Proper ideal $I$ of commutative ring with unity $R$, is unique maximal ideal of $R$ if $u$ is unit of $R\ \forall u \in I^{c} \cap R$.
$R$ : commutative ring with unity, $I$ : proper ideal of R.
Assume that $a \in R$ is unit $\forall a \notin I$.
Let's first show that $I$ should be maximal ideal of $R$.
We already know that ideal with unit could never be proper ideal. So we can conclude that $I$ never contains any unit element in $R$. Since that every element outside of $I$ is unit, any ideal of $R$ which properly containing $I$, must have unit of $R$. Which implies $I$ is maximal ideal by definition.
Next, we need to prove that $I$ is the unique maximal ideal of $R$. Obviously, any proper ideal of $R$ must be in $I$. Therefore, $I$ is the only maximal ideal.
Assume that $a \in R$ is unit $\forall a \notin I$.
Let's first show that $I$ should be maximal ideal of $R$.
We already know that ideal with unit could never be proper ideal. So we can conclude that $I$ never contains any unit element in $R$. Since that every element outside of $I$ is unit, any ideal of $R$ which properly containing $I$, must have unit of $R$. Which implies $I$ is maximal ideal by definition.
Next, we need to prove that $I$ is the unique maximal ideal of $R$. Obviously, any proper ideal of $R$ must be in $I$. Therefore, $I$ is the only maximal ideal.
댓글
댓글 쓰기