Find all elements in ideal of $\Bbb Q$ generated by $J$ as follows.

Question

Let $\Bbb Q$ be a field of all rational numbersa and its ideal $J = \langle \Bbb Z \cup \{\frac{1}{2}\} \rangle$.

What are the all elements in J? Is it like this?

$J = \{z + \frac{q}{2} + \frac{n}{2} \mid z,n \in \Bbb Z, q \in \Bbb Q \}$.

Answer 1

A field $K$ never has nontrivial ideals. Indeed as an ideal $I$ is closed under multiplication with field elements and any nonzero element in a field is invertible we find $0 \neq k \in I \Rightarrow 1=k^{-1} k \in I \Rightarrow j = jk^{-1}k \in I {\forall j \in K} \Rightarrow K=I$

This shows that any ideal in a field $K$ is either $0$ or $K$.