📂Abstract Algebra

Ideals Containing a Unit

Theorem¹

• [1]: If an ideal $I$ of a ring $R$ with unity $1$ contains a unit, then $I = R$
• [2]: A field $F$ has no ideals other than $\left\{ 0 \right\}$ and $F$.

Explanation

Theorem [1] says that an ideal becomes the whole ring merely by containing a unit, and it is a lemma frequently used in proofs by contradiction. Moreover, since unity is a unit, it also guarantees that any decent ideal does not contain $1$. For example, the ideals of $\mathbb{Z}$ are things like $$ n \mathbb{Z} = \left\{ \cdots , -2n , -n , 0 , n , 2n , \cdots \right\} $$ and the moment $1$ is included, it becomes $1 \mathbb{Z} = \mathbb{Z}$ itself.

Theorem [2] means that a field cannot have a proper Nontrivial ideal. This effectively suggests that the ideal is a concept exclusive to rings.

Proof

[1]

Let $u$ be one of the units contained in $I$.

If we set $r : = u^{-1}$, then $u^{-1} u = 1$, and since $r I \subset I$, $r \cdot u = 1 $ is also contained in $I$. Since $r I \subset I$ for every $r \in R$ by the definition of an ideal, $r \cdot 1 = r \in I$ and $R \subset I$

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[2]

Since $F$ is a field, every element other than $0$ is a unit, and by Theorem [1], any ideal other than $\left\{ 0 \right\}$ becomes $F$.

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