Simple exercise about ideals of integers

Question

Let $V$ is an ideal of $\mathbb Z$ and $7\mathbb Z \subseteq V \subseteq \mathbb Z$

Show that $V=\mathbb Z$ or $V=7\mathbb Z$

I know it is so simple but I'm confused. Thanks for any help.

My tryings :

Since $V$ is an ideal $V=\{vk | k \in \mathbb Z\}$ ($v\in \mathbb Z$) and since $7$ is a prime $gcd(v,7)=1$ or $gcd(v,7)=7$.

Answer 1

$V/7\mathbb{Z}$ is an ideal of $\mathbb{Z}/7\mathbb{Z}$ which is a field. Then $V/7\mathbb{Z}=0$ or $V/7\mathbb{Z}=\mathbb{Z}/7\mathbb{Z}$. In other words $V=7\mathbb{Z}$ or $V=\mathbb{Z}$.

Answer 2

Assume $V$ is an ideal of $\mathbb Z$ containing $7 \mathbb Z$.

Then either $\bbox [yellow]{V=7\mathbb Z \;\text {or}}$ there is an element $n\in V$ not in $7 \mathbb Z$.

Note in the latter case $\gcd(n,7)=1$ (because if $\gcd(n,7)=7$ then $n \in 7\mathbb Z).$

But then by Bezout's lemma, there exist $a$ and $b$ such that $na+7b=1,$

so, because $V$ is an ideal containing $n$ and $7$, it contains $na$ and $7b$, and hence $1$, so $\bbox[yellow]{V \;\text{contains} \;\mathbb Z.}$

Answer 3

$\mathbb{Z}$ is an abelian group so by the third isomorphism theorem, $|\mathbb{Z}:V| |V:7\mathbb{Z}| = |\mathbb{Z}:7\mathbb{Z}| = 7$ which is prime, so $|V:7\mathbb{Z}| = 1$ or $7$ (corresponding to $7\mathbb{Z}$ and $\mathbb{Z}$ respectively)

Answer 4

$7\mathbb Z \subseteq V \subseteq \mathbb Z$

Show that $7\mathbb Z= V$ or $V= \mathbb Z$

Is the same as saying, show that $7\mathbb Z$ is a maximal ideal of $\mathbb Z$

And the maximal ideals in $\mathbb Z$ are of the form $p\mathbb Z$ where $p$ is a prime number.