The center of a ring is a subring

The center of a ring R is Z(R) = \{ z \in R \ |\ zr = rz\ \mathrm{for\ all}\ r \in R \}. Prove that Z(R) is a subring of R and that if R has a 1, then 1 \in Z(R). Prove also that the center of a division ring is a field.

Note first that 0 \in Z(R) since 0 \cdot r = 0 = r \cdot 0 for all r \in R; in particular, Z(R) is nonempty. Next, if x,y \in Z(R) and r \in R, then (x-y)r = xr - yr = rx - ry = r(x-y). By the subgroup criterion, Z(R) \leq R. Moreover, xyr = xry = rxy, so that xy \in Z(R); by definition, Z(R) is a subring.

If R has a 1, then by definition, 1 \cdot x = x \dot 1 = x for all x \in R. Thus 1 \in Z(R).

Now let R be a division ring, and consider Z(R). If x \in Z(R), note that by the cancellation law, the inverse of x is unique; denote it by x^{-1}. Clearly, (r^{-1})^{-1} = r. Since (ab)(b^{-1}a^{-1}) = 1, we have (ab)^{-1} = b^{-1}a^{-1}. Now let r \in R. x^{-1}r^{-1} = (rx)^{-1} = (xr)^{-1} = r^{-1}x^{-1}, and since r^{-1} is arbitrary in R, x^{-1} \in Z(R). Thus Z(R) is a commutative division ring- that is, a field.

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