# 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.

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.

• Put $a=1_S$ in the equality $(1_R-1_S)\ast a=0$. Then either $1_R=1_S$ or $1_S$ is zero divisor. – Janko Bracic Jan 27 '15 at 7:02
• Thank you, @JankoBracic Does this mean that $1_R$ can also be a zero divisor if we let $a = 1_R$? EDIT: Wait, nevermind. That wouldn't make sense -- then the ring R wouldn't have unity. – August Jan 27 '15 at 7:06
• @JankoBracic: Nice comment, worthy of being written up as an answer, in my humble opinion. Cheers! – Robert Lewis Jan 27 '15 at 7:15
• @RobertLewis In my opinion the most of the work has already be done by August. – Janko Bracic Jan 27 '15 at 7:16
• @JankoBracic: yes, I see your point, but you added the essential detail noting $1_S(1_S - 1_R) = 0$, which follows from $1_S^2 = 1_S$. Also, questions need answers! But it looks like Federico beat us to the punch! Salud! – Robert Lewis Jan 27 '15 at 7:19

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.