Quotient ring local if Ring is local

Question

I want to show: If $R$ is local and $I\neq R$ an ideal, then $R/I$ is also local.

We already know: A Ring $R$ is local if and only if $R-R^{\times} = \{r\in R \, | r \notin R^\times \}$ is an ideal. We also know that if $I$ is an maximal Ideal, $R/I$ is a field.

I have no idea how to show that other than using the above Lemma.

Answer 1

The easiest characterization of local rings here is that a ring is local if and only if it has exactly one maximal ideal. Let $M$ be the maximal ideal of $R$ and $f:R\to R/I$ the surjection. If $R/I$ has a maximal ideal $M'\neq f(M) $, then $f^{-1}(M')\neq M$ is a maximal ideal of $R$, which is a contradiction.

Answer 2

By definition, $R$ is a local ring if and only if it has a unique maximal ideal $\mathfrak{m}.$ The lattice isomorphism theorem for commutative rings states that the quotient homomorphism $\pi:R\to R/I$ induces an inclusion preserving correspondence between ideals of $R$ containing $I$ and ideals of $R/I$.

Every ideal $I\subseteq R$ is contained in a (unique) maximal ideal. $R$ is local so that the only maximal ideal is $\mathfrak{m}$. So, $I\subseteq \mathfrak{m}$. The lattice isomorphism tells us that the ideals of $R/I$ are all contained in $\pi(\mathfrak{m})$. $\widetilde{\mathfrak{m}}=\pi(\mathfrak{m})$ is a maximal ideal in $R/I$ (check this) so that it is also the unique maximal ideal in $R/I$ since any other maximal ideal is contained in it by the lattice isomorphism theorem.

So, $R/I$ has unique maximal ideal $\pi(\mathfrak{m})$ and hence is local.