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($R$,+,$\cdot$) is a ring with 1. We define $R^\times$ := { r $\in$ R : r is invertible with respect to multiplication}.

(i) Show that ($R^\times$, $\cdot$) is a group ( the so called uniform-group of $R$). Uniform group definition: The uniform-group from a ring with 1 is defined as the set of all multiplicative inverse elements).

(ii) Let $K$ be a field. Find $K^\times$. My solution: (i) $R^\times$ := { r $\in$ R : r is invertible with respect to multiplication} $\Rightarrow$ $\exists$ r$^{-1}$ $\cdot$ r = 1 = r $\cdot$ r$^{-1}$. Let r, s, t $\in$ $R^\times$ s.t there exists $\tilde{r}$, $\tilde{s}$, $\tilde{t}$ $\in$ $R$. r$\cdot$$\tilde{r}$ = $\tilde{r}$$\cdot$r = s$\cdot$$\tilde{s}$ = $\tilde{s}$$\cdot$s = t$\cdot$$\tilde{t}$ = $\tilde{t}$$\cdot$t = 1. Furthermore, we have 1 $\in$ $R^\times$ and for r $\in$ $R^\times$ is $\tilde{r}$ an inverse element.

My idea was to create enough elements to prove that those elements fulfil the group axioms. (closure under multiplication, associativity, identity element and inverse element)

(ii) I honestly don't know what I really need to prove or show here but I came up with this solution so far: $K^\times$ := {k $\in$ K : k is invertible with respect to the multiplication and addition}.

I think that my solutions are wrong and I don't know how to continue at this point. Any hints guiding to the right direction I very much appreciate.

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  • $\begingroup$ Read your solution out loud: "$R^\times$ is, by item (i), the set of all elements of $r$ which are invertible with respect to multiplication. This implies that there exists the inverse of $r$ multiplied by $r$ is equal to $1$, which in turn is equal to $r$ times the inverse of $r$." Doesn't make sense, which should point out that this is not going too well... $\endgroup$ – Luiz Cordeiro May 19 '18 at 2:53
  • $\begingroup$ Any question that says "show that A is a B" is trying to see if you understand the definition of a B. Look up the definition of a B. Here you need to verify the group axioms. You know multiplication is associative, so that is easy. Is there an identity? Are there inverses? You haven't defined what $A^\times$ is based on $A$. $\endgroup$ – Ross Millikan May 19 '18 at 2:57
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Apart from associativity (which you already have for free because $(R,+\cdot)$ is a ring), none of the group axioms require three elements. What are you trying to prove with $r,s$ and $t$? Also "create enough elements" is irrelevant. There are groups with very few elements (all the way down to a single element), so as long as you have $1$, and show that that's the identity element, you don't need more elements. You just need that for all elements you can find, the corresponding products and inverses also exist.

  1. Closure: show that if $r$ and $s$ are both invertible, then $rs$ is invertible too (construct the inverse of $rs$ using the inverses of $r$ and $s$)
  2. Associativity we get for free
  3. Identity element. Show that $1$ fulfills the requirement for the identity element
  4. Inverses. Show that if $r$ is invertible, then so is $r^{-1}$ (what is $(r^{-1})^{-1}$?)
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Just stay calm and look at the definitions:

Definition: $R^\times=\left\{r\in R:r\text{ is invertible with respect to multiplication}\right\}$.

In (ii), we are taking a particular case: $R=K$ is a field

Look at the definitions again: What is special about a field?

Definition: A field $K$ is a commutative ring with unit such that every $r\neq 0$ has a multiplicative inverse. (And we usually also assume that $0\neq 1$.)

Ok, so item (ii) is asking: Find $K^\times$ if $K$ is a field.

Look at the definitions and the answer shouldn't be hard. Think about a particular case if it makes it easier to visualize: what if $K=\mathbb{Q}$? $\mathbb{R}$?..

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