Let g be a finite abelian group and suppose that g acts on...

Name: let g be a finite abelian group and suppose that g acts on the polynomial ring s kx1 ldots xn via characters ie suppose that for each i 1 ldots n there exists a homomorphism alphai g righ 22594
Uploaded: 2023-09-21T02:14:23-08:00
Duration: 10 min 15 s
Channel: Sri K
Description: let g be a finite abelian group and suppose that g acts on the polynomial ring s kx1 ldots xn via characters ie suppose that for each i 1 ldots n there exists a homomorphism alphai g righ 22594

Let \( G \) be a finite abelian group, and suppose that \( G \) acts on the polynomial ring \( S = k[x_1, \ldots, x_n] \) via characters, i.e., suppose that for each \( i = 1, \ldots, n \) there exists a homomorphism \( \alpha_i : G \rightarrow k^times \), with \( k^times \) being the multiplicative group of \( k \), such that \( g \cdot x_i = \alpha_i(g) x_i \) for all \( g \in G \). Show that the invariants of the action of the group \( G \) are generated by those monomials \( \prod_{i=1}^n x_i^{a_i} \) whose exponent vectors \( (a_1, \ldots, a_n) \) belong to the kernel of an application of \( \mathbb{Z}^n \) to a certain finite abelian group. Conclude that the ring of invariants \( S^G \) is isomorphic to the field of rational functions in \( n \) variables.

Transcript

00:01 Hello, let's have a look at the question.

00:04 So the question state let n be a positive integer and fn be the set of all polynomials in n variables.

00:13 So we have fn which is the set of all polynomials in n variables and we have variables as x1 x2 up to xn and sn denotes the symmetric group on n letters.

00:42 So here we have taken let r is defined as sn multiplied with fn defined on fn be the map defined by nu multiplied with f at x1 x2 up to xn is equals to f at x new 1 comma x new 2 and this will be continuing up to x new n for all new belongs to sn and f belongs to fn.

01:13 Now in the a part here we have to commute 1 4 3 2 multiplied with x1 square x2 plus x4 square x3.

01:26 So here new is equals to 1 4 3 2 this means that we will have 1 replaced with 4 4 replaced with 3 3 replaced with 2 and 2 replaced with 1.

01:44 So therefore we have x new x square new 1 x new 2 plus x square new 2 and x new 3 will be now written as x square 4 x1 plus x square 3 x3 so this is the required answer now for the b part here we have to show that star is a group action on fn.

02:19 So first of all for star to be a group action then we have a that e multiplied with f x1 x2 up to xn is equals to f at x1 x2 up to xn and b new star into beta function with x is equals to new beta multiplied with x where new and beta belongs to sn and x is equals to xn x2 and up to xn so let us see if the a part is true.

03:12 So here e multiplied with f x1 x2 up to xn can be written as f x e at 1 x e at 2 up to x e at n so now here e is the identity permutation and here we can write this as f x1 x2 up to xn since e1 is equals to 1 e2 is equals to 2 and so on next in the b part...

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