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

Question
Commutative Algebra
asked 2020-12-03
Show that \(\displaystyle{\left(\mathbb{Z}_{{6}}+_{{6}},\times_{{6}}\right)}\) is a commutative ring. Is \(\displaystyle{\left(\mathbb{Z}_{{6}}+_{{6}},\times_{{6}}\right)}\) a field?

Answers (1)

2020-12-04
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.
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