Localization at a prime ideal in $\mathbb{Z}/6\mathbb{Z}$ How can we compute the localization of the ring $\mathbb{Z}/6\mathbb{Z}$ at the prime ideal $2\mathbb{Z}/\mathbb{6Z}$? (or how do we see that this localization is an integral domain)?

One simple way to compute this is to exploit the universal property of localization. By definition $\rm\ L\ =\ \mathbb Z/6_{\:(2)}\ =\ S^{-1}\ \mathbb Z/6\ $ where $\rm\ S\ =\ \mathbb Z/6 \ \backslash\ 2\ \mathbb Z/6\ =\ \{\bar 1, \bar 3, \bar 5\}\:.\: $ Hence, since the natural map $\rm\ \mathbb Z/6\ \to\ \mathbb Z/2\ $ maps $\rm\:S\:$ to units, by universality  it must factor through $\rm\:L\:,\ $ i.e. $\rm\ \mathbb Z/6\ \to\ L\ \to\ \mathbb Z/2\:.\ $ Thus either $\rm\ L =  \mathbb Z/6\ $ or $\rm\ L = \mathbb Z/2\:.\: $ But $\rm\ \bar 3\in S,\ \ {\bar3}^{-1}\not\in \mathbb Z/6\:,\ $ so $\rm\:L \ne \mathbb Z/6\:.\ $ Thus we infer $\rm\:L = \mathbb Z/2\:. $
