----- > [!proposition] Proposition. ([[division ring iff ideals are {0} and R]]) > A [[ring]] $R$ is a [[division ring]] if and only if its only (two-sided) [[ideal|ideals]] $\{ 0 \}$ and $R$. ^proposition > [!proposition] Corollary. > A [[commutative ring]] $R$ is a [[field]] if and only if its only [[ideal|ideals]] are $\{ 0 \}$ and $R$. ^proposition > [!proof]- Proof[](field.md)sion ring iff ideals are {0} and R]]) > ~$\to$. Let $R$ be a [[division ring]]. As always, $\{ 0 \}$ is an [[ideal]] of $R$. So let $I \neq \{ 0 \}$ be a nontrivial [[ideal]]; take $a \in I$. $a$ is a [[unit]]; obtain $b \in R$ such that $ab=1=ba$. Now $bI=I$ by definition if [[ideal]], meaning that $ab=1 \in I$. But recall (in any context) that $1 \in I \iff I=R$. So we have that if $I \neq \{ 0 \}$ then $I=R$. > $\leftarrow$. Conversely suppose the only ideals are $\{ 0 \}$ and $R$. Let $a \in R$ be arbitrary. Since the left-ideal $Ra$ equals $R$, $1 \in Ra$ and therefore $1=ra$ for some $r \in R$, so $a$ is a left-unit. Likewise since the right-ideal $aR$ equals $R$, $1 \in aR$ and $a$ is a right-unit. Thus $a$ is a [[unit]] and since it was arbitrary we conclude $R$ is a [[division ring]]. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```