---- > [!definition] Definition. ([[center of a ring]]) > The **center** of a [[ring]] $R$ is the [[subring]] consisting of the elements $a$ such that $ar=ra$ for all $r \in R$. ^definition > [!note] Remark. > How does this interact with the [[center of a group|center]] of the underlying [[abelian group]]? Well, the [[group|group's]] center equals $R$ itself! So it makes sense to just talk about multiplication here. ^note > [!justification] > We should check that the center $Z$ is indeed a [[subring]]. Clearly $0 \in Z$ since $0r=0=r0$ for all $r \in R$, and $1 \in Z$ since $1r=r=r_{1}$ for all $r \in R$. And closure under addition holds: given $a,b \in Z$ and $r$ arbitrary we see $(a+b)r=ra+rb=rb+ra=r(b+a)$ so that $a+b \in Z$. And closure under additive inverse holds: if $a \in R$ then $\big( a+(-a) \big)r=ar + (-a)r$ while $r\big( a+(-a) \big)=ra+r(-a)=ar + r(-a)$ (using $a \in Z$). Hence $ar+(-a)r=ar+r(-a)$ and [[cancellation law for groups]] yields $(-a)r=r(-a)$ so that $-a \in Z$. All that is left to show is closure under multiplication: let $a,b \in Z$; then excessive use of associativity yields $ab(r)=a(br)=a(rb)=(ar)b=(ra)b=(r)ab.$ > This finishes the proof. ^justification ---- #### ---- #### 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 > ```