# Show that (ZZ_6 +_6, xx_6) is a commutative ring. Is (ZZ_6 +_6, xx_6) a field?

Show that $\left({\mathbb{Z}}_{6}{+}_{6},{×}_{6}\right)$ is a commutative ring. Is $\left({\mathbb{Z}}_{6}{+}_{6},{×}_{6}\right)$ a field?
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Anonym

Since addition modulo n and multiplication modulo n of integers are commutative . So, the ring $\left({\mathbb{Z}}_{6}{+}_{6},{×}_{6}\right)$ is a commutative ring.
Suppose $a,b\in {\mathbb{Z}}_{6}$
Then, $ab=ba=m$ , where m is the reminder obtained when ab, ba is divided by 6 and $0\le m\le 5$
ab $\equiv$ $m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}6,ba$ $\equiv$ $m\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}6$
Also, $a+b=b+a=n$ , where n is the reminder obtained when $a+b,b+a$ is divided by 6 and $0\le n\le 5$
$a+b$ $\equiv$ $n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}6,b+a$ $\equiv$ $n\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}6$
Hence, ${\mathbb{Z}}_{6}$ is a commutative ring
A ring is a field if its every element has an inverse.
As $2,3\in {\mathbb{Z}}_{6}$
but $2\cdot 3=6\phantom{\rule{0.667em}{0ex}}\mathrm{mod}\phantom{\rule{0.167em}{0ex}}\phantom{\rule{0.167em}{0ex}}6=0$
So, 2 and 3 are zero divisors and do not have inverse .
So, $\left({\mathbb{Z}}_{6}{+}_{6},{×}_{6}\right)$ is not a field.