If $\mathrm{char}(R)=\mathrm{char}(k)=0$, then $R$ contains a copy of $\mathbb{Q}$

Question

Suppose $R$ is a complete local ring with unique maximal ideal $m$ and let $k$ be the residue field $R/m$. It is given that the characteristic of $k$ is zero. So it implies that $\mathrm{char}(R)$ is zero also. From here we can conclude two things.

$1.$ $R$ contains one copy of $\mathbb{Z}$.

$2.$ $\mathbb{Z}\cap m=0$.

But from here how can we conclude that

$1.$ $R$ contains a copy of $\mathbb{Q}$.

$2.$ $R$ contains a maximal subfield, say $L$, using Zorn's Lemma.

Actually I was reading the proof of Cohen's Structure Theorem, where all these steps come, but I'm stuck to prove the above two results.

Any help or hints would be highly appreciated.

Thanks in advance.

Answer 1

The first question has been answered by Tsemo Aristide already. You only have to prove that primes $p \in {\mathbb Z} \subseteq R$ are invertible, because that implies that every non-zero element of ${\mathbb Z}$ is invertible in $R$. If such a $p$ is not invertible, then the ideal $pR$ is contained in the unique maximal ideal ${\frak m}$ of $R$, so $p \in {\frak m}$, so $p = 0$ in $k = R/{\frak m}$, contradicting the assumption that $k$ has characteristic $0$.

For the second question, the intuition is to start with the subfield ${\mathbb Q}$ of $R$. If this is not a maximal subfield of $R$, then pick a larger subfield and keep repeating this (transfinitely many times, if needed). Ultimately, take the union of all the subfields you've found and that is then, by construction, a maximal subfield.

This intuition is formalized through Zorn's Lemma. Consider the collection ${\cal P}$ of all subfields of $R$, ordered by set-inclusion. This collection is not empty, because it contains ${\mathbb Q}$. Every chain $C \subseteq {\cal P}$ has an upper bound in ${\cal P}$, namely $\bigcup C$. Therefore, by Zorn's Lemma, ${\cal P}$ contains a maximal element.

Answer 2

Hint: Let $p\in \mathbb{Z}\subset R$ not invertible, the ideal $I_p$ generated by $p$ is contained in the maximal ideal. It implies that $p\in m$ contradiction.