Metamath Proof Explorer


Theorem evls1scasrng

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, 13-Sep-2019)

Ref Expression
Hypotheses evls1scasrng.q Q=SevalSub1R
evls1scasrng.o O=eval1S
evls1scasrng.w W=Poly1U
evls1scasrng.u U=S𝑠R
evls1scasrng.p P=Poly1S
evls1scasrng.b B=BaseS
evls1scasrng.a A=algScW
evls1scasrng.c C=algScP
evls1scasrng.s φSCRing
evls1scasrng.r φRSubRingS
evls1scasrng.x φXR
Assertion evls1scasrng φQAX=OCX

Proof

Step Hyp Ref Expression
1 evls1scasrng.q Q=SevalSub1R
2 evls1scasrng.o O=eval1S
3 evls1scasrng.w W=Poly1U
4 evls1scasrng.u U=S𝑠R
5 evls1scasrng.p P=Poly1S
6 evls1scasrng.b B=BaseS
7 evls1scasrng.a A=algScW
8 evls1scasrng.c C=algScP
9 evls1scasrng.s φSCRing
10 evls1scasrng.r φRSubRingS
11 evls1scasrng.x φXR
12 6 ressid SCRingS𝑠B=S
13 12 eqcomd SCRingS=S𝑠B
14 9 13 syl φS=S𝑠B
15 14 fveq2d φPoly1S=Poly1S𝑠B
16 5 15 eqtrid φP=Poly1S𝑠B
17 16 fveq2d φalgScP=algScPoly1S𝑠B
18 8 17 eqtrid φC=algScPoly1S𝑠B
19 18 fveq1d φCX=algScPoly1S𝑠BX
20 19 fveq2d φSevalSub1BCX=SevalSub1BalgScPoly1S𝑠BX
21 eqid SevalSub1B=SevalSub1B
22 eqid Poly1S𝑠B=Poly1S𝑠B
23 eqid S𝑠B=S𝑠B
24 eqid algScPoly1S𝑠B=algScPoly1S𝑠B
25 crngring SCRingSRing
26 6 subrgid SRingBSubRingS
27 9 25 26 3syl φBSubRingS
28 6 subrgss RSubRingSRB
29 10 28 syl φRB
30 29 11 sseldd φXB
31 21 22 23 6 24 9 27 30 evls1sca φSevalSub1BalgScPoly1S𝑠BX=B×X
32 20 31 eqtrd φSevalSub1BCX=B×X
33 2 6 evl1fval1 O=SevalSub1B
34 33 a1i φO=SevalSub1B
35 34 fveq1d φOCX=SevalSub1BCX
36 1 3 4 6 7 9 10 11 evls1sca φQAX=B×X
37 32 35 36 3eqtr4rd φQAX=OCX