What is $\mathbb{Z}^*_2$

Question

What is $\displaystyle{\mathbb{Z}^*_2}$?

Do I understand correctly it's the odd numbers? It would appear it includes both positive and negative.

$\displaystyle{\mathbb{Z}^*_2}=2\mathbb{Z}-1$

Sorry - to clarify. I was told $\mathbb{Z}_p^*$ is the group of units of the ring of p-adics, and that this meant the set of numbers not divisible by $p$.

So this question is to test whether I understand correctly. The explanation given would seem to imply that $\mathbb{Z}_2^*$ is the set of 2-adic integers not divisible by $2$, so the odd numbers. Is the explanation given to me incorrect, or did I misinterpret it?

Thanks in advance.

Answer 1

For any prime $p$, the $p$-adic integers can be thought of as a sequence of integers $\alpha=\{a_0,a_1,\cdots\}$ satisfying $a_n\equiv a_{n-1}\pmod {p^n}$ for all $n≥1$.

These form a ring under component-wise addition and multiplication. We'll call that ring $\mathbb Z_p$.

With $p=2$, your question is equivalent to:

Claim: $\alpha$ is a unit in $\mathbb Z_p$ if and only if $a_0\not \equiv 0 \pmod p$.

Proof:

Suppose $\alpha$ is a unit. Then there is a $p$-adic integer $\beta=\{b_n\}$ with $\alpha\times \beta =1 $ In particular we have $a_0\times b_0\equiv 1 \pmod p$ so $a_0$ can't be divisible by $p$.

Now suppose that $a_0 \not \equiv 0 \pmod p$. Then we can find $b_0$ with $a_0\times b_0\equiv 1 \pmod p$. By the definition of the $p$-adic integers we know that $a_i\not \equiv 0 \pmod p\;\forall i$ as well, so we can always solve $a_n\times b_n\equiv 1 \pmod {p^{n+1}}$ and it is easy to see that the $b_n$ so constructed define a $p$-adic integer $\beta$ with $\alpha \times \beta =1$.

Answer 2

Generally, we denote

$$\mathbb Z_n:=\mathbb Z/n\mathbb Z,$$

and when $A$ is a ring,

$$A^\times:=\{a\in A,\ a\text{ is invertible in }A\}.$$

So

$$\mathbb Z_n^\times=\{k\in \mathbb Z/n\mathbb Z, \ \gcd(k,n)=1\}.$$

In your case,

$$\mathbb Z_2^\times=\{k\in \mathbb Z/2\mathbb Z, \ \gcd(k,2)=1\}=\{1\}.$$