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

Question
Commutative Algebra
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?

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.

### Relevant Questions

Suppose that R and S are commutative rings with unites, Let PSJphiZSK be a ring homomorphism from R onto S and let A be an ideal of S.
a. If A is prime in S, show that $$\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}={\left\lbrace{x}\in{R}{\mid}\phi{\left({x}\right)}\in{A}\right\rbrace}$$ is prime $$\displaystyle\in{R}$$.
b. If A is maximal in S, show that $$\displaystyle\phi^{{-{{1}}}}{\left({A}\right)}$$ is maximal $$\displaystyle\in{R}$$.
Let R be a commutative ring with unit element .if f(x) is a prime ideal of R[x] then show that R is an integral domain.
Suppose that a and b belong to a commutative ring and ab is a zero-divisor. Show that either a or b is a zero-divisor.
Show that a ring is commutative if it has the property that ab = ca implies b = c when $$\displaystyle{a}\ne{0}$$.
If R is a commutative ring with unity and A is a proper ideal of R, show that $$\displaystyle\frac{{R}}{{A}}$$ is a commutative ring with unity.
Suppose that R is a ring and that $$\displaystyle{a}^{{2}}={a}$$ for all $$\displaystyle{a}\in{R}{Z}$$. Show that R is commutative.
If A and B are ideals of a commutative ring R with unity and A+B=R show that $$\displaystyle{A}\cap{B}={A}{B}$$