Center of a ring is a subring that contains identity, but what happens in the case of ring of all Even integers?

Question

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?

Answer 1

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.)

Answer 2

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 $Z(R)=R$.