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I want to find the kernel of $f: \mathbb{Z} \rightarrow \mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/18\mathbb{Z}$ where $f(n) = (3n \bmod 7, 3n \bmod 18)$

However, it seems to me that I will not be able to $1 \bmod 18$ in any way. Is it because the identity is $(0 \bmod 7, 0 \bmod 18)$? I did read this in the solution manual that one of my colleague students wrote. But I am not sure if he/she was right. Can someone confirm this or explain what the kernel is?

Thank you,

V.

V. Voorspel
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    And yes, the identity is $(0 \bmod 7, 0 \bmod 18)$ instead of $1\mod 18$. – Cave Johnson Jun 27 '17 at 14:31
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    Can you be a little more clear in your question? Is this a ring homomorphism or a group homomorphism? (not that it necessarily matters but you can definitely check ring homomorphisms differently). Furthermore if group, is your operation addition then? Because at least with ring your addition and multiplication are defined and clear as to which is which – Charlie Tian Jun 27 '17 at 14:32
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    https://en.wikipedia.org/wiki/Ring_homomorphism

    in either case, the kernel is the preimage of 0, as the unity is not taken as the identity, but only the 0 value

     – Charlie Tian Jun 27 '17 at 14:32

2 Answers2

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The kernel is, by definition, the set $\ker f \subset \mathbb{Z}$ of numbers $n$ that satisfy $f(n)=0$. Note that $0$ here means the neutral element with respect to addition. In $\mathbb{Z}/7\mathbb{Z} \times \mathbb{Z}/18\mathbb{Z}$ this neutral element is $(0 \mod 7, 0 \mod 18)$.

Now, $f(n)=0$ means

$$\begin{cases} 3n = 0 \mod 7 \\ 3n = 0 \mod 18 \end{cases}$$

Which means

$$\begin{cases} 3n = 7k \\ n = 6l \\ \end{cases}$$

for some $k$ and $l$.

So, $n$ must be divisible by both $7$ and $6$. Hence the kernel is ${\rm lcm}(7,6) \mathbb{Z} = 42\mathbb{Z}$.

Bill Dubuque
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lisyarus
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We have $\ker f = \ker f_1 \cap \ker f_2$ where $$ f_1(n) = 3n \bmod 7 $$ $$ f_2(n) = 3n \bmod 18 $$

Now, $\ker f_1 = 7\mathbb Z$ and $\ker f_2 = 6\mathbb Z$.

Therefore, $\ker f = 7\mathbb Z \cap 6\mathbb Z = lcm(7,6)\mathbb Z = 42 \mathbb Z$.

lhf
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