Submodule Criterion for Rings without 1

Question

In section 10.1 (page 342) of Dummit & Foote's Abstract Algebra, they state the following:

Proposition 1. (The Submodule Criterion) Let $R$ be a ring and let $M$ be an $R$-module. A subset $N$ of $M$ is a submodule of $M$ if and only if

(1) $N \neq \emptyset$, and

(2) $x + ry \in N$ for all $r \in R$ and for all $x, y \in N$.

This proposition states that $R$ is simply a ring, not that it has identity. In the next definition, they specifically state "a commutative ring with identity." In the definition of a module at the beginning of the section, they emphasize that $R$ is "not necessarily commutative nor with 1." However, the proof proceeds as follows:

Proof: If $N$ is a submodule, then $0 \in N$ so $N \neq \emptyset$. Also $N$ is closed under addition and is sent to itself under the action of elements of $R$. Conversely, suppose (1) and (2) hold. Let $r = -1$ and apply the subgroup criterion (in additive form) to see that $N$ is a subgroup of $M$. In particular, $0 \in N$. Now let $x = 0$ and apply hypothesis (2) to see that $N$ is sent to itself under the action of $R$. This establishes the proposition.

Clearly, by letting $r = -1$, this proof assumes that $R$ has identity.

Now, consider $R = 4\mathbb{Z}$, which is a ring without identity. Then, $4 \mathbb{Z}$ acts on the set $M = \mathbb{Z}$ by $r \cdot m = rm$, that is, by normal multiplication. Clearly, distributivity and associativity are inherited, so $\mathbb{Z}$ is a left $4\mathbb{Z}$-module.

Take $N = 4\mathbb{Z} \cup (3 + 4\mathbb{Z})$, that is, the subset of integers congruent to $3$ or $0$ modulo $4$. Certainly, $N \neq \emptyset$. If $x, y \in N$, then $x + ry \equiv x \pmod{4}$, for any $r \in R$, so that $x + ry \in N$.

Therefore, by Proposition 1, $N$ should be a $4\mathbb{Z}$-submodule of $\mathbb{Z}$. But $N$ is not an abelian group because $3 \in N$ but $-3 \equiv 1 \pmod{4}$ is not in $N$.

I have checked the errata, but I did not see this listed. I have also checked other textbooks, but it can be difficult to determine when they are assuming identity or not, since there is often a sentence somewhere earlier in the book that says all rings are assumed unital. I have also found the following alternative definition:

Definition. Let $N$ be a subset of the $R$-module $M$. Then $N$ is a submodule of $M$ if and only if the following hold:

(1) $N \neq \emptyset$, and

(2) $rx + sy \in N$ for all $r, s \in R$ and $x, y \in N$.

However, with this definition, similar counterexamples can still be found.

Questions:

• Am I correct in my assessment that this is an error?
• Is there a "Submodule Criterion" that works for rings $R$ without identity?

Answer 1

I think $N$ should be closed under all the structure specified by $M$, that is $N$ is a $+$-subgroup of $M$ and $N$ is closed under scalar multiplication. In total, we should have

1. $N\neq \emptyset$
2. $N-N\subseteq N$
3. $R\cdot N \subseteq N$

Unless I'm forgetting something important...

Answer 2

After a bit of thought, I think you are correct, this is an oversight as I believe the authors intended the ring to be unital - as clearly evidenced by their proof and their restriction that all the exercises in the chapter should be assumed to be unital rings. For your second question, I believe a minimal requirement would be to use your second definition and add a 3rd condition that the group action by $R$ is faithful on $N$, or that the kernel of the ring action on $N$ is empty - so that we can guarantee the additive inverse of all elements in $N$ are also in $N$ (I will let you think about that if it is not immediate) and thus can use the additive subgroup criterion to finish the proof. I can't think of a weaker condition than that. Nice observation.