Group homomorphisms and kernels of complex numbers

Question

Consider the additive group $$ \mathbb{Z}[\sqrt{-13}]:= \{n+m·\sqrt{-13} \mid n, m ∈ \mathbb{Z}\} $$ and a map $$ f: (\mathbb{Z}[\sqrt{-13}],+) → (\mathbb{Z}_{11}, +) $$ defined by $f(n+m \sqrt{-13}):= \overline {n+3m} ∈ \mathbb{Z}_{11}$.

• Prove that $f$ is a group homomorphism.

• Then prove that $$ \ker(f)= \left\{11n +\left(8+\sqrt{−13}\right)m \mid n,m \in \mathbb{Z}\right\} $$

• Show that $(\mathbb{Z}[\sqrt{-13}], +)/\ker(f) \cong \mathbb{Z}_{11}$.

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For the first part, this is what I have so far:

Since \begin{align*} (n+m \sqrt{-13}) + (s+t \sqrt{-13}) &= n + s + m\sqrt{-13} + t\sqrt{-13}\\ &= (n+s) + (m+t)\sqrt{-13} \end{align*} and \begin{align*} (\overline {n+3m}) + (\overline{s+3t}) &= \overline {n}+ \overline {s} + \overline {3m} + \overline {3t} = \overline {n+s} + \overline {3m + 3t}\\ &= \overline {n+s} + \overline {n+s + 3m + 3t} = \overline {(n+3m) + (s +3t) }, \end{align*} we see that $f$ is a homomorphism.

But I am really unsure about the kernel. Any hints would be greatly appreciated.

Answer 1

Remember that $\ker f = \{n + m\sqrt{-13}\in\Bbb Z[\sqrt{-13}]\mid f(n + m\sqrt{-13}) = n + 3m = 0\}$. To show that $\ker f$ is the set you described (call it $S$), we show inclusions in both directions.

Let $n + m\sqrt{-13}\in\ker f$. Then $f(n + m\sqrt{-13}) = n + 3m = 0$. That is, $n + 3m\equiv 0\pmod{11}$, so for some $k\in\Bbb Z$, $n + 3m = 11k$. Then $n = 11k - 3m$, so $n + m\sqrt{-13} = 11k - 3m + m\sqrt{-13} = 11(k -m) + (8 + \sqrt{-13})m$. This is visibly of the form above, so $\ker f\subseteq S$.

Conversely, take an element $11n + (8+ \sqrt{-13})m\in S$. Applying $f$, we have $$ f(11n + (8+ \sqrt{-13})m) = 11n + 8m +3m = 11(n+m), $$ and $11(n+m)\equiv 0\pmod{11}$, so $S\subseteq\ker f$.

Thus, it follows that $S = \ker f$.

To show the isomorphism, use the first isomorphism theorem, which says that for a homomorphism of groups $f: G\to H$, $G/\ker f\cong\operatorname{im}(f)$. You will have your result if you show that $\operatorname{im}(f) = \Bbb Z_{11}$; i.e., that $f$ is surjective. For this, consider elements of the form $n + 0\sqrt{-13}$. Can you show that every element of $\Bbb Z_{11}$ arises as the image of some $n\in\Bbb Z$ under $f$?

Answer 2

Idea:

For the kernel, you want all those elements $\alpha=n+m\sqrt{-13} \in \mathbb{Z}[\sqrt{-13}]$ such that it maps to the identity (which in this case is $\bar{0}$) of $\mathbb{Z}_{11}$. So you want \begin{align*} f(\alpha) & = \bar{0}\\ n+3m & \equiv 0 \pmod{11}\\ n & \equiv -3m \pmod{11}\\ n & \equiv 8m \pmod{11} \end{align*} This shows $\alpha=n+m\sqrt{-13}=8m+11k+m\sqrt{-13}=11k+(8+\sqrt{-13})m$, where $m,k \in \mathbb{Z}$.