Since addition modulo n and multiplication modulo n of integers are commutative . So, the ring \(\displaystyle{\left(\mathbb{Z}_{{6}}+_{{6}},\times_{{6}}\right)}\) is a commutative ring.

Suppose \(\displaystyle{a},{b}\in\mathbb{Z}_{{6}}\)

Then, ab=ba=m , where m is the reminder obtained when ab, ba is divided by 6 and \(\displaystyle{0}\le{m}\le{5}\)

ab \(\displaystyle\equiv\) m mod 6 , ba \(\displaystyle\equiv\) m mod 6

Also, a+b=b+a=n , where n is the reminder obtained when a+b, b+a is divided by 6 and \(\displaystyle{0}\le{n}\le{5}\)

a+b \(\displaystyle\equiv\) n mod 6 , b+a \(\displaystyle\equiv\) n mod 6

Hence, \(\displaystyle\mathbb{Z}_{{6}}\) is a commutative ring

A ring is a field if its every element has an inverse.

As \(\displaystyle{2},{3}\in\mathbb{Z}_{{6}}\)

but 2*3=6 mod 6=0

So, 2 and 3 are zero divisors and do not have inverse .

So, \(\displaystyle{\left(\mathbb{Z}_{{6}}+_{{6}},\times_{{6}}\right)}\) is not a field.

Suppose \(\displaystyle{a},{b}\in\mathbb{Z}_{{6}}\)

Then, ab=ba=m , where m is the reminder obtained when ab, ba is divided by 6 and \(\displaystyle{0}\le{m}\le{5}\)

ab \(\displaystyle\equiv\) m mod 6 , ba \(\displaystyle\equiv\) m mod 6

Also, a+b=b+a=n , where n is the reminder obtained when a+b, b+a is divided by 6 and \(\displaystyle{0}\le{n}\le{5}\)

a+b \(\displaystyle\equiv\) n mod 6 , b+a \(\displaystyle\equiv\) n mod 6

Hence, \(\displaystyle\mathbb{Z}_{{6}}\) is a commutative ring

A ring is a field if its every element has an inverse.

As \(\displaystyle{2},{3}\in\mathbb{Z}_{{6}}\)

but 2*3=6 mod 6=0

So, 2 and 3 are zero divisors and do not have inverse .

So, \(\displaystyle{\left(\mathbb{Z}_{{6}}+_{{6}},\times_{{6}}\right)}\) is not a field.