TITLE: Subring which is not an ideal?
QUESTION [9 upvotes]: 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?

REPLY [2 votes]: 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.