Ideals

We say a proper ideal \( I \) of \( R \), is a maximal ideal if for all ideals \( J \) such that \( I \subseteq J \), then either \( J = I \) or \( J = R \).

The factor ring \( R / I \) is a field iff \( I \) is a maximal ideal in \( R \).

(forward case): By contrapositive; suppose \( I \) is not maximal in \( R \), ie there exists some ideal \( J \) such that \( I \subset J \subset R \). We show that there is a non-invertible element in \( R/I \). Let:

\[ K = J \backslash I \]

Note \( 1 \notin J \) (else \( J = R \)) and \( K \) is nonempty and contains some nonzero element, say \( b \). Now consider \( b + I \in R / I \). Let:

\[ S = \{ (r + I)(b + I) = (rb) + I : r \in R \} \]

Now suppose for some \( r \in R \):

\begin{align*} (rb) + I = 1 + I & \implies b' + I = 1 + I \\ & \implies \exists c \in b' + I \subset J \ s.t. \ c = 1 \end{align*}

Which is a contradiction as \( J \) is a proper ideal. Hence \( b + I \) has no inverse in \( R/I \), hence \( R/I \) is not a field.

(backward case): By contrapositive; note \( R/I \) is a field iff its only ideals are \( \{ 0 + I \} \) and \( R/I \). Now suppose \( R/I \) is not a field, ie there exists some nontrivial proper ideal \( J \) which contains some nonzero element (along with \( 0 + I \)). We show then that it cannot be the case that \( I \) is maximal.

\begin{align*} J &= \{ b_1 + I, \ b_2 + I, ... \}\\ &= \{ \{ b_1 + a : a \in I \}, \ \{b_2 + a : a \in I \}, ... \} \end{align*}

We claim the following is a proper ideal of \( R \) containing \( I \):

\[ J' = \bigcup\limits_{\forall i} (b_i + J) \]

We already know \( J' \) contains all elements of \( I \) (including \( 0 \)) as \( J \) contains \( 0 + I = I \). So we first show \( J' \) is closed under multiplication from \( R \). since \( J \) is an ideal of \( R/I \) we know that for all \(r \in R, \ b_i + I \in J \):

\[ (r + I)*(b_i + I) = b_j + I \]

For some \( b_j + I \in J \). So then:

\[ \implies rb_i \in b_j + I \subset J' \]

Hence \( J' \) is closed under multiplication from \( R \). We show \( J' \) is closed under addition. Let \( b_i, b_j \in J \), and consider:

\[ b_i + b_j \in (b_i + I) + (b_j + I) \]

But since \( J \) is an ideal, it is closed under addition and so:

\[ (b_i + I) + (b_j + I) = (b_k + I) \subset J' \\ \implies b_i + b_j \in J' \]

For some \( k \). It remains to show \( J' \) is proper, suppose it's not for a contradiction:

\begin{align*} J' = R &\implies 1 \in J' \\ &\implies (1 + I) \in J \\ &\implies J = R/I \end{align*}

Which is a contradiction since \( J \) is proper. So we have show \( J' \) is a proper ideal containing \( I \), which is therefore not maximal. This concludes the backward case.

We say an proper ideal \( I \) of a commutative ring \( R \) is prime if, for any two elements \( a, b \in R \) such that \( ab \in I \) we must have \( a \in I \lor b \in I \).