In our introductory course on groups, we defined cosets and quotient groups in the following way.

Let $N\trianglelefteq G$ be a normal subgroup of a group $(G,\,\cdot\,)$. Then the quotient group $G/N$ contains all the cosets of $N$ with respect to the elements of $G$, i.e. $G/N=\{Ng:g\in G\}$.

Later on, we generalised this idea to rings:

Let $I$ be an ideal of a ring $(R, +, \,\cdot\,)$. Then the quotient ring $R/I$ is defined $R/I = \{I+r:r\in R\}$.

My question is: why do we take the `additive' cosets of $R$ in this case? i.e. does $R/I$ always contain cosets of the form $I+r$, and not $I\cdot r$?

I reasoned that perhaps we take the $+$ cosets since $(R,+)$ forms a group, but $(R,\,\cdot\,)$ does not. But what if we have a field?

  • 1
    $\begingroup$ For the same reason we take the kernel of a ring homomorphism to be the kernel of it as an additive group homomorphism. Ideals and cosets are fibres of ring homomorphisms. $\endgroup$ – Angina Seng May 31 '17 at 20:20
  • 1
    $\begingroup$ $(R,\cdot)$ doesn't form a group if $R$ is a field either (think about $0$). $\endgroup$ – Alex Provost May 31 '17 at 20:32
  • $\begingroup$ @AlexProvost Correct, but we can take $R^\times = R\setminus\{0\}$. $\endgroup$ – Luke Collins May 31 '17 at 20:33
  • 1
    $\begingroup$ @LukeCollins Yes, but I don't see how this is relevant to your original question. $\endgroup$ – Alex Provost May 31 '17 at 20:38

Say I is an ideal in a ring R.

Then a definition of a quotient ring with multiplicative subsets

R/I := {Ir : r in R}

would not make much sense, since this would be equal to just {I} (since I is an ideal).

| cite | improve this answer | |
  • $\begingroup$ Of course! I overlooked that. By absorption right? $\endgroup$ – Luke Collins May 31 '17 at 20:51
  • 1
    $\begingroup$ yes, by definition of ideal $\endgroup$ – FWE May 31 '17 at 20:54

This addresses the question in the comments about taking cosets of (subgroups of) $R^\times$.

A ring is a group with a multiplication. A quotient ring is a quotient group with a multiplication. Since the group structure on a ring is additive, it only makes sense to look at additive cosets.

There is, however, another construction you can use if you want a multiplicative quotient. You take $G$ to be a subgroup of the unit group $R^\times$. Then, form the set $\{rG : r \in R\}$. Multiplication works fine:

$$ (rG)(sG) = (rs)G $$

but in order for addition to make sense you need to have a multi-valued addition:

$$ (rG) + (sG) = \{ tG : t = gr + hs \text{ for some } g, h \in G\}. $$

This structure is called a hyperring.

The definition of addition is equivalent to the following:

$$ (rG) + (sG) = \{ (r' + s')G : r' \in rG, s' \in sG \}. $$


Let $R = \mathbf{Q}$ be the rationals and take $G = \mathbf{Q}^\times$. Then $R/G$ consists of two elements: $0 := 0G$ and $1 := 1G$. The multiplication is as you would expect:

$$ \begin{array}{c|cc} * & 0 & 1 \\\hline 0 & 0 & 0 \\ 1 & 0 & 1 \end{array} $$

Addition, is a bit "odd". Note that if $x \in \mathbf{Q}$ then $0 + x = x$ so it makes sense that $0G + xG = xG$. On the other hand, if you look at the coset $1G = \mathbf{Q}^\times$ then you could have, $1 + 1 = 2 \in \mathbf{Q}^\times$ or $1 + (-1) = 0$. So in $R/G$, $1 + 1 = \{0, 1\}$. This gives you

$$ \begin{array}{c|cc} + & 0 & 1 \\\hline 0 & 0 & 1 \\ 1 & 1 & \{0,1\} \end{array} $$

People have started to call $\mathbf{Q}/\mathbf{Q}^\times$ the Krasner hyperfield. It is also equal to $k/k^\times$ where $k$ is any field with at least 3 elements.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.