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 2


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).

  • $\begingroup$ Any ideal with $1$ coincides with whole ring? $\endgroup$
    – Leox
    Commented Apr 20, 2015 at 12:25
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    $\begingroup$ @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. $\endgroup$ Commented Apr 20, 2015 at 12:29
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    $\begingroup$ @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. $\endgroup$ Commented Apr 20, 2015 at 12:33
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    $\begingroup$ Tangential question but is a ring with a "local" $1$ necessarily isomorphic to a ring with an "actual" $1$? (Where the "local" $1$ is mapped to the "actual" $1$.) $\endgroup$
    – user157227
    Commented Apr 20, 2015 at 13:55
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    $\begingroup$ @user157227 No, if the element only acts as a $1$ on a certain subset, then you cannot send it to an element that acts as a $1$ on all elements. But it is always possible to add an extra element which becomes a "global" $1$. $\endgroup$ Commented Apr 20, 2015 at 14:01

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.

  • $\begingroup$ 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.) $\endgroup$
    – quid
    Commented Apr 20, 2015 at 12:21

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