Invariant Rings: Application I

Section 8.5 Invariant Rings: Application I

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“No taxation without representation!”
―James Otis

Historically, commutative algebra has roots in classical questions of algebraic and geometric flavors, including the following natural question:

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Question 8.58.

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Given a (finite) set of symmetries, consider the collection of polynomial functions that are fixed by all of those symmetries. Can we describe all the fixed polynomials in terms of finitely many of them?

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To make this precise, let

G

be a group acting on a ring

R .

The main case we have in mind is when

R = k [x_{1}, \dots, x_{d}]

and

k

is a field; we let

G

act trivially on

k,

and the action respects the sum and product in the ring:

g \cdot (\sum_{a} c_{a} x_{1}^{a_{1}} \dots x_{d}^{a_{d}}) = \sum_{a} c_{a} (g \cdot x_{1})^{a_{1}} \dots (g \cdot x_{d})^{a_{d}} .

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We are interested in the set of elements that are {\bf invariant} under the action,

R^{G} := {r \in R | g (r) = r for all g \in G} .

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Note that

R^{G}

is a subring of

R .

Indeed, given

r, s \in R^{G},

then

r + s = g \cdot r + g \cdot s = g \cdot (r + s) and r s = (g \cdot r) (g \cdot s) = g \cdot (r s) for all g \in G,

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since each

g

is a homomorphism. Note also that if

G = ⟨ g_{1}, \dots, g_{t} ⟩,

then

r \in R^{G}

if and only if

g_{i} (r) = r

for

i = 1, \dots, t .

The question above can now be rephrased as follows:

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Question 8.59.

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Given a finite group

G

acting on

R = k [x_{1}, \dots, x_{d}],

R^{G}

a finitely generated

k

-algebra?

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Note that

R^{G}

is a

k

-subalgebra of

R .

Even though

R

is a finitely generated

k

-algebra, this does not guarantee a priori that

R^{G}

is a finitely generated

k

-algebra --- recall \Cref{inf gen subalg} [provisional cross-reference: cite], where we saw a subalgebra of a finitely generated algebra which is nevertheless not finitely generated.

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Example 8.60. Root of Unity Invariants.

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Consider the group with two elements

G = {e, g} .

To define an action of

G

R = k [x],

we need only to define

g \cdot x,

since

e

is the identity and

g

acts linearly. Consider the action of

G

R = k [x]

given by

g \cdot x = - x,

g \cdot f (x) = f (- x) .

Suppose that the characteristic of

k

is not 2, so

- 1 \neq 1 .

Write

f = a_{n} x^{n} + a_{n - 1} x^{n - 1} + \dots + a_{0} .

We have

g \cdot x^{i} = (- x)^{i} = (- 1)^{i} x^{i},

g \cdot f = (- 1)^{n} a_{n} x^{n} + (- 1)^{n - 1} a_{n - 1} x^{n - 1} + \dots + a_{0},

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which differs from

f

unless for each odd

i,

a_{i} = 0 .

That is, [ R^G = { f R | }.] Any such

f

is a polynomial in

x^{2},

so we have

R^{G} = k [x^{2}] .

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In particular,

R^{G}

is a finitely generated

k

-algebra.

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Example 8.61. Veronese Subring.

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Generalize the last example as follows: let

k

be a field with a primitive

d

th root of unity

ζ,

and let

G = ⟨ g ⟩ ≅ C_{d}

act on

R = k [x_{1}, \dots, x_{n}]

via

g \cdot x_{i} = ζ x_{i}

for all

i .

Then

R^{G} = {f \in R | every term of f has degree a multiple of d} = k [{monomials of degree d}] .

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This what is known as the Veronese subring of

R

of degree

d .

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Example 8.62. Standard Representation of the Symmetric Group.

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Let

S_{n}

be the symmetric group on

n

letters acting on

R = k [x_{1}, \dots, x_{n}]

via

σ (x_{i}) = x_{σ (i)} .

For example, if

n = 3,

then

f = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}

is invariant, while

g = x_{1}^{2} + x_{1} x_{2} + x_{2}^{2} + x_{3}^{2}

is not, since swapping 1 with 3 gives a different polynomial.

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You may recall the Fundamental Theorem of Symmetric Polynomials says that every element of

R^{S_{n}}

can be written as polynomial expression in the elementary symmetric polynomials

\begin{array}{r} e_{1} \amp = x_{1} + \dots + x_{n} \\ e_{2} \amp = \sum x_{i} x_{j} \\ ⋮ \\ e_{n} \amp = x_{1} x_{2} \dots x_{n} . \end{array}

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More precisely,

R^{S_{n}} = k [e_{1}, \dots, e_{n}] .

For example,

f

above is

e_{1}^{2} - 2 e_{2} .

In fact, any symmetric polynomial can be written like so in a way, so

R^{S_{n}}

is a free

k

-algebra. So even though we have infinitely many invariant polynomials, we can understand them in terms of only finitely many of them, which are fundamental invariants.

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Proposition 8.63. Group Ring and Automorphisms Mod-Fin.

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Let

k

be a field,

R

be a finitely generated

k

-algebra, and

G

a finite group of automorphisms of

R

that fix

k .

Then

R^{G} \subseteq R

is module-finite.

Proof.

By , integral and algebra-finite implies module-finite, so we will show that

R

is algebra-finite and integral over

R^{G} .

First, since

k \subseteq R^{G}

and

R

is generated finitely generated

k

-algebra, it is generated by the same finite set as an

R^{G}

-algebra as well. Thus

R^{G} \subseteq R

is algebra-finite.

To show that

R^{G} \subseteq R

is integral, let us first extend the action of

G

R

R [t]

trivially, meaning that we will let

G

fix

t .

Given

r \in R,

consider the polynomial

F_{r} (t) := \prod_{g \in G} (t - g \cdot r) \in R [t] .

Now

G

fixes

F_{r} (t),

since for each

h \in G,

h \cdot F_{r} (t) = h \prod_{g \in G} (t - g \cdot r) = \prod_{g \in G} (h \cdot t - (h g) \cdot r) = F_{r} (t)

Thus,

F_{r} (t) \in (R [t])^{G} .

Notice that

(R [t])^{G} = R^{G} [t],

since

g (a_{n} t^{n} + \dots + a_{0}) = a_{n} t^{n} + \dots + a_{0} ⟹ (g \cdot a_{n}) t^{n} + \dots + (g \cdot a_{0}) = a_{n} t^{n} + \dots + a_{0} .

Therefore,

F_{r} (t) \in R^{G} [t] .

The leading term (with respect to

t

) of

F_{r} (t)

t^{| G |},

F_{r} (t)

is monic. On the other hand, one of the factors of

F_{r} (t)

(t - r),

F_{r} (r) = 0 .

Therefore,

r

satisfies a monic polynomial with coefficients in

R^{G},

and thus

R

is integral over

R^{G} .

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Theorem 8.64. Noether’s Finiteness Theorem.

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Let

k

be a field,

R

be a polynomial ring over

k,

and

G

be a finite group acting

k

-linearly on

R .

Then

R^{G}

is a finitely generated

k

-algebra.

Proof.

Observe that

k \subseteq R^{G} \subseteq R,

that

k

is noetherian,

k \subseteq R

is algebra-finite, and

R^{G} \subseteq R

is module-finite. The desired result is now a corollary of the Artin-Tate Lemma.

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