Let S be the ring of rational numbers in $[0,1)$ with operations defined by $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc \:(mod \: bd)}{bd}$ and $\frac{a}{b}\times\frac{c}{d}=\frac{ac \:(mod \: bd)}{bd}$. Basically its a way to make a ring out of proper fractions. I've never seen this ring before, and I was lead to it by considering the quotient ring $R/(x)$ where $R=x\mathbb{Q}[x]+\mathbb{Z}\subset\mathbb{Q}[x]$. Is S isomoprhic to anything more familiar?

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    $\begingroup$ I am sure you're right, but I don't understand this: two polynomials in R are congruent if they have the same constant term and the difference of the coefficients of the degree 1 terms is an integer. Doesn't this mean the quotient ring is more complicated? $\endgroup$ – SihOASHoihd Jun 27 '17 at 16:36
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    $\begingroup$ $(S,+)$ is just $\Bbb{Q/Z}$, but multiplication does not behave well in this structure. $\endgroup$ – Crostul Jun 27 '17 at 16:39
  • $\begingroup$ There are many ways to understand this, and you are more likely to get further explanations if you don't accept an answer so quickly (many folks don't read questions that already have accepted answers). First/quick answers are rarely best answers, so generally you will gain deeper insight by delaying acceptance to encourage other answers. $\endgroup$ – Bill Dubuque Jun 27 '17 at 16:40
  • $\begingroup$ @SihOASHoihd You're right! It is more complicated.. $\endgroup$ – rschwieb Jun 27 '17 at 16:50

It's not a ring. For instance, $$\frac{1}{2}\cdot\left(\frac{1}{2}+\frac{1}{2}\right)=\frac{1}{2}\cdot0=0$$ but $$\frac{1}{2}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}.$$


As Eric Wofsey has already shown, this does not describe a ring.

To go back to your motivating example of $R/(x)$ with $R=\mathbb Z+x\mathbb Q[x]$, we can also clarify what that should look like.

Your description in the comments is apt: two things are equal if they have the same constant term, and if their 'linear' term is equal modulo $\mathbb Z$. That suggests a ring structure on $\mathbb Z\oplus \mathbb Q/\mathbb Z$.

Now, there is a well known ring structure on $\mathbb Z\oplus \mathbb Q/\mathbb Z$: it is called the idealization of $\mathbb Q/\mathbb Z$ by $\mathbb Z$. The sum is given by

$(n,q\mod{\mathbb {Z}})+(m,p\mod{\mathbb {Z}})=(n+m, q+p\mod{\mathbb {Z}})$ and the product is $(n,q\mod{\mathbb {Z}})(m,p\mod{\mathbb {Z}})=(nm,np+mq\mod{\mathbb {Z}})$.

You should be able to verify that the homomorphism $\sum a_ix^i\mapsto (a_0, a_1\mod \mathbb Z)$ from $R\to \mathbb Z\oplus \mathbb Q/\mathbb Z$ has kernel $(x)$.

I can't be sure what led you to the candidate ring in your question, but it "sounds" like you were considering the subset with constant terms zero. In this idealization, $\mathbb Q/\mathbb Z$ does become a ring, but the multiplication is uniformly zero.

  • $\begingroup$ I came up with this: $R/(x)=\{\overline a x + b \mid \overline a\in \mathbb{Q}/\mathbb{Z} \:and\: b \in \mathbb{Z} \}$ with operations $\overline a x + b \times \overline c x + d = \overline {(ac+bd)} x + bd$ and $\overline a x + b + \overline c x + d = \overline {(a+c)} x + (b+d)$. Is this the ring your describing? $\endgroup$ – SihOASHoihd Jun 27 '17 at 17:29
  • $\begingroup$ @SihOASHoihd Look at the operations you've proposed: they're the same as the ones I gave, just written with the overline to indicate mod instead of the $\mod$. $\endgroup$ – rschwieb Jun 27 '17 at 19:43
  • $\begingroup$ This is sort of interesting: $\mathbb{Q}/\mathbb{Z}$ is not a ring with multiplication $\overline{a}\overline{b}=\overline{ab}$ since the multiplication is not well-defined (as Eric Wolfsey pointed out), i.e., $\overline{a_{1}x}\neq \overline{a_{2}x}$ in general, where $a_{1}\equiv a_{2}\:(mod\:\mathbb{Z})$. But if $a_{1},a_{2}\in \mathbb{Q}$ with $a_{1}\equiv a_{2}\:(mod\:\mathbb{Z})$ and $x\in \mathbb{Z}$, then $\overline{a_{1}x}$ $\it{is}$ equal to $\overline{a_{2}x}$ which implies that $(\overline{a}x+b)(\overline{c}x+d)$ is well defined. $\endgroup$ – SihOASHoihd Jun 27 '17 at 20:37
  • $\begingroup$ The multiplication in $R/(x)$ seems quit natural to me, so it is kinda of surprising that it turned out to be a ring even though $\mathbb{Q}/ \mathbb{Z}$ isn't. $\endgroup$ – SihOASHoihd Jun 27 '17 at 20:48
  • $\begingroup$ @SihOASHoihd The idealization is very natural: it takes any abelian group and makes it an ideal in a ring with identity. $\endgroup$ – rschwieb Jun 27 '17 at 21:46

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