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I've recently begun reading Miles Reid's notes on Commutative Algebra. On page $19$ he gives a definition for the ideal of a ring that I am unfamiliar with:

An ideal of a ring $A$ is a subset $I \subset A$ such that $0 \in I$, and $$ af + bg \in I \quad \text{for all } a,b \in A \text{ and } f,g \in I $$

The definition I have always used is the following:

An ideal of a ring $A$ is a subset $I \subset A$ such that $I$ is an additive subgroup of $A$ and $$ ri \in I \text{ and } ir \in I \quad \text{for all } i \in I \text{ and } r \in R $$

I initially figured that the definitions were equivalent, but I can't seem to prove that an ideal defined by Reid produces an ideal according to the second definition with which I'm more familiar. There are two specific issues I'm having.

First, assuming that $af + bg \in I$ it is clear that $I$ respects multiplication on the left by elements in $A$, because if $af + bg \in I$ then clearly $a\cdot 0 + bg \in I \implies bf \in I$. However, unless the ring is commutative I can't see how to also show that $gb \in I$.

Second, I can't prove from the first definition that $I$ is an additive subgroup of $A$ unless I assume that $A$ is a ring with unity, because I can't show that it is closed under addition. On the other hand, if $1 \in A$ then clearly $af + bg \in I \implies 1\cdot f + 1\cdot g \in I$.

$\textbf{Question:}$ Does Reid's definition for an ideal match the usual definition only when $A$ is a commutative ring with unity, or am I missing something?

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    $\begingroup$ You are not missing anything. Reid's definition works for commutative rings: indeed it comes from a book of commutative algebra. $\endgroup$ – Crostul Apr 30 '17 at 22:21
  • $\begingroup$ @Crustul Yeah, I know he assumes all rings are commutative with $1$. I've convinced myself that it's true in those cases, but since he said that this was the definition of an ideal I was wondering if it was indeed true for more general cases. Are you saying that I'm right and it only works for commutative rings with $1$? $\endgroup$ – wgrenard Apr 30 '17 at 22:26
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Yes, you are right. Reid consider only commutative rings with identity, and his definition of ideal needs that to be correct. In this situation both definitions are equal. Second definition is broader and can be applied to noncomutative rings or rings without unity (or both). It is standard mathematical practice to assume in definition as much as you need and nothing else - the checklist of properties is small and determining if some object fulfills it is shorter. In this case by taking any subset $S$ of a ring $A$ we already know that all elements could commute and $1\in A$ so we can examine if $S$ is:

  1. additive group
  2. closed under multiplication

in one handy formula $$0\in S \land \forall_{a,b\in A}\forall_{f,g\in S}: af+bg\in S$$ instead of $$0\in S \land \forall_{f,g\in S}: f-g\in S$$ $$\forall_{a\in A}\forall_{f\in S}: af\in S \land fa\in S$$

Here is example that fulfills only first definition: $A:=\{k\in \mathbb{Z}:2|k\},\ S:=\{4k\in \mathbb{Z}:k\neq-1 \}$. Note, that $S$ is not a group under addition, therefore $A$ do need the identity.

In my opinion your confusion is caused by taking one or another as the definition of the ideal. If one definition is broader than the other, why it is not called e.g. "noncommutative ideal"? I think it is convention to make text shorter and more readable. The same goes for ring: if every ring considered in a book is commutative with identity, it could be referred simply as "ring". Rule of thumb is to sticks to given definitions throughout book, but comparing different formulations is valuable too.

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  • $\begingroup$ Thanks for the example. That is just what I was hoping to see in an answer, and what I couldn't seem to come up with myself. $\endgroup$ – wgrenard May 1 '17 at 2:55

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