Field that contains $(\mathbb{Z}/p^n\mathbb{Z})^*$

Question

Let $p$ be an odd prime. There exists a field $F$ that contains an isomorphic copy of $(\mathbb{Z}/p^n\mathbb{Z})^*$ as a multiplicative subgroup?

Clearly if $n=1$ we can take $F=\mathbb{F}_p$, but for $n>1$ ?

Answer 1

I assume $p$ is an odd prime.

The order of $(\mathbb{Z} / p^n)^*$ is $(p-1)p^{n-1}$. You can decompose it as the direct sum of the subgroup of $(p-1)$-th roots of unity and the group of elements that are $1 \bmod p$.

We can define the logarithm function on the latter group as

$$ \log(1 + x) \equiv \sum_{i=1}^{\infty} (-1)^{i+1} \frac{x^i}{i} \pmod{p^n}$$

This is really a finite sum, because by assumption $p \mid x$, and for sufficiently large $i$ we have $p^i / i \equiv 0 \pmod{p^n}$. And well-defined mod $p^n$, because the power of $p$ in the numerator is always at least as much as the power of $p$ in the denominator.

By doing the tedious calculations with power series, you can show $\log$ is a homomorphism from the multiplicative group of elements that are $1 \bmod p$ to the additive group of elements that are $0 \bmod p$.

Incidentally, you can define its inverse $\exp$ in the same way, as the series

$$ \exp(x) \equiv \sum_{i=0}^{\infty} \frac{x^i}{i!} \pmod{p^n} $$

Thus, $(\mathbb{Z} / p^n)^*$ is cyclic, and thus can be embedded in any field with a primitive $p^n(p-1)$-th root of unity.

Answer 2

According to Theorem 22 in section 1.6 in this note by Pete L. Clark, the group $(\mathbb Z/p^n\mathbb Z)^*$ is cyclic, when $p$ is an odd prime. Hence we can answer your question affirmatively:
Just take the algebraic closure $K$ of $\mathbb Q,$ then $K$ contains all solutions of the equation $$x^{p^n-p^{n-1}}-1=0,$$ which is a cyclic group of the same order as $(\mathbb Z/p^n\mathbb Z)^*,$ and so $K$ suffices for our prupose.
P.S. It suffices to take, for a given $n,$ the splitting field of the equation in question over $\mathbb Q.$
Hope this helps.