I'm experimenting around with ring ideals (perhaps ideals is always for rings, so when speaking of ideals we always refer to these ring subsets?), and my book gives me the definition that an ideal $I$ is a subset of a ring $R$, for which the elements are closed under addition and $ra\in I$ for all $r\in R$ and $a\in I$.

Wikipedia say that an ideal is an additive subgroup which I clearly see, but I suspect that $I$ is also a subring to $R$. I cannot find any info on this in my book, but I tried to apply the "subring criterion" theorem (let $S$ be a subset of a ring $R$): (i) additive and multiplicative closures, (ii) if $a\in S \implies -a \in S$ and (iii) $S$ contains the identity.

This is what I did:

(i) addition follows from the definition, but for multiplication let $a,b \in I$ then for any $r_1,r_2 \in R$ we get $r_1ar_2b = (r_1ar_2)b \in I$ because $r_1ar_2 \in R$. Same argument for $a$, thus closure for multiplication.

(ii) This holds from definition of a ring.

(iii) $I$ is an additive subgroup hence $0\in I$.

So, does my "proof" hold? Is it really true that an ideal is always a subring? If it is not true, ca anyone illustrate some counter example?

Best regards,

  • 3
    $\begingroup$ Your "proof" holds if and only if you do not require that rings have a multiplicative identity. $\hspace{1.28 in}$ $\endgroup$
    – user57159
    Sep 21, 2013 at 9:09
  • $\begingroup$ Ricky: I'm sorry I'm a bit slow today, but could you explain further what you mean? If the ring is a commutative ring, it might not hold? $\endgroup$
    – hsalin
    Sep 21, 2013 at 9:12
  • 2
    $\begingroup$ No. He means that e.g. $2\mathbb Z$ is a sub*rng* of $\mathbb Z$ not a subring (when rings are required to have a multiplicative unit). $\endgroup$
    – kahen
    Sep 21, 2013 at 9:13
  • $\begingroup$ Unitary rings are the rule in the Bourbaki schol not necessarily by most algebraists. $\endgroup$ Sep 21, 2013 at 15:11
  • $\begingroup$ Your definition of ideal is incorrect: you need to require that it is a additive subgroup of $R$ (so it contains $0$ and is closed under taking inverses), not just that it is closed under addition. By your definition, the empty set would be an ideal in any ring. $\endgroup$ Nov 26, 2016 at 2:03

1 Answer 1


If your book defines (like most books do I think) rings as unitary rings, in other words requires them to contain a neutral element for multiplication, then ideals are almost never subrings. Indeed, for unitary rings one requires subrings of$~R$ to contain the element $1\in R$ (rather than some element that is neutral for mulitplication restricted to the subring), and an ideal that contains $1$ must contain its multiples by elements of $R$, which gives all of $R$; the only ideal of $R$ that is a subring is $R$ itself.

If on the other hand your book says nothing of neutral elements for multiplication, then it deals with a larger category of structures that are often called "rngs", to indicate the (possible) lack of $1$. Then indeed all ideals are subrngs.

Finally I note that sometimes a non-trivial ideal can be considered a ring in itself by choosing a different element to be $1$ (this was the reason for my parenthesised remark above). For instance in $\def\Z{\mathbf Z}\Z/10\Z$ you can make the ideal of "even" classes into a ring (using the same addition and multiplication) by electing the class of $6$ to be the neutral element for multiplication. This situation arrives typically in rings that can be decomposed as a product of rings; in the current example $\Z/10\Z\equiv (\Z/5\Z)\times(\Z/2\Z)$ by the Chinese remainder theorem. The factors in such a decomposition are naturally quotients of the original ring, but also correspond to an ideal of that ring (namely the kernel of projecting to the other factor(s)); such an ideal is not a subring, but can be made into a ring by choosing the projection of $1$ onto the ideal as new neutral element.

  • $\begingroup$ Thank you for a very structured and explaining answer! This helped a lot. $\endgroup$
    – hsalin
    Sep 21, 2013 at 11:57

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