Let R be a ring with unity $1_R$ and let S (with unity $1_S$) be a subring of R. Prove that either $1_S=1_R$ or $1_S$ is a zero divisor of R.

Question

Let R be a ring with unity $1_R$ and let S (with unity $1_S$) be a subring of R. Prove that either $1_S=1_R$ or $1_S$ is a zero divisor of R.

My attempt:

Let $a \in S$. Then $1_S*a = a$ and this a exists in R so that $1_R*a=a$

Then

$1_S*a=1_R*a$

$1_S*a - 1_R*a = 0$

$(1_S-1_R)*a=0$

Then

1) $1_S=1_R$

2) a = 0

3) $1_S-1_R \neq 0$ and $a \neq 0$

Then I'm not sure what to do. I don't even know if this is right. Can somebody please help?

Thank you.

Answer 1

As the commenters have noted, you do have the key idea for an answer, although in the post you never seem to realize you have it, and the 1) 2) 3) sequence at the end is not really going anywhere.

The closest you get is this: $(1_S-1_R)\ast a=0$.

For $a=1_S$, you get

$$(1_S-1_R)\ast 1_S=0$$

On one hand, it could be that $1_S=1_R$. On the other hand, if $1_S\neq 1_R$, then this is a product of two nonzero elements of the ring which is zero, so both pieces are zero divisors, and that means $1_S$ is a zero divisor.