Metamath Proof Explorer


Theorem splysubrg

Description: The symmetric polynomials form a subring of the ring of polynomials. (Contributed by Thierry Arnoux, 15-Jan-2026)

Ref Expression
Hypotheses splyval.s S = SymGrp I
splyval.p P = Base S
splyval.m M = Base I mPoly R
splyval.a A = d P , f M x h 0 I | finSupp 0 h f x d
splyval.i φ I V
splysubrg.r φ R Ring
Assertion splysubrg Could not format assertion : No typesetting found for |- ( ph -> ( I SymPoly R ) e. ( SubRing ` ( I mPoly R ) ) ) with typecode |-

Proof

Step Hyp Ref Expression
1 splyval.s S = SymGrp I
2 splyval.p P = Base S
3 splyval.m M = Base I mPoly R
4 splyval.a A = d P , f M x h 0 I | finSupp 0 h f x d
5 splyval.i φ I V
6 splysubrg.r φ R Ring
7 1 2 3 4 5 6 splyval Could not format ( ph -> ( I SymPoly R ) = ( M FixPts A ) ) : No typesetting found for |- ( ph -> ( I SymPoly R ) = ( M FixPts A ) ) with typecode |-
8 eqid f M d A f = f M d A f
9 1 2 3 4 5 mplvrpmga φ A S GrpAct M
10 coeq2 d = e x d = x e
11 10 fveq2d d = e f x d = f x e
12 11 mpteq2dv d = e x h 0 I | finSupp 0 h f x d = x h 0 I | finSupp 0 h f x e
13 fveq1 f = g f x e = g x e
14 13 mpteq2dv f = g x h 0 I | finSupp 0 h f x e = x h 0 I | finSupp 0 h g x e
15 12 14 cbvmpov d P , f M x h 0 I | finSupp 0 h f x d = e P , g M x h 0 I | finSupp 0 h g x e
16 4 15 eqtri A = e P , g M x h 0 I | finSupp 0 h g x e
17 5 adantr φ d P I V
18 oveq2 f = g d A f = d A g
19 18 cbvmptv f M d A f = g M d A g
20 eqid I mPoly R = I mPoly R
21 6 adantr φ d P R Ring
22 simpr φ d P d P
23 1 2 3 16 17 19 20 21 22 mplvrpmrhm φ d P f M d A f I mPoly R RingHom I mPoly R
24 2 3 8 9 23 fxpsubrg Could not format ( ph -> ( M FixPts A ) e. ( SubRing ` ( I mPoly R ) ) ) : No typesetting found for |- ( ph -> ( M FixPts A ) e. ( SubRing ` ( I mPoly R ) ) ) with typecode |-
25 7 24 eqeltrd Could not format ( ph -> ( I SymPoly R ) e. ( SubRing ` ( I mPoly R ) ) ) : No typesetting found for |- ( ph -> ( I SymPoly R ) e. ( SubRing ` ( I mPoly R ) ) ) with typecode |-