**Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.7**

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.

Solution: 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, $1 \cdot x = x \cdot 1 = x$ 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 $r \in R$, we have $$x^{-1}r^{-1} = (rx)^{-1} = (xr)^{-1} = r^{-1}x^{-1},$$ 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.