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Exercise: Give an example of a subring of a finite commutative ring $R$, that is not an ideal of $R$.

I recently learned the following: Let $2^\Omega$ be the power set of an arbitrary set $\Omega$. For $A, B \in 2^\Omega$ we define $A \oplus B := (A\setminus B) \cup (B\setminus A)$ and $A\odot B := A \cap B$. Then $(2^\Omega, \oplus, \odot)$ is a commutative ring.

This ring has the following properties:

  • Zero corresponds to $\emptyset$.
  • One corresponds to $\Omega$.
  • The additive inverse of $A \in 2^\Omega$ is $A$.

    • Now let's take $\Omega = \{a, b, c\}$. Define $M := \bigl\{\emptyset, \{a, b\}, \{c\}, \Omega\bigr\}$. Then $(M, \oplus, \odot)$ is a subring of $(2^\Omega, \oplus, \odot)$, but not an ideal: $$\{a\} \odot \{a, b\} = \{a\} \not\in M \, .$$

    Is that correct?

    Are there easier examples?

    Ystar
    • 2,866

    2 Answers2

    13

    Your example looks correct.

    An easier example (at least notationally) seems to be: Let $R = \mathbb{Z}_3 \oplus \mathbb{Z}_3$. We see that $S = \{(0, 0), (1, 1), (2, 2)\}$ is a subring isomorphic to $\mathbb{Z}_3$, but $(1, 2)*(1, 1) = (1, 2)$, which is not an element of $S$, so it cannot be an ideal.

    A much easier example (allowing the ring to be infinite): Try the ring (really a field) $\mathbb{Q}$ and the integers $\mathbb{Z}$. Clearly $\mathbb{Z}$ is a subring of $\mathbb{Q}$, but it is not an ideal of $\mathbb{Q}$ (which has only two ideals, $0$ and itself).

    Of course I overlooked the simplest example: Let $R = \mathbb{Z}_2 \oplus \mathbb{Z}_2$ and take $S = \{(0, 0), (1, 1)\}$. Then $S$ is clearly a ring (as in the first example), while $(1, 0)*(1, 1) = (1, 0) \notin S$. This is your example is in the original post, just in a much easier to recognize form.

    cbishop
    • 639
    2

    You could try the ring of dual numbers $\mathbb{F}_2[x]/x^2$ where $\mathbb{F}_2$ is the field with two elements. It has four elements and the subset $\{0,1\}\cong {\mathbb{Z}}/2$ is a subring but not an ideal.