Find all elements $m$ in $\mathbb{Z}_{42}$ such that $(12)+(m)$ is a proper ideal of $\mathbb{Z}_{42}$.

Question

This doesn't really sound too hard, and my guess is that it's all the even numbers in $\mathbb{Z}_{42}$ since if $m$ is any odd number it would seem that $(12) + (m) = \mathbb{Z}_{42}$. The system used to check for answers marks it wrong and I can't see why I'm wrong. Any tips appreciated.

Answer 1

Hint. $(12)+(m)$ is not a proper ideal of $\mathbb{Z}_{42}$ if and only if $(12)+(m)=(1).$

Answer 2

Suppose you have a $a,b,c,p$ such that $p$ is a prime divisor of $a$,$b$, and $c$. Then $p$ will also divide every element of $(a)+(b)$. Thus, $(a)+(b)$ is a (not necessarily proper) subset of $(p)$. Since $(p)$ is a proper subset of $\mathbb{Z}_c$, $(a)+(b)$ is also a proper subset of $\mathbb{Z}_c$. Thus, in the problem you present, it is at least sufficient (I haven't addressed necessity) condition to find a number that shares a prime factor with 12 and 42. 2 is one such number, but 3 is as well.

Another way of putting it is that if $gcd(a,b)$ is not a unit in $\mathbb{Z}_c$, then $(a)+(b)$ is a proper ideal.