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:

$$K=J\mathrm{\setminus }I$$

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 $r\in R$:

$$\begin{array}{rl}(rb)+I=1+I& ⟹b^{\prime }+I=1+I\\ & ⟹\mathrm{\exists }c\in b^{\prime }+I\subset Js.t.c=1\end{array}$$

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.

$$\begin{array}{rl}J& =\{b_1+I,b_2+I,...\}\\ & =\{\{b_1+a:a\in I\},\{b_2+a:a\in I\},...\}\end{array}$$

We claim the following is a proper ideal of $R$ containing $I$:

$$J^{\prime }=\bigcup _{\mathrm{\forall }i}(b_i+J)$$

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

$$(r+I)\ast (b_i+I)=b_j+I$$

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

$$⟹r{b}_i\in b_j+I\subset J^{\prime }$$

Hence $J^{\prime }$ is closed under multiplication from $R$. We show $J^{\prime }$ 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:

$$\begin{gathered} (b_i+I)+(b_j+I)=(b_k+I)\subset J^{\prime } \\ ⟹b_i+b_j\in J^{\prime } \end{gathered}$$

For some $k$. It remains to show $J^{\prime }$ is proper, suppose it's not for a contradiction:

$$\begin{array}{rl}{J}^{\prime }=R& ⟹1\in J^{\prime }\\ & ⟹(1+I)\in J\\ & ⟹J=R/I\end{array}$$

Which is a contradiction since $J$ is proper. So we have show $J^{\prime }$ 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 $a\in I\vee b\in I$.