Problem Statement
Prove that $a + bi$ and $a' + b'i$ represent the same cosets in $\mathbb{Z}[i]/(5)$ iff $a \equiv a' (\text{mod 5})$ and $b \equiv b' (\text{mod 5})$
Definitions
$\mathbb{Z}[i] = \{ a + bi \mid a,b, \in \mathbb{Z} \}$
$(5) = \{5a + 5bi \mid a,b \in \mathbb{Z} \}$
Relevant Facts
Lemma: Let $G$ be a group, $H < G$ a subgroup and for all $g_1,g_2 \in G$ the following are equivalent:
- $g_1, g_2$ belong to the same left coset of $H$
- $g_1 H = g_2 H$
- $g_2^{-1}g_2 \in H$
My Attempt
This being an 'if an only if' statement we need to prove the conclusion from the predicate ($\Rightarrow$) and the predicate from the conclusion ($\Leftarrow$). However, given the nature of my misunderstanding I would like to focus solely on the forward statement and address some questions I have there that I believe will carry over to the reverse statement. To that end:
$\Rightarrow$ Suppose $(a+bi),(a'+b'i)$ belong to the same coset of $\mathbb{Z}[i]/(5)$ Then by the above Lemma that means that:
- $(a+bi) + (5) = (a'+b'i) + (5)$
- $-(a+bi)+(a'+b'i) = (-a-bi)+(a'+b'i) = (a'-a) + (b'-b)i \in \mathbb{Z}[i]/(5)$
Now, elements of the quotient group $\mathbb{Z}[i]/(5)$ have the form: $$z + (5) = (c+di) + (5f + 5gi) = (c+5f) + (d + 5g)i,\ \text{where}\ c,d,f,g,5 \in \mathbb{Z}$$ Which can be written more succinctly as $$ k + ni,\ \text{where}\ k,n \in \mathbb{Z}$$ Since $(c + 5f), (d + 5g) \in \mathbb{Z}$
Now the statement $t \equiv t' (\text{mod}\ 5)$ implies that $t-t' = 5s,\ s \in \mathbb{Z}$
So we have: $$(a'-a) + (b'-b)i = 5q + (5s)i = k + ni$$ If $a \equiv a' (\text{mod}\ 5)$ and $b \equiv b' (\text{mod}\ 5)$
Questions/Concerns
My main concern here is that it seems that the inclusion $(a'-a) + (b'-b)i \in \mathbb{Z}[i]/(5)$ doesn't really rely on the fact that $a,b$ are congruent modulo $5$ to their primed counterparts. Rather, as long as $a,a',b,b' \in \mathbb{Z}$ then we have our inclusion. Is it the case that I've messed up the form for an arbitrary element of $\mathbb{Z}[i]/(5)$? I wanted to pull the $5$'s out of each factor in the resultant sum: $(c+5f) + (d + 5g)i = 5(\frac{c}{5} + f) + 5(\frac{d}{5} + g)i$ as that would have the same sort of form as the implied expression of congruence modulo $5$ i.e., $t \equiv t' (\text{mod}\ 5) \implies t-t' = 5s,\ s \in \mathbb{Z}$ however the sums in the parentheses $(\frac{c}{5} + f)$ and $(\frac{d}{5} + g)$ would, in general, not be integers, where the divisors $5s$ are necessarily integers.
What am I missing here? Or am I on the complete wrong track for this part of the proof?
I think your problem is simply a misapplication of the lemma you quote, the condition that $a+bi + (5) = a'+b'i + (5)$ is equivalent to the statement that $-(a+bi)+(a'+b'i) \in (5)$ (this lemma is sometimes referred to as the "coset criterion"). This should make your analysis much more straightforward.
Note that if the version of the lemma you were applying were true, it would force every element of the quotient ring to be identified. This is a good "sanity check" when you're doing a proof to see if you've made a mistake.