The set $(E, +, .)$ of even integers forms a commutative ring without unit element( multiplicative identity), with usual rules of addition and multiplication operation.

Next, the $center$ of a ring R is {${z\in R :zr=rz,\forall r\in R}$} (i,e. the set of all elements which commute with every element of R).

I have problem to prove that the center of a ring is a subring that contains the identity.

Since the ring of even integers is commutative, so every element of this ring commutes with other. So I think that its center should be the ring itself. But how does it contain unit element?


2 Answers 2


The statement "the center of a ring is a subring that contains the identity" is only true for rings with unit. Indeed, the statement does not even make sense for a ring without unit, because there is no such thing as "the identity". (Note that many authors assume a unit as part of the definition of "ring", possibly including whoever posed this problem. Although, if you do include a unit in the definition of "ring", then containing the identity element should be included in your definition of "subring", so the phrasing is still odd.)

  • $\begingroup$ I made the title of question "center of ring is subring that contains identity" which i need to prove. Now in the case of ring of even integers, what would be the center? Is it the ring itself or it is meaningless to define center in the case of even integers? $\endgroup$ Commented Jan 4, 2018 at 20:40
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    $\begingroup$ It is the ring itself. The statement you are asked to prove is not correct, though, if your rings are not required to have a unit. $\endgroup$ Commented Jan 4, 2018 at 20:55

The definition of center of a ring doesn't mention the identity: $$ Z(R)=\{z\in R:zr=rz, \forall r\in R\} $$ Closure under subtraction and multiplication is easy to prove.

It is also a fact that, if $R$ has an identity $1$, then $1\in Z(R)$, simply because $$ 1r=r=r1 $$ by definition.

If a ring $R$ (with or without identity) is commutative, then obviously $Z(R)=R$.


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