Let $R$ be the smallest subring of $\mathbb Q$ (the field of rational numbers) that contains $3/10$ ($R$ doesn't have to be a unital ring). Does $1 \in R$?

Is the desired smallest subring this one: $R_1=\{\frac{3a}{10^b},a,b \in \mathbb Z \}$ or this one: $R_2=\{\frac{3^ca}{10^b},a,b,c \in \mathbb Z \}$? Either case, $1 \notin R_i$, because $3\nmid10^b$, correct?

  • $\begingroup$ Isn't $R_1=R_2$ since you can put powers of $3$ into $a$? $\endgroup$ – Cheerful Parsnip Feb 24 '17 at 20:11
  • $\begingroup$ I was wondering the same thing! $\endgroup$ – John Feb 24 '17 at 20:14
  • $\begingroup$ Oh wait, actually they are not the same. $R_2$ allows $c=0$, giving $a/10^b$. $\endgroup$ – Cheerful Parsnip Feb 24 '17 at 20:18

Your $R_1$ is not a ring because it is not closed under multiplication. Your $R_2$ contains $1$ (for $a=1, b=c=0$). However, if you modify $R_2$ to $\{\frac{3a}{10^c}\mid a\in\Bbb Z, c\in\Bbb N\,\}$, you are fine. Even if this should turn out not to be the smallest such ring, it is sufficient to know that it is a sub-ring containing $\frac3{10}$ and not containing $1$; the smallest can only contain less elements.

  • 5
    $\begingroup$ Why is $\mathbb R_1$ not closed under multiplication? $\endgroup$ – Henning Makholm Feb 24 '17 at 19:59
  • $\begingroup$ Actually, if my $R_1$ is not a ring, then your $R_2$, shouldn't be also not a ring? $\endgroup$ – John Feb 24 '17 at 20:11
  • $\begingroup$ Like you, I originally thought that the minimal ring needs $c\in\mathbb N$, not $c\in\mathbb Z$. But in fact that is an illusion $-$ they define the same object. $\endgroup$ – TonyK Feb 24 '17 at 20:24
  • $\begingroup$ Very strange final comments: 1) "if this should turn out not to be the smallest such ring"; it turns out to be the smallest one; 2) "the smallest can only contain less elements"; I wonder what this could mean for infinite rings like these. $\endgroup$ – user26857 Feb 25 '17 at 0:17

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