Proving that $\mathbb{Z}/128\mathbb{Z}$ has exactly one maximal ideal?

Question

I would like the prove that $\mathbb{Z}/128\mathbb{Z}$ has exactly one maximal ideal. I believe this has to do with the fact that $128 = 2^7$, but I'm a little lost on everything else here.

I'm strongly inclined to believe the Chinese Remainder Theorem fits in here, but breaking $128$ down to a multiple of $2$ doesn't seem to lend itself to the CRT because $2$ is obviously not relatively prime with itself, which is a requirement of the CRT.

The fact that it needs to be an ideal of a quotient further confuses me, but I suspect an understanding of this point is what is holding me back from getting closer to a solution. Additionally, I'm concerned with the "exactly one" part, but I'm not sure if this results from proving something about any maximal ideals of the ring.

Answer 1

Hint: The ideals of $\mathbb Z/128\mathbb Z$ correspond to the ideals of $\mathbb Z$ that contain $128\mathbb Z$. There is only one maximal ideal in $\mathbb Z$ that contains $128\mathbb Z$.

Answer 2

Do you know maximal ideals are prime?

In your ring, $\overline 2^7=0$, so that any prime ideal of the ring necessarily contains $\overline 2$. But the ideal $(\overline 2)$ is clearly maximal.

In particular, all of your maximal ideals contain the ideal $(\overline 2)$ and so...

Answer 3

Hint: If $\phi\colon R\to S$ is a ring epimorphism, then a subset $I\subseteq S$ is an ideal if and only if $\phi^{-1}[I]$ is an ideal, and it is maximal if and only if $\phi^{-1}[I]$ is maximal.

Alternative Hint: Note that ${\mathbf Z}/128\mathbf Z$ has cyclic additive group and characteristic 128. Notice that this implies that every ring quotient also has cyclic additive group and characteristic dividing 128. This leaves only one possibility of a quotient which is a field.

Answer 4

$\Bbb Z$ is a principal ideal domain, meaning that any ideal is generated by one element. The same is true for the quotient $\Bbb Z/128\Bbb Z$.

Now every non-unit generates a non-trivial ideal (like the whole ring or $(0)$). A non-unit needs to share som factors with $2^7$ (otherwise it would be invertible), so any non-unit is an even number. Let $I=(2k)$, then $I\subseteq (2)$ and $I\subseteq (k)$. Can you find a $k$, such that $(2)\subset(k)\neq \Bbb Z/128\Bbb Z$?