Show that $f^{-1}(\langle0\rangle)$ is not a maximal ideal of $\mathbb{Z}$.

Question

Let $f\colon \mathbb{Z} \to \mathbb{Q}$ be a ring homomorphism. Show that $f^{-1}(\langle0\rangle)$ is not a maximal ideal of $\mathbb{Z}$.

Answer 1

Hint: $\mathbb{Z}/\mathrm{ker}(f)$ is isomorphic to the image of $f$. What can $\mathrm{ker}(f)$ be?

Answer 2

If $f(n)=0$ then $nf(1)=0$ and so $f(1)=0$ thus $f(x)=0$ for all $x$.

Answer 3

Hint:

$f^{-1}(0)=\ker f\;$ , so if this is a maximal ideal, then $\;\Bbb Z/\ker f\le \Bbb Q\;$ is a field ...but what are the only possible subfields of the rationals?

Answer 4

Let $R$ and $S$ be any two rings, commutative or not, unital or not, and let $\phi:R \to S$ be a homomorphism. Even if $\phi$ doesn't map $1_R \mapsto 1_S$, it must preserve idempotents; that is, if $e \in R$ satisfies $e^2 = e$, then $\phi(e) = \phi(e^2) = (\phi(e))^2$. Applying this notion to the homomorphism $f:\Bbb Z \to \Bbb Q$, we see that $f(1_{\Bbb Z}) = (f(1_{\Bbb Z}))^2$; since $\Bbb Q$ is a field, this forces $f(1_{\Bbb Z}) = 0$ or $f(1_{\Bbb Z}) = 1_{\Bbb Q}$. In the former case, $f(\Bbb Z) = \{0\}$, in the latter $f(\Bbb Z) = \Bbb Z \subset \Bbb Q$, that is $\Bbb Z$ considered (via an isomorphic copy or whatever) as a subset of $\Bbb Q$. In the former case, $\ker f = \Bbb Z$, in the latter $\ker f = \{0\}$. In neither case is $\ker f$ a maximal ideal in $\Bbb Z$. QED.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!

Answer 5

Consider $f(1)$. If you require the condition $f(1_{\mathbb{Z}})=1_{\mathbb{Q}}$, then what must $f(n)$ be for any other given $n\in\mathbb{Z}$? When can $f(n)$ be equal to $0_{\mathbb{Q}}$?

If you do not require the condition $f(1_{\mathbb{Z}})=1_{\mathbb{Q}}$, then either $f(1)=0$ in which case what must $f(n)$ be for any other given $n\in\mathbb{Z}$? or $f(1)\neq 0$ in which case again what must $f(n)$ be for any other given $n\in\mathbb{Q}$?

Are the subsets $\{0_{\mathbb{Q}}\}$ or $\mathbb{Q}$ maximal ideals of $\mathbb{Q}$?