My proof for $|R|=1$ if and only if $1=0$ where $R$ is a ring with unity.

Question

I'm reading Dummit's book. (Abstract Algebra, 3ed, p224) Now I want to prove the following. Suppose $R$ is a ring with 1. Then, $|R|=1$($R$ is a trivial ring, [or the zero ring in the book]) if and only if $1=0$.

$(\Rightarrow)$ : Since $0\in R$, $1\in R$ and $|R|=1$, it is immediate that $1=0$.

$(\Leftarrow)$ : Suppose that $a\in R$. Then $a=a\cdot1=a\cdot0\stackrel{(*)}=0$. Thus $R=\{0\}$.

But how can I justify $(*)$? Does that follow immediately from the definition of a ring with unity?

Or, is there another proof for $(\Leftarrow)$?

That's what I wanted to know. Actually I've seen it before, but forgot now. Thanks. — govin, Apr 13 '21 at 10:48
Should I close this thread? or... What should I do? Is the question too trivial? — govin, Apr 13 '21 at 10:57
@govin The question is ok as it stands. However, this is a standard fact and I there are already answers somewhere on this side (for example here). So make sure to search for those first the next time. — mrtaurho, Apr 13 '21 at 11:56

score 0 · Accepted Answer · answered Apr 13 '21 at 11:54

The comments already answer the question. However, I will just post this answer to close the thread.

We know that $0+0=0$ as $0$ is the additive neutral element and that $a(0+0)=a0+a0$ by distributivity. Putting these together gives us the following:

$$a0=a(0+0)=a0+a0 \implies (a0+a0)-a0=a0-a0 \implies a0=0$$

We know that $a0+(-a0)=a0-a0=0$ as every element $r$ in a ring has an additive inverse and one can show (and I would advise you doing so) that this inverse is in fact given by $-r=(-1)r$.

My proof for $|R|=1$ if and only if $1=0$ where $R$ is a ring with unity.

1 Answers1