$\mathbb {Z}_{84}/(7) \cong \mathbb {Z}_{7}$

Question

Prove $\mathbb {Z}_{84}/(7) \cong \mathbb {Z}_{7}$ using each of the three isomorphism theorems for rings.

For the first isomorphism theorem I defined a homomorphism $\phi: \mathbb {Z}_{84} \to \mathbb {Z}_{7}$ defined by $\phi(x+84\mathbb {Z}):=x+7\mathbb {Z}$ and found that $ker(\phi) = (7)$ and said that $\phi$ is clearly surjective. Is this thinking correct?

Moving onto the other two isomorphism theorems I'm having trouble with these.

For the second theorem, I think that using the homomorphism $\phi(x+7\mathbb {Z}):=x+7\mathbb {Z}$ and $S = \mathbb{Z}/84\mathbb{Z}$ and $I = 7\mathbb{Z}$ might work, but I'm having trouble showing $S+I=\mathbb{Z}$ and $S ∩ I = 7\mathbb{Z}$

For the third theorem, I think that using the homomorphism $\phi(x+7\mathbb {Z}):=x+84\mathbb {Z} + 7\mathbb {Z} $ and $R = \mathbb{Z} $ and $I=84\mathbb{Z}$ and $J = 7\mathbb{Z}$ might work, but I'm having trouble showing $J/I=7\mathbb{Z}$

Also would a general form of this $\mathbb {Z}_{m}/(n) \cong \mathbb {Z}_{n}$ if $n|m$?

Answer 1

Your answer for the first isomorphism theorem is correct, but you would have to check that $\phi$ is well-defined upto choice of representative.

For the second isomorphism theorem, $S$ must be a subring of $\mathbb{Z}$; $\mathbb{Z}/84\mathbb{Z}$ is not a subring of $\mathbb{Z}$.

For the third theorem, $J/I$ is an ideal of $\mathbb{Z}/84\mathbb{Z}$ not $\mathbb{Z}$, and in fact $J/I = 7\mathbb{Z}/84\mathbb{Z} \simeq 7(\mathbb{Z}/84\mathbb{Z})$. The result is then an immediate application of the third isomorphism theorem.

Make sure you have all your rings and ideals straight!