Let π : A → A/I be the natural projection, π preserves maximals ideals.

Question

Let A be a ring and I an ideal of A. Let π : A → A/I be the natural projection. Claim the correspondence induced by π between ideals of A which contain I and ideals of A/I preserves maximal ideals.

I know that there is a 1-1 correspondence between the ideals J containing $I$ and the ideals $\bar J$ of A/I. However, I don't know how to use this to prove above claim. Can you give some hints? Many thanks

Answer 1

Use the third isomorphism theorem.

Let $G$ be a group, $H\subset K$ two normal subgroups of $G$. There is a canonical isomorphism: \begin{align*} (G/H)/(K/H)&\longrightarrow G/K\\ (xH)K/H&\longmapsto xK \end{align*}

For rings, it should be written in additive notation. In any case, the above theorem implies that if $\mathfrak m$ is a maximal ideal of $A$ which contains $I$, then $\mathfrak m/I$ is a maximal ideal of $A/I$, since both quotient rings are fields.

Answer 2

Maximal ideals are by definition maximal elements in the poset of ideals of $A$. The correspondence actually gives you an isomorphism between the poset of ideals in $A/I$ and the poset of ideals in $A$ containing $I$.

Show that an isomorphism of posets preserves maximal elements and you win.

Answer 3

if $M$ is maximal in $A/I$, the quotient $(A/I)/M$ is a field, $A/p^{-1}(M)\rightarrow (A/I)/M$ is injective, thus $A/p^{-1}(M)$ is also a field since it is a subring of a field.

Answer 4

For answering to this question it is necessary and sufficient that you verify for all ideals $I$, $J$, and $K$ of $R$ with $I\subseteq J$ we have $ \frac{J}{I}\subseteq \frac{K}{I}$ if and only if $J\subseteq K$.