Find all ideals in $\mathbb{Z}$ and also in $\mathbb{Z}_{10}$

Question

I have thought that the even numbers form an ideal in $\mathbb{Z}$ and also the $\mathbb{Z}_n$ where n is a natural, besides {0} and $\mathbb{Z}$ itself I am well? What others are there? The ideals in $\mathbb{Z}_{10}$ are {0,2,4,6,8}, {0}, and $\mathbb{Z}_{10}$, but I can not find any more.

Answer 1

The ideals of $\mathbb{Z}$ must be subgroups, and the only additive subgroups of $\mathbb{Z}$ are those generated by any given element of $\mathbb{Z}$, i.e. for each $n \in \mathbb{Z}$ there is a subgroup $(n)$ whose elements are the multiples of $n$, and in fact these exhaust all possible subgroups.

To see that $(n)$ is a subgroup, notice if $an, bn \in (n)$, then $an+bn = (a+b)n \in (n)$, and if $an \in (n)$, then $-an$ is also a multiple of $n$, so $-an \in (n)$.

To see that these are all the subgroups, notice that if $H \subset \mathbb{Z}$ is a subgroup, if $H = \{0\}$, we have $H = (0)$ and so this case is covered. Otherwise, let $n \in H$ be minimal among all nonzero integers in $H$. We can select $n$ to be positive as $H$ contains additive inverses for all of its elements. So then certainly all multiples of $n$ are in $H$ as $n+n+\cdots + n \in H$ for as many summands as you might want. Similarly for $-(n+n+\cdots + n)$.

Now suppose $z \in H$ with $z$ not a multiple of $n$. Then by the division algorithm in $\mathbb{Z}$, we have $z = qn+r$ for some $q,r \in \mathbb{Z}$ with $0 \le r < n$. But then $z-qn \in H$ as $qn \in H$, and $z-qn=r$, but we select $n$ minimal among all (positive) integers in $H$, a contradiction. So the only elements of $H$ are multiples of $n$, i.e. $H = (n)$.

Further, almost by definition, $(n)$ is also an ideal in $\mathbb{Z}$, as if $kn \in (n)$ and $m \in \mathbb{Z}$, we must have $kn\cdot m = (km)\cdot n \in (n)$.

Similarly for $\mathbb{Z}_{k} = \mathbb{Z}/(k)$ for any integer $k$.