# Are ideals also rings?

I am learning about rings and ideals. But I am confused about something. My book (Gallian) says that an ideal of a ring by definition is a subring. But I have talked to other people who insist that an ideal is not a ring itself. I am confused about this. According to Wikipedia an ideal isn't necessarily a subring. But maybe it follows from the definition in Wikipedia that an ideal is a subring?

So my question is: is an ideal a ring?

## 2 Answers

The answer is both yes and no, so this will take a bit of elaboration.

There are two ways to define a ring. One of them require the existence of a $1$, the other does not.

Let's start with the one that does. In this case, a subring is required to contain the $1$ from the larger ring, and hence no proper ideal can be a subring (as any ideal containing $1$ will be the entire ring).

If we removed the requirement that the subring contained the original $1$, then the answer would be "sometimes", since the ideal might or might not have a "local" $1$ (I invite you to try some small examples of rings to find examples of either case).

If we instead take the case where a ring need not have a $1$, then it follows straight from the definitions that any ideal will also be a subring.

Here is a general way to get a proper non-trivial ideal which has a "local" $1$: Take any $x\in R$ such that $x^2 = x$ (such $x$ are called idempotent) and such that $x\neq 0$ and $x\neq 1$ (which need not exist, but will for example exist in any ring of the form $R_1\times R_2$). Then the ideal generated by $x$ will have $x$ as its "local" $1$ (I am assuming the ring to be commutative, or at least $x$ to be central here).
Conversely, if we have such an $x$ then the ring will be the direct product of the ideals generated by $x$ and $1-x$ (note that $1-x$ is also idempotent).

• Any ideal with $1$ coincides with whole ring? – Leox Apr 20 '15 at 12:25
• Yes, since any $a\in R$ can be written as $1a$ or $a1$, so if the ideal contains $1$ then it also contains $a$ (whether it is a left-, right- or twosided ideal). – Tobias Kildetoft Apr 20 '15 at 12:28
• @Leox But note that this is when the $1$ is the same as for the whole ring. It is possible to have an element behave like a $1$ for a proper ideal but not for the entire ring. – Tobias Kildetoft Apr 20 '15 at 12:29
• yes, but in this case any ideal of a ring be trivial one – Leox Apr 20 '15 at 12:32
• @Leox I am not sure what you mean by trivial here. It is certainly possible to have a non-trivial ideal which is a ring with unity and the unit just happens to be different from the one of the original ring. – Tobias Kildetoft Apr 20 '15 at 12:33

In general, an ideal is a ring without unity - i.e. without a multiplicative identity - even if the ring it is an ideal of has unity. For example $\mathbb{2Z} \subset \mathbb{Z}$ is an ideal but $\mathbb{2Z}$ is not a ring with unity. So if you require your rings to have unity - and a lot of the time one does - then an ideal is in general not a ring. This is where the disagreement comes.

Sometimes the ideal can have a different identity (e.g., as quid points out in the comments, $\mathbb{Z}^2$ has $\mathbb{Z}\times {0}$ as an ideal, which is a ring with unity). But the only way the ideal can have the same multiplicative identity - and so be a sub-ring-with-identity - is if it is the whole ring.

• Thanks for following up I think it is fine now. (I delete the original comment an will detete that one if I don't forger about it.) – quid Apr 20 '15 at 12:21