1

Let $R$ be a finite commutative ring with unity, and $I$ be a proper ideal of $R$ (i.e. $I\neq R$). Is always $R/I$ isomorphic to some subring $S$ of $R$?

I know:

  • Quotient group need not be isomorphic to any subgroup.
  • For a finite abelian group, any quotient group is isomorphic to some subgroup.

I wonder if the similar holds for the ring case. However, I could not find an explicit answer to this question. Any idea, comment, or reference will be very helpful.

actcon
  • 29
  • If $I=R$ then $R/I$ is the (fine, a) zero ring, but if $R$ itself is not the zero ring then $R$ has no subring isomorphic to the zero ring (since if $1_R\not=0_R$ then ${0_R}$ is not a sub-ring of $R$ in the sense of rings with unity). So you'll at least want $I$ to be a proper ideal. – Noah Schweber Dec 02 '21 at 06:03
  • 1
    Separately, note that a finite group may be expandable to a ring in different ways (see here). So we can't forget the multiplicative structure, apply the result for groups, an then bring the multiplicative structure back to get a subring (even if we only want a "subrng"). – Noah Schweber Dec 02 '21 at 06:09

1 Answers1

3

The ring $\mathbb{Z}/4\mathbb{Z}$ has no proper subrings, but has quotient $\mathbb{Z}/2\mathbb{Z}$.

Note the ideal $(2)\subseteq \mathbb{Z}/4\mathbb{Z}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}$ as an additive group. However its multiplicative structure is different, as $2\times 2 = 0$ in $\mathbb{Z}/4\mathbb{Z}$, but $1\times 1=1$ in $\mathbb{Z}/2\mathbb{Z}$.

So in one case the additive generator squares to the additive identity, but in the other case it squares to the additive generator.

tkf
  • 11,563