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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.

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    $\begingroup$ 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. $\endgroup$ – Janko Bracic Jan 27 '15 at 7:02
  • $\begingroup$ 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. $\endgroup$ – August Jan 27 '15 at 7:06
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    $\begingroup$ @JankoBracic: Nice comment, worthy of being written up as an answer, in my humble opinion. Cheers! $\endgroup$ – Robert Lewis Jan 27 '15 at 7:15
  • $\begingroup$ @RobertLewis In my opinion the most of the work has already be done by August. $\endgroup$ – Janko Bracic Jan 27 '15 at 7:16
  • $\begingroup$ @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! $\endgroup$ – Robert Lewis Jan 27 '15 at 7:19
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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.

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