{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

College Algebra Exam Review 431

College Algebra Exam Review 431 - 441 9.6 SYMMETRIC...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 441 9.6. SYMMETRIC FUNCTIONS I Proof. Exercise 9.6.2. Proposition 9.6.2. The ﬁeld K.x1 ; : : : ; xn / of rational functions is Galois over the ﬁeld K S .x1 ; : : : ; xn / of symmetric rational functions, and the Galois group AutK S .x1 ;:::;xn / .K.x1 ; : : : ; xn // is Sn . Proof. By Exercise 9.6.1, Sn acts on K.x1 ; : : : ; xn / by ﬁeld automorphisms and K S .x1 ; : : : ; xn / is the ﬁxed ﬁeld. Therefore, by Proposition 9.5.5 the extension is Galois, with Galois group is Sn . I We deﬁne a distinguished family of symmetric polynomials, the elementary symmetric functions as follows: 0 .x1 ; : : : ; xn / D 1 1 .x1 ; : : : ; xn / D 2 .x1 ; : : : ; xn / D x1 C x2 C C xn X xi xj 1Äi <j Än ::: X D k .x1 ; : : : ; xn / x i1 xik 1Äi1 <i2 < <ik Än ::: D n .x1 ; : : : ; xn / We put j .x1 ; : : : xn / x1 x2 xn D 0 if j > n. Lemma 9.6.3. .x x1 /.x Dx D n n X 1x x2 /. n1 . 1/k C kx /.x 2x nk n2 xn / C . 1/n n ; k D0 where k is short for k .x1 ; : : : ; xn /. Proof. Exercise 9.6.4. I ...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online