Show that $4\mathbb{Z}$ is maximal in $2\mathbb{Z}$.

Question

Show that $4\mathbb{Z}$ is maximal in $2\mathbb{Z}$.

We will use the fact that $4\mathbb{Z}$ maximal $\iff 2\mathbb{Z}/4\mathbb{Z}$ is a field. $$2\mathbb{Z}/4\mathbb{Z} = \{k + 4\mathbb{Z} : k \in 2\mathbb{Z}\}$$ We are going to determine the number of elements in this set. Let $k\in 2\mathbb{Z}$. $$k \equiv 0 \mod4 \implies k+4\mathbb{Z} = 0 + 4\mathbb{Z}$$ $$k \equiv 1 \mod4 \implies 4|(k-1) \implies k = 4r+1 \implies 2\not|k$$ $$k \equiv 2\mod 4 \implies 2+4\mathbb{Z} \in 2\mathbb{Z}/4\mathbb{Z}$$ $$k \equiv 3 \mod4 \implies 4|(k-3) \implies k = 4r+3 \implies 2\not|k$$ So the set has $2$ elements. This means it is a field since $0 +4\mathbb{Z}$ is the zero, and $2+4\mathbb{Z}$ is the one. i.e every non-zero element is invertible. Infact we have $$2\mathbb{Z}/4\mathbb{Z} \cong\mathbb{Z}_{2}$$

There are most likely an easier way to show the isomorphism, but I can't think of it.

Answer 1

Suppose there exists an ideal $A$ in $2 \mathbb Z$ such that $4\mathbb Z \subseteq A \subseteq 2\mathbb Z.$ Since, $I$ is an ideal of $2 \mathbb Z,$ therefore, it must be of the form $2m \mathbb Z$ for some $m \in \mathbb Z.$

So, we have $2\cdot 2 \mathbb Z \subseteq 2m \mathbb Z\subseteq 2 \mathbb Z.$ So, $I$ must either be $4\mathbb Z$ or $2\mathbb Z.$ Hence, $4 \mathbb Z$ is maximal.

Answer 2

The fact that any subgroup of $2\Bbb{Z}$ has to be of the form $2n\Bbb{Z}$ for some $n\in\Bbb{N}$, implies that $4\Bbb{Z}$ is maximal in $2\Bbb{Z}$.