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Àëãåáðà II, îñåííèé ñåìåñòð 2014 ã.
Çàäà÷è äëÿ ñåìèíàðà 4.
ôàêóëüòåò ìàòåìàòèêè, ÍÈÓ ÂØÝ
(à) Íàéäèòå ìíîãî÷ëåí ÷åòâ¼ðòîé ñòåïåíè, êîðíÿìè êîòîðîãî ÿâëÿþòñÿ
êâàäðàòû êîìïëåêñíûõ êîðíåé ìíîãî÷ëåíà
Çàäà÷à 1.
x4 + 2x3 − x + 3.
(á) Ïóñòü η ïåðâîîáðàçíûé êîðåíü ñòåïåíè k èç åäèíèöû, à f (x) ∈ C[x] ìíîãî÷ëåí. Äîêàæèòå, ÷òî ìíîãî÷ëåí f (x)f (ηx) · · · f (ηk−1x) ïðåäñòàâèì â âèäå h(xk ) äëÿ
íåêîòîðîãî ìíîãî÷ëåíà h(x), è êîðíè ìíîãî÷ëåíà h ýòî â òî÷íîñòè k-ûå ñòåïåíè
êîðíåé ìíîãî÷ëåíà f .
(à) Äîêàæèòå, ÷òî äèñêðèìèíàíò ìíîãî÷ëåíà P (t) = (t − x1) · · · (t − xn)
ðàâåí êâàäðàòó îïðåäåëèòåëÿ Âàíäåðìîíäà
Çàäà÷à 2.
n−1 n−1
x1
x
n−2 2n−2
x1
x2
.
...
∆(x1 , . . . , xn ) := ..
x
x2
1
1
1
. . . xnn−1 . . . xn−2
n . . . ... .
. . . xn ...
1 (á) Ïóñòü f (x1, . . . , xn) êîñîñèììåòðè÷åñêèé ìíîãî÷ëåí îò x1,. . . , xn. Äîêàæèòå
÷òî, f (x1, . . . , xn) = ∆(x1, . . . , xn)g(x1, . . . , xn), ãäå g(x1, . . . , xn) ñèììåòðè÷åñêèé
ìíîãî÷ëåí.
(â) Ïóñòü h(x1, . . . , xn) ñèììåòðè÷åñêèé ìíîãî÷ëåí îò x1,. . . , xn, ïðè÷¼ì
h(x1 , x1 , x3 , . . . , xn ) = 0.
Äîêàæèòå ÷òî, h(x1, . . . , xn) = ∆2(x1, . . . , xn)g(x1, . . . , xn), ãäå g(x1, . . . , xn) ñèììåòðè÷åñêèé ìíîãî÷ëåí.
Äàíà 3 × 3-ìàòðèöà A, è èçâåñòíî, ÷òî tr A = tr A2 = tr A3 = 1. Íàéäèòå
õàðàêòåðèñòè÷åñêèé ìíîãî÷ëåí ìàòðèöû A.
Äëÿ âñåõ íàòóðàëüíûõ ÷èñåë m è n âû÷èñëèòå
(à) äèñêðèìèíàíò ìíîãî÷ëåíà xn + xn−1 + . . . + 1;
(á) ðåçóëüòàíò ìíîãî÷ëåíîâ xn + xn−1 + . . . + 1 è xm + xm−1 + . . . + 1.
Ïóñòü R îáëàñòü öåëîñòíîñòè, è f, g ∈ R[t]. Äîêàæèòå, ÷òî Res(f, g)
âñåãäà ëåæèò â èäåàëå êîëüöà R[t], ïîðîæä¼ííîì f è g.
Çàäà÷à 3.
Çàäà÷à 4.
Çàäà÷à 5.
2
(a) Find a polynomial of degree 4 whose roots coincide with the 4-th powers
of the complex roots of the polynomial
Problem 1.
x4 + 2x3 − x + 3.
(á) Let η be a primitive k-th root of unity, and f (x) ∈ C[x] a polynomial. Prove that
the polynomial f (x)f (ηx) · · · f (ηk−1x) can be represented as h(xk ) for some polynomial
h(x), and the roots of h coincide with the k -th powers of the roots of f .
(a) Prove that the discriminant of the polynomial P (t) = (t − x1) · · · (t − xn)
is equal to the square of the Vandermonde determinant
Problem 2.
n−1 n−1
x1
x
n−2 2n−2
x1
x2
.
...
∆(x1 , . . . , xn ) := ..
x
x2
1
1
1
. . . xn−1
n . . . xn−2
. . . n... .
. . . xn ...
1 (b) Let f (x1, . . . , xn) be an alternating polynoial in x1,. . . , xn. Prove that f (x1, . . . , xn) =
∆(x1 , . . . , xn )g(x1 , . . . , xn ), where g(x1 , . . . , xn ) is a symmetric polynomial.
(c) Let h(x1, . . . , xn) be a symmetric polynomial in x1,. . . , xn such that
h(x1 , x1 , x3 , . . . , xn ) = 0.
Prove that h(x1, . . . , xn) = ∆2(x1, . . . , xn)g(x1, . . . , xn), where g(x1, . . . , xn) is a symmetric
polynomial.
A 3 × 3-matrix A satises tr A = tr A2 = tr A3 = 1. Find the characteristic
polynomial of A.
For all positive integer m and n, compute
(a) the discriminant of xn + xn−1 + . . . + 1;
(b) the resultant of xn + xn−1 + . . . + 1 and xm + xn−1 + . . . + 1.
Let R be an integral domain, and f, g ∈ R[t]. Show that Res(f, g) always
lies in the ideal of R[t] generated by f and g.
Problem 3.
Problem 4.
Problem 5.
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