📂Abstract Algebra

Units of an Ideal

Theorem ¹

• [1]: A ring $R$ with a unit element $1$ whose ideals $I$ have a unit is $I = R$
• [2]: A field $F$ is $\left\{ 0 \right\}$, not having any ideals other than $F$.

Explanation

Summary [1] is an auxiliary theorem frequently used in proofs by contradiction, stating that having a unit in an ideal makes it whole, and from the point that the unit element is a unit, it guarantees that a proper ideal does not contain $1$. For instance, the ideals of $\mathbb{Z}$ are $$ n \mathbb{Z} = \left\{ \cdots , -2n , -n , 0 , n , 2n , \cdots \right\} $$ and the moment $1$ is included, it becomes $1 \mathbb{Z} = \mathbb{Z}$ itself.

Summary [2] implies that a field cannot possess proper nontrivial ideals. This essentially suggests that the concept of ideals is exclusive to rings.

Proof

[1]

Let’s say one of the units contained in $I$ is $u$.

If we set $r : = u^{-1}$, then $u^{-1} u = 1$ and, since $r I \subset I$, $r \cdot u = 1 $ is also included in $I$. From the definition of ideals, since it was $r I \subset I$ for all $r \in R$, $r \cdot 1 = r \in I$ and $R \subset I$

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[2]

Since $F$ is a field, all elements other than $0$ are units, and by Summary [1], any ideals other than $\left\{ 0 \right\}$ become $F$ itself.

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