjis an isomorphism. Deduce
(using a result from the previous problem set) that if the Pjare irreducible then
F[X]/(P) is a product of fields. [Yes, the usual Chinese Remainder Theorem can
be phrased in such terms as well, but linear algebra makes the polynomial version
simpler.]
ii) Now suppose F=Rand the Pjare irreducible. Show that Ajis isomorphic with
Ror Caccording as deg(Pj) = 1 or 2. Let h·,·i be the pairing on Adefined by
hf, gi= tr(f g ), that is, the trace of the multiplication-by-f g map on the finite-
dimensional real vector space A. Prove that this pairing is nondegenerate, and that
its signature is (r, s) where sis the number of jfor which Pjis quadratic and r−s
is the number of jfor which Pjis linear.
iii) Suppose that moreover each Pjis linear and none equals X, so that the roots are
all real and nonzero (as happens if Pis the characteristic polynomial of an invertible
Hermitian matrix, though if we have such a matrix we already know how to count
the roots of each sign.) Construct a p olynomial for which the signature formula of
part (ii) yields the counts of positive and negative roots of P.
The formula hf, gi= tr(f g) yields a pairing on F[X]/(P) for any Fand P. It is non-
degenerate if and only if Phas no repeated factors; this will be easier to see once we
have developed some more field algebra, but for now you can easily check it if Pfactors
completely in F[X] (i.e. has all roots in F).
Next some classical product formulas for determinants:
2. For elements x1, x2, . . . , xnof any field F, let V(x1, x2, . . . , xn) be the n×nmatrix
whose (i, j) entry is xj−1
i. Find a homomorphism Tfrom the group (F , +) to the
group of upper triangular n×nmatrices over F, such that
V(x1+t, x2+t, . . . , xn+t) = V(x1, x2, . . . , xn)T(t)
for all tand xi. Use this to derive inductively the formula Qn−1
i=1 Qn
j=i+1(xj−xi)
for the Vandermonde determinant ∆(x1, x2, . . . , xn) = det V(x1, x2, . . . , xn). What
is the determinant of the n×nmatrix whose (j, k) entry is Pn
i=1 xj+k−2
i?
0The “j” in “Pontrjagin” is pronounced as a “y”.
1According to Wikipedia’s “Stinking badges” page. Yes, Wikipedia has a page on “Stinking badges”!
