Is it possible in some ring $1=0$?

Question

Consider a ring $R$ and the ideal $I$ in $R$ and the quotient ring $R/I$. With defining $(r+I)(s+I)=(rs+I)$ for any $r,s \in R$, I concluded that $0=I$ and $1=I$ in $R/I$ thus $0=1$ in $R/I$ :

$$I+(r+I)=r+(I+I)=r+I \implies I=0$$ and $$(I)(r+I)=(0+I)(r+I)=0r+I=I \implies I=1.$$

1- Is that possible that identity of two operations $\times$ and $+$ in a ring be the same?

2- Am I right with the calculations above?

Do a Google search for the trivial ring. – John Douma Nov 09 '15 at 02:17 — John Douma, Nov 09 '15 at 02:17
Your conclusion that $I=1$ is incorrect. – Noah Schweber Nov 09 '15 at 02:27 — Noah Schweber, Nov 09 '15 at 02:27

score 2 · Answer 1 · answered Nov 09 '15 at 02:19

2

The only ring in which $0=1$ is the trivial ring, the ring with 1 element. Proof: Let $x\in R$. Then $1x=x=0x=(0+0)x=0x+0x$, hence $0x=0=1x=x$, hence all $x\in R$, $x=0$. Some authors don't allow this to be a ring, but most do.

answered Nov 09 '15 at 02:19

Alan

16,895

score 0 · Answer 2 · answered Nov 09 '15 at 02:18

0

There's the trivial ring, which has one element. In that case $0=1$.

answered Nov 09 '15 at 02:18

Everybody has mentioned the trivial group. As in the question 2 and the calculations in the OP, in the quotient ring also 1=0. Am I right? Thank you. – Nov 09 '15 at 02:28

Is it possible in some ring $1=0$?

2 Answers2

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