Element of Unital Commutative Algebra Invertible iff not Contained in Maximal Ideal

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Theorem

Let $R$ be a ring.

Let $A$ be an unital commutative $R$-algebra.

Let $a \in A$.


Then $a$ is invertible if and only if there does not exist a maximal ideal $M$ with $a \in M$.


Proof

Necessary Condition

We prove the contrapositive.

Then, by Proof by Contraposition we will done.

We will prove that if $a$ is in a maximal ideal then $a$ is not invertible.

Let $M$ be a maximal ideal with $a \in M$.

Aiming for a contradiction, suppose there exists $b \in A$ with $a b = {\mathbf 1}_A$.

Since $M$ is an ideal, we have $a b \in M$, hence ${\mathbf 1}_A \in M$.

This contradicts that $M$ is proper.

$\Box$

Sufficient Condition

We will prove that if $a$ is not invertible then there exists a maximal ideal $M$ with $a \in M$.

Let:

$I = \set {a m : m \in A}$

We show that $I$ is a proper ideal containing $a$.

That $a \in I$ is immediate since $a {\mathbf 1}_A \in A$.

Let $a x, a y \in I$ and $\lambda, \mu \in R$.

Then we have:

$\mu a x + \lambda a y = a \paren {\mu x + \lambda y} \in I$

So $I$ is a submodule of $A$.

Further, if $a x \in I$ and $b \in A$, we have:

$\paren {a x} b = a \paren {x b} \in I$

So $I$ is a right ideal.

Since $A$ is commutative, $I$ is an ideal.

From Ideal of Unital Algebra is Modular, $I$ is modular.

Further, since $a$ is invertible there exists no $m \in A$ such that $a m = {\mathbf 1}_A$.

Hence ${\mathbf 1}_A \not \in \set {a m : m \in A}$.

From Proper Modular Ideal of Algebra is Contained in Maximal Ideal, there then exists a maximal ideal $M$ with $I \subseteq M$.

Then $a \in M$.

$\blacksquare$