# Identity element of a subring is the same as the identity element of the ring

I am studying the first chapter of "Introduction to Commutative Algebra" by M. F. Atiyah and I. G. MacDonald. The book defined a subring as follows:

A subset $$S$$ of a ring $$A$$ is a subring of $$A$$ if $$S$$ is closed under addition and multiplication and contains the identity element of $$A.$$

I want to prove that if $$S$$ has identity $$1_S$$ then $$1_S = 1_A,$$ I do not know how to prove this, can anyone help me please?

• Do you know that the identity is unique? Commented Sep 13, 2022 at 3:20
• Well, don't we have $1_S=1_S\cdot 1_A=1_A$? Commented Sep 13, 2022 at 3:21
• Yeah @paulblartmathcop I know that the identity is unique Commented Sep 13, 2022 at 3:22
• @pipe oh is this the proof? but I have seen here questions like this math.stackexchange.com/questions/1616674/… so I got confused Commented Sep 13, 2022 at 3:25
• I think it'll work, with your definition. The linked examples don't contain $1_A$. Commented Sep 13, 2022 at 3:37

A ring can't have two identities, since $$1_1=1_1\cdot 1_2=1_2$$.