Does R/N Form a Ring with Unity if N is a Proper Ideal of R?

[Fantini]

Good afternoon. Here is the problem:

Show that if $R$ is a ring with unity and $N$ is an ideal of $R$ such that $N \neq R$, then $R/N$ is a ring with unity.

My answer: Consider the homomorphism $\phi: R \to R/N$. Given $r \in R$ we have that $\phi(r) = r + N = \phi(1 \cdot r) = \phi(r \cdot 1) = (1+N)(r+N) = (r+N)(1+N)$, therefore $1+N$ is the unity of $R/N$.

I appreciate the help. Cheers! (Yes)

P.S.: I am assuming the usual operations concerning factor rings:

$(a+N) + (b+N) = (a+b) + N$ and $(a+N)(b+N) = (ab) + N$.

 

[caffeinemachine]

Good afternoon. Here is the problem:

Show that if $R$ is a ring with unity and $N$ is an ideal of $R$ such that $N \neq R$, then $R/N$ is a ring with unity.

My answer: Consider the homomorphism $\phi: R \to R/N$. Given $r \in R$ we have that $\phi(r) = r + N = \phi(1 \cdot r) = \phi(r \cdot 1) = (1+N)(r+N) = (r+N)(1+N)$, therefore $1+N$ is the unity of $R/N$.

I appreciate the help. Cheers! (Yes)

P.S.: I am assuming the usual operations concerning factor rings:

$(a+N) + (b+N) = (a+b) + N$ and $(a+N)(b+N) = (ab) + N$.

Looks right to me. Only thing, there wasn't any need to invoke a homomorphism here. You could just simply show that $(1+N)(r+N)=(r+N)(1+N)=r+N$ for all r in R. Well, there ain't much to show though.

 

[Deveno]

the only tricky part is that 1+N might actually = N, but this means that 1 is in N,

in which case we still have a multiplicative identity, but:

R/N = R/R = {0}, which is a "silly" ring. this is why we insist N ≠ R at the outset.