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- Feb 29, 2012

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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.

__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 $R/N$.__

**My answer:**I appreciate the help. Cheers!

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

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