The center of a ring

Zero and one are central: for every $a\in A$ distributivity gives $a\cdot 0=a\cdot (0+0)=a\cdot 0+a\cdot 0$, and cancelling $a\cdot 0$ in the additive group yields $a\cdot 0=0$, while $0\cdot a=(0+0)\cdot a=0\cdot a+0\cdot a$ similarly yields $0\cdot a=0$; hence $0\cdot a=0=a\cdot 0$ and $0\in Z(A)$. Likewise $1\cdot a=a=a\cdot 1$ for every $a$ by the identity law, so $1\in Z(A)$.

The center is closed under addition: if $z,z^{\mathrm{\prime }}\in Z(A)$ and $a\in A$, then $(z+z^{\mathrm{\prime }})\cdot a=z\cdot a+z^{\mathrm{\prime }}\cdot a=a\cdot z+a\cdot z^{\mathrm{\prime }}=a\cdot (z+z^{\mathrm{\prime }})$ by the two distributive laws, so $z+z^{\mathrm{\prime }}\in Z(A)$.

The center is closed under multiplication: if $z,z^{\mathrm{\prime }}\in Z(A)$ and $a\in A$, then $(z{z}^{\mathrm{\prime }})\cdot a=z\cdot (z^{\mathrm{\prime }}\cdot a)=z\cdot (a\cdot z^{\mathrm{\prime }})=(z\cdot a)\cdot z^{\mathrm{\prime }}=(a\cdot z)\cdot z^{\mathrm{\prime }}=a\cdot (z\cdot z^{\mathrm{\prime }})$, using associativity of multiplication together with the centrality of $z$ and then of $z^{\mathrm{\prime }}$; hence $z{z}^{\mathrm{\prime }}\in Z(A)$.

The center is commutative: for $z,z^{\mathrm{\prime }}\in Z(A)$, centrality of $z$ evaluated at $a=z^{\mathrm{\prime }}$ gives $z{z}^{\mathrm{\prime }}=z^{\mathrm{\prime }}z$, so multiplication in $Z(A)$ is commutative.

The equivalence $A$ commutative $\iff$ $Z(A)=A$ holds: if $A$ is commutative then $za=az$ for all $z,a\in A$ by [F3], so every element of $A$ is central and $Z(A)=A$; conversely if $Z(A)=A$, then for arbitrary $x,y\in A$ the element $x$ lies in $Z(A)$ and hence $xy=yx$, so $A$ is commutative by [F3].

The center is closed under additive inverses: if $z\in Z(A)$ and $a\in A$, then $(-z)\cdot a+z\cdot a=(-z+z)\cdot a=0\cdot a=0$ by step 1.1, so $(-z)\cdot a$ is the additive inverse of $z\cdot a$ and therefore equals $-(z\cdot a)$ by uniqueness of additive inverses in $(A,+,0)$; symmetrically $a\cdot (-z)=-(a\cdot z)$. Since $z$ is central, $-(z\cdot a)=-(a\cdot z)$, so $(-z)\cdot a=a\cdot (-z)$ and $-z\in Z(A)$.

By steps 1.1, 1.2, 2.1 and 1.3 the subset $Z(A)$ contains $1$ and is closed under addition, additive inverses and multiplication, so it is a subring of $A$ by [F2]; in particular it is a ring in its own right, with the addition, multiplication, zero and identity inherited from $A$.

Steps 1.1-1.3 and 2.1 supply the closure conditions of the center with its identity, step 3.1 assembles them into the statement that $Z(A)$ is a subring of $A$, step 1.4 shows that this subring is commutative, and step 1.5 gives the asserted equivalence between commutativity of $A$ and $Z(A)=A$; every element considered lies in $A$ and no choice principle is used.