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How to prove that the $p$-adic units can be written as $$\mathbb{Z}_p^\times \cong \mu_{p-1}\times(1 + p\mathbb{Z}_p) \cong \mathbb{Z}/(p-1)\mathbb{Z}\times\mathbb{Z}_p$$ where $\mu_n$ is the $n$-th roots of unity in $\mathbb{Z}_p$?

Here $p>2$ is a prime number.

Any hint or link would be helpful.

Besides, it looks strange: why the units of $\mathbb{Z}_p$ is isomorphic to some guy which looks bigger than $\mathbb{Z}_p$?

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    To answer your second question, the isomorphism you’re worrying about, from $\Bbb Z_p^+$ into the units, shrinks things, that’s all. – 2017-01-16
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    The first is just a way to factor units as roots of unity multiplied with a 1-unit. The roots of unity are in $\mathbb{Z}_p$ by Hensel's lemma, and there is a root of unity in each of the cosets $j+p\mathbb{Z}_p$. – 2017-01-16
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    I think the second follows from exponentiation of $1+p$. What is the set $(1+p)^{\mathbb{Z}_p}$? – 2017-01-16
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    Analogously, $\mathbb{R}^\times \cong \mu_2 \times \mathbb{R}_{>0} \cong \mathbb{Z} / 2\mathbb{Z} \times \mathbb{R}$ and $\mathbb{C}^\times \cong S^1 \times \mathbb{R}_{>0} \cong \mathbb{R}/2 \pi\mathbb{Z} \times \mathbb{R}$ – 2017-01-16

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This must be covered in almost every text on the $p$-adic numbers; I think the book of Gouvêa is the the best of these.

And your statement is not quite true: for $p=2$, $\mu_{p-1}$ is trivial all right, but the other part is isomorphic to $\{\pm1\}\times(1+4\Bbb Z_2)$. My tale below omits the story for $p=2$, and you can fill this in yourself.

First, you can consider the units, $\Bbb Z_p^\times$, and reduce them modulo $p$ to the multiplicative group of $\Bbb F_p\cong\Bbb Z/p\Bbb Z$. It’s cyclic of order $p-1$, as I’m sure you know. So we have an exact sequence: $$ 0\longrightarrow K\longrightarrow\Bbb Z_p^\times\longrightarrow\Bbb F_p^\times\longrightarrow0\,; $$ if you’re unfamiliar with the notation of exact sequences this merely says that there’s a surjective map from the middle term to the one to its right, with kernel equal to the one to its left.

What’s the kernel? It’s the units that go to $1$ in the field $\Bbb F_p$, in other words $1+p\Bbb Z_p$. To prove that $\Bbb Z_p^\times$ is the direct product of the two things to either side of it, it’s enough to show that there’s a homomorphism from $\Bbb F_p^\times$ into it whose image hits $K$ only in the identity. This is the fun part:

You can find $(p-1)$-th roots of unity in $\Bbb Z_p$ either by a routine application of any version of Hensel’s Lemma that you like, or, my favorite method, take an element of $\Bbb F_p^\times$, lift it to any element of $\Bbb Z_p$ that goes to it modulo $p$, and take successive $p$-th powers: $x\mapsto x^p\mapsto(x^p)^p\mapsto\cdots$ etc. I’ll leave it to you to show that this is a good convergent sequence, and its limit is clearly a suitable root of unity.

Showing that the multiplicative group $1+p\Bbb Z_p$ is isomorphic to the additive group $\Bbb Z_p^+$ is rather less fun. To tell you the truth, when I first saw the logarithmic argument, I didn’t like it, but I now think it’s the best one. You have to convince yourself that as long as the series for $\log(1+x)$ that you saw in Calculus is convergent, then $\log\bigl[(1+x)(1+y)\bigr]=\log(1+x)+\log(1+y)$, and for $x\in p\Bbb Z_p$, the series is convergent. And the values of the log are all in $p\Bbb Z_p\cong\Bbb Z_p$ (as additive groups, of course), and fill out that group.

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    Very helpful. Thanks. – 2017-01-16