Metamath Proof Explorer


Theorem evlsscasrng

Description: The evaluation of a scalar of a subring yields the same result as evaluated as a scalar over the ring itself. (Contributed by AV, 12-Sep-2019)

Ref Expression
Hypotheses evlsscasrng.q Q = I evalSub S R
evlsscasrng.o O = I eval S
evlsscasrng.w W = I mPoly U
evlsscasrng.u U = S 𝑠 R
evlsscasrng.p P = I mPoly S
evlsscasrng.b B = Base S
evlsscasrng.a A = algSc W
evlsscasrng.c C = algSc P
evlsscasrng.i φ I V
evlsscasrng.s φ S CRing
evlsscasrng.r φ R SubRing S
evlsscasrng.x φ X R
Assertion evlsscasrng φ Q A X = O C X

Proof

Step Hyp Ref Expression
1 evlsscasrng.q Q = I evalSub S R
2 evlsscasrng.o O = I eval S
3 evlsscasrng.w W = I mPoly U
4 evlsscasrng.u U = S 𝑠 R
5 evlsscasrng.p P = I mPoly S
6 evlsscasrng.b B = Base S
7 evlsscasrng.a A = algSc W
8 evlsscasrng.c C = algSc P
9 evlsscasrng.i φ I V
10 evlsscasrng.s φ S CRing
11 evlsscasrng.r φ R SubRing S
12 evlsscasrng.x φ X R
13 6 ressid S CRing S 𝑠 B = S
14 13 eqcomd S CRing S = S 𝑠 B
15 10 14 syl φ S = S 𝑠 B
16 15 oveq2d φ I mPoly S = I mPoly S 𝑠 B
17 5 16 syl5eq φ P = I mPoly S 𝑠 B
18 17 fveq2d φ algSc P = algSc I mPoly S 𝑠 B
19 8 18 syl5eq φ C = algSc I mPoly S 𝑠 B
20 19 fveq1d φ C X = algSc I mPoly S 𝑠 B X
21 20 fveq2d φ I evalSub S B C X = I evalSub S B algSc I mPoly S 𝑠 B X
22 eqid I evalSub S B = I evalSub S B
23 eqid I mPoly S 𝑠 B = I mPoly S 𝑠 B
24 eqid S 𝑠 B = S 𝑠 B
25 eqid algSc I mPoly S 𝑠 B = algSc I mPoly S 𝑠 B
26 crngring S CRing S Ring
27 6 subrgid S Ring B SubRing S
28 10 26 27 3syl φ B SubRing S
29 6 subrgss R SubRing S R B
30 11 29 syl φ R B
31 30 12 sseldd φ X B
32 22 23 24 6 25 9 10 28 31 evlssca φ I evalSub S B algSc I mPoly S 𝑠 B X = B I × X
33 21 32 eqtrd φ I evalSub S B C X = B I × X
34 2 6 evlval O = I evalSub S B
35 34 a1i φ O = I evalSub S B
36 35 fveq1d φ O C X = I evalSub S B C X
37 1 3 4 6 7 9 10 11 12 evlssca φ Q A X = B I × X
38 33 36 37 3eqtr4rd φ Q A X = O C X