Non-trivial examples for non-ideal subset of commutative rings closed under addition and multiplication by ring elements

Question

In a commutative ring $(R,+,\cdot)$ (not necessarily having a multiplicative identity), an ideal $I$ is a subset $I \subseteq R$ such that

1. $I \neq \emptyset$
2. $\forall x,y \in I: x-y \in I$
3. $\forall x \in I, r \in R: r\cdot x \in I$

I am now looking for subsets $N \subseteq R$ which are not ideals but satisfy

1. $N \neq \emptyset$
2. $\forall x,y \in N: x\mathop{\color{red}+}y \in N$
3. $\forall x \in N, r \in R: r\cdot x \in N$

(in other words: I am trying to better understand why/when these two definitions are not equivalent[*])

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What I already know:

• A trivial example is the ring $(\mathbb{Z},+,\ast)$, where $\ast$ is the trival multiplication $x\ast y := 0$ for all $x,y \in \mathbb{Z}$ and the set $N := \mathbb{Z}_{\geq 0}$.
• If $(R,+,\cdot)$ has a multiplicative identity, the two definitions are equivalent as then $$x-y= x+(-y) = x+(-1)\cdot y \in N$$
• In fact, it suffices if for any element $y \in N$ there exists an element $r \in R$ and $n \in \mathbb{N}^\ast$ with $r\cdot y = n\cdot y := \underbrace{y + \dots + y}_{n \text{ times}}$ since, then, we have $$x - y = x + (-r)\cdot y + (n-1)\cdot y \in N$$

The last observation, in particular, seems to exclude most of the non-unital rings I know of (like essentially everything involving $\mathbb{Z}$ (e.g. $2\mathbb{Z}$) or function spaces like $C_0(\mathbb{R})$)

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[*] Why am I interested in that? Because I was giving a lecture on ideals today where I accidentally used the second definition as definition for ideals. When I corrected myself, students wanted to know why the two definitions aren't equivalent -- and I wasn't able to come up with any good counter examples.

Answer 1

Here is a (nontrivial?) example: $R=x\mathbb Z[x]$, then the following is not an ideal $$\mathbb Nx + xR=\{a_1x+a_2x^2+\cdots+a_nx^n\in R\mid a_1\in\mathbb N\}$$

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The above construction has some universality. The OP's observation is very sharp. Let's call any $N$ that satisfies the second group of conditions a "deal" (instead of "ideal"). We can show

A rng $R$ contains a deal that is not an ideal iff $\exists x\in R$, $x$ has infinite order as an element in the group $(R, +)$ and $$\mathbb Nx\cap Rx = \{0\}\Leftrightarrow \mathbb Zx\cap Rx=\{0\}$$

"$\Rightarrow$" is proved by the OP.

"$\Leftarrow$" $: \mathbb Nx + Rx$ is a deal that is not an ideal.

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In terms of pedagogy, I think we should not define the ideal in this way, instead just combine the first two conditions to say $I$ is an additive subgroup of $R$, which is much more intuitive.

The counter example $(\mathbb Z, +, *)$ is "good" by all means. That is, when the multiplication is trivial, the difference between a subgroup and a sub-semigroup shows up.

If the students are not happy, we can present it in a "fancy" way:

$$R=\{\begin{pmatrix} 0 & a \\ 0 & 0 \end{pmatrix}\mid a\in\mathbb Z\}$$