Every Maximal Ideal of a Commutative Ring is a Prime Ideal

Prime Ideal Problems and Solution in Ring Theory in Mathematics

Problem 351

Let $R$ be a commutative ring with unity.
Then show that every maximal ideal of $R$ is a prime ideal.

 
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We give two proofs.

Proof 1.

The first proof uses the following facts.

  • Fact 1. An ideal $I$ of $R$ is a prime ideal if and only if $R/I$ is an integral domain.
  • Fact 2. An ideal $I$ of $R$ is a maximal ideal if and only if $R/I$ is a field.

Let $M$ be a maximal ideal of $R$. Then by Fact 2, $R/M$ is a field.
Since a field is an integral domain, $R/M$ is an integral domain. Thus by Fact 1, $M$ is a prime ideal.

Proof 2.

In this proof, we solve the problem without using Fact 1, 2.
Let $M$ be a maximal ideal of $R$.

Seeking a contradiction, let us assume that $M$ is not a prime ideal.

Then there exists $a, b\in R$ such that the product $ab$ is in $M$ but $a \not \in M$ and $b \not \in M$.

Consider the ideal $(a)+M$ generated by $a$ and $M$.
Since the ideal $(a)+M$ is strictly larger than $M$, we must have $R=(a)+M$ by the maximality of $M$.

Since $1\in R=(a)+M$, we have
\[1=ra+m,\] where $r\in R$ and $m\in M$.

Similarly, we have $R=(b)+M$ and
\[1=sb+n,\] where $s\in R$ and $n\in M$.

From these equalities, we obtain
\begin{align*}
1&=1\cdot 1\\
&=(ra+m)(sb+n)\\
&=rsab+ran+msb+mn.
\end{align*}

Since $ab, m, n$ are elements in the ideal $M$, the last expression is in $M$.

This yields that $1\in M$, and hence $M=R$. Since a maximal ideal is a proper ideal by definition, this is a contradiction.
Thus, $R$ must be a prime ideal.

Prime ideals are Not Necessarily Maximal

We just have shown that every maximal ideal is a prime ideal.

The converse, however, is not true.

That is, some prime ideals are not maximal ideals.

See the post ↴
Examples of Prime Ideals in Commutative Rings that are Not Maximal Ideals
for examples of rings and prime ideals that are not maximal ideals.


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