Theorem rlmlmod

Description: The ring module is a module. (Contributed by Stefan O'Rear, 6-Dec-2014)

	Ref	Expression
Assertion	rlmlmod	$$⊢R\in \mathrm{Ring}\to \mathrm{ringLMod}\left(R\right)\in \mathrm{LMod}$$

Proof

Step	Hyp	Ref	Expression
1		rlmval	$$⊢\mathrm{ringLMod}\left(R\right)=\mathrm{subringAlg}\left(R\right)\left({\mathrm{Base}}_R\right)$$
2		eqid	$$⊢{\mathrm{Base}}_R={\mathrm{Base}}_R$$
3	2	subrgid	$$⊢R\in \mathrm{Ring}\to {\mathrm{Base}}_R\in \mathrm{SubRing}\left(R\right)$$
4		eqid	$$⊢\mathrm{subringAlg}\left(R\right)\left({\mathrm{Base}}_R\right)=\mathrm{subringAlg}\left(R\right)\left({\mathrm{Base}}_R\right)$$
5	4	sralmod	$$⊢{\mathrm{Base}}_R\in \mathrm{SubRing}\left(R\right)\to \mathrm{subringAlg}\left(R\right)\left({\mathrm{Base}}_R\right)\in \mathrm{LMod}$$
6	3 5	syl	$$⊢R\in \mathrm{Ring}\to \mathrm{subringAlg}\left(R\right)\left({\mathrm{Base}}_R\right)\in \mathrm{LMod}$$
7	1 6	eqeltrid	$$⊢R\in \mathrm{Ring}\to \mathrm{ringLMod}\left(R\right)\in \mathrm{LMod}$$