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The question is Consider $\mathbb Z \left[\frac{1}{10}\right] = \{\frac{a}{10} \mid a\in\mathbb Z\}$ under addition and multiplication. Is it a ring? Is it commutative?

is the set really just all the elements:

$$\frac{1}{10}=\frac{1}{10},\space \frac{1}{10}=\frac{2}{10}, \space\space \frac{1}{10}=\frac{3}{10},\space \space\frac{1}{10}=\frac{4}{10},\ldots$$

which if that's the case, that would be all the elements are really just the set (from cross multiplying) of the form: $10 = 10a \to 1 = a$ Which doesent make any sense to me. Once I know what it looks like to add two elements in the set whether or not is a ring should be straightforward.

Michael Hardy
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K. Gibson
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  • What do you mean when you write "$\frac{1}{10} = \frac{2}{10}$"? – Eric Towers Feb 23 '20 at 02:47
  • By set notation ${\frac a{10}|a \in \mathbb Z}$ is indeed just the set ${....., \frac {-4}{10}, \frac {-3}{10}, \frac {-2}{10},\frac {-1}{10}, \frac 0{10}, \frac 1{10}, \frac 2{10},\frac 3{10}....} = {....., -\frac {2}{5}, -\frac {3}{10},-\frac {1}{5},-\frac {1}{10}, 0, \frac 1{10}, \frac 1{5},\frac 3{10}....} $. But your "list of all the elements" $\frac 1{10}=\frac 2{10}$ and $\frac 1{10}=\frac 3{10}$ makes utterly no sense at all. – fleablood Feb 23 '20 at 03:49
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    What is $k$ if $\frac 3{10}\times \frac 7{10} = \frac k{10}$? – fleablood Feb 23 '20 at 03:52
  • @fleablood Okay I think I was just confused by the notation. But now that you clarified it. I see where it wouldn't make sense. – K. Gibson Feb 23 '20 at 04:08
  • @K.Gibson : As soon as I got the notification that you'd accepted my answer, I looked at it again and wondered why I wrote what I did. So I changed it. – Michael Hardy Feb 23 '20 at 04:33

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If you start with $\mathbb Z$ and adjoin $1/10$ and then close under addition, subtraction, and multiplication, then you get all fractions of the form $a/10^k$ for $k=0,1,2,3,\ldots.$ This is a ring since it's closed under addition, subtraction, and multiplication.

(Here I'm taking $\mathbb Z[a]$ to mean adjoin $a$ and close under addition, subtraction, and multiplication, whereas $\mathbb Z(a)$ would mean do that and also close under division.)

ok, Ignore the following:

The underlying set of this structure is $\left\{ 0, \pm\frac1{10}, \pm\frac2{10}, \pm\frac3{10},\ldots \right\}.$

Since $\mathbb Z$ includes negative integers and $0,$ the corresponding fractions need to be there.

I wonder why you wrote $\text{“}\frac 1 {10}=\cdots\text{''}$ with each member of the set.

Note, however, that this set is not closed under multiplication, since, for example, $\frac 2{10} \times\frac3 {10} = \frac 6{100},$ and this is not $\frac a{10}$ for any $a\in\mathbb Z.$

Michael Hardy
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  • Right: The localized ring is the smallest ring containing the set. – Rushabh Mehta Feb 23 '20 at 02:59
  • can you clarify more about the fraction in your answer why is it $\frac{1}{10^k}$ – K. Gibson Feb 23 '20 at 20:56
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    @K.Gibson : Because rings are closed under multiplication: $$ \underbrace{\frac\bullet{10} \times \cdots\times\frac\bullet{10}}_\text{$k$ factors} {} = \frac\bullet {10^k} $$ where each $\text{“}\bullet\text{''}$ represents some member of $\mathbb Z. \qquad$ – Michael Hardy Feb 24 '20 at 21:10