# In a ring, are cosets always additive?

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?

• 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. – Angina Seng May 31 '17 at 20:20
• $(R,\cdot)$ doesn't form a group if $R$ is a field either (think about $0$). – Alex Provost May 31 '17 at 20:32
• @AlexProvost Correct, but we can take $R^\times = R\setminus\{0\}$. – Luke Collins May 31 '17 at 20:33
• @LukeCollins Yes, but I don't see how this is relevant to your original question. – 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).

• Of course! I overlooked that. By absorption right? – Luke Collins May 31 '17 at 20:51
• yes, by definition of ideal – 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:

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

Example:

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.