Intereting Posts

Show that this sum is an integer.
Image of complex circle under polynomial
If $|\lbrace g \in G: \pi (g)=g^{-1} \rbrace|>\frac{3|G|}{4}$, then $G$ is an abelian group.
set of Kolmogorov-random strings is co-re
Kaplansky's theorem of infinitely many right inverses in monoids?
How to prove that $(A\cup B) \cap C \subseteq A\cup (B\cap C)$
How do I find the Laurent series for $\frac{1}{z^2 – 4}$ at $z = 2$?
Composite functions and one to one
Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square.
Are quartic minimal polynomials over $\mathbb{Q}$ always reducible over $\mathbb{F}_p$?
Calculate $\lim_{n\to\infty}(\sqrt{n^2+n}-n)$.
Showing range is countable
Can $(X_1,X_2) \cap (X_3,X_4)$ be generated with two elements from $k$?
A limit and a coordinate trigonometric transformation of the interior points of a square into the interior points of a triangle
$\operatorname{Aut}(S_4)$ is isomorphic to $S_4$

1/ How to count the number of elements of $\mathbb{Z}[i]/(1+2i)^n$?

2/ How to write $\mathbb{Z}[i]/(1+2i)^n$ as direct sum of cyclic groups (in view of the structure theorem of finite abelian groups)?

- Contraction of maximal ideals in polynomial rings over PIDs
- Prove that the Gaussian Integer's ring is a Euclidean domain
- Is it true that an ideal is primary iff its radical is prime?
- For a ring $R$ with a single proper ideal $I$, show that $I$ is prime
- Why are maximal ideals prime?
- Irreducible elements in $\mathbb{Z} $

- Prove that a ring $R$ with no non-trivial right ideals and $aR=0$ has $|R|=p$ prime
- Does there exist a unital ring whose underlying abelian group is $\mathbb{Q}^*$?
- A doubt about lower nil radical while proving 2-primality of ring.( Baer-McCoy Radical)
- Subring of a finitely generated Noetherian ring need not be Noetherian?
- Jacobson radical of $R$, where $R$ is domain
- Is the group of units of a finite ring cyclic?
- Finitely many prime ideals lying over $\mathfrak{p}$
- Why are the only division algebras over the real numbers the real numbers, the complex numbers, and the quaternions?

Write $(1+2i)^n=u+iv$, $u,v\in\mathbb Z\setminus\{0\}$. Since $\mathbb Z[i]\simeq\mathbb Z\times\mathbb Z$ (as $\mathbb Z$-modules), we have $$\mathbb Z[i]/(u+iv)\simeq \mathbb Z\times\mathbb Z/\langle(u,v),(-v,u)\rangle.$$ The Smith Normal Form of the matrix $\left(\begin{matrix}u&v\\-v&u \end{matrix}\right)$ is $\left(\begin{matrix}1&0\\0&u^2+v^2 \end{matrix}\right)$. (It is not hard to show that $\gcd(u,v)=1$.) Thus we get $$\mathbb Z[i]/(u+iv)\simeq \mathbb Z/(u^2+v^2)\mathbb Z.$$ This shows that $|\mathbb Z[i]/(u+iv)|=u^2+v^2$, and therefore $$u^2+v^2=N(u+iv)=N((1+2i)^n)=N(1+2i)^n=5^n.$$

Here’s one approach you could take. Note that $(1+2i)^n(1-2i)^n=5^n$ and $(1+2i)^n$ and $(1-2i)^n$ are coprime in $\mathbb{Z}[i]$, so by the Chinese remainder theorem, $$\mathbb{Z}[i]/5^n\cong \mathbb{Z}[i]/(1+2i)^n\times \mathbb{Z}[i]/(1-2i)^n$$ as rings. But $$\mathbb{Z}[i]/(1+2i)^n\cong \mathbb{Z}[i]/(1-2i)^n$$ as rings (though not as $\mathbb{Z}[i]$-modules) by sending $i$ to $-i$, and it is easy to see that the underlying abelian group of $\mathbb{Z}[i]/5^n$ is $\mathbb{Z}/5^n\times\mathbb{Z}/5^n$ (since an element of $\mathbb{Z}[i]/5^n$ is just a number $a+bi$ where both $a$ and $b$ are taken mod $5^n$). If you have any decomposition of $\mathbb{Z}/5^n\times\mathbb{Z}/5^n$ as a product of two isomorphic abelian groups, then both factors must be isomorphic to $\mathbb{Z}/5^n$ (this follows easily from the classification of finite abelian groups, for instance). Thus the additive group of $\mathbb{Z}[i]/(1+2i)^n$ must be isomorphic to $\mathbb{Z}/5^n$.

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