This is how the Wikipedia article on subring defines the subring test

The subring test states that for any ring $R$, a nonempty subset of $R$ is a subring if it is closed under addition and multiplication, and contains the multiplicative identity of $R$.

When you follow the link for the subring test, it is stated as follows

In abstract algebra, the subring test is a theorem that states that for any ring, a nonempty subset of that ring is a subring if it is closed under multiplication and subtraction. Note that here that the terms ring and subring are used without requiring a multiplicative identity element.

My question is, is the first statement of subring test correct? This is also how a subring is "defined" in Atiyah-Macdonald. It seems incorrect to me as $\mathbb{R}_+$ satisfies those conditions and is not a subring unless I am missing something.

Looking at the responses I feel I should further clarify my question. Closure under subtraction and multiplication (with the added provision that the given subset contain the identity depending on how you define your rings), guarantees a subring, as in the second statement. I am comfortable with this statement as I know that closure under subtraction for a subset of a group (written additively) gives a subgroup. My question is whether the first statement is correct - is closure under addition and multiplication enough?

  • $\begingroup$ @yoyo: In either case, my question is, is closure under addition and multiplication, enough to guarantee a subring as in the first statement. $\endgroup$
    – Florian J
    May 22, 2011 at 15:56
  • 1
    $\begingroup$ natural numbers are not a subring of the integers despite being closed under addition and multiplication $\endgroup$
    – user10575
    May 22, 2011 at 17:40

3 Answers 3


Yes, you're right: the version of the subring test found in the wikipedia article on "subring" was faulty, whereas the article subring test has a correct statement.

I just edited the first wikipedia article to read as follows:

"The subring test states that for any ring R, a subset of R is a subring if it contains the multiplicative identity of R and is closed under subtraction and multiplication.'

I hope all will agree that this is an appropriate statement.

There is a slight discrepancy in that the article on subrings explicitly assumes we are working in the category of rings (in which we have a multiplicative identity which all the homomorphisms much respect), whereas the article on the subring test works in the category of rngs (i.e., there may not be a multiplicative identity and even if there is it need not be preserved by homomorphisms). In the category of rngs, one should state the nonemptiness explicitly, whereas in the category of rings it is guaranteed by the presence of the multiplicative identity.

If anyone has further ideas for improving either of these two articles, please let me know. Or rather, please go ahead and implement them -- be bold, as they say on that other site -- but it would be nice to come back here and tell us what you've done.

  • 1
    $\begingroup$ Thanks for the response and for editing the wikipedia article. $\endgroup$
    – Florian J
    May 23, 2011 at 0:39

As your counterexample shows, the first statement is incorrect. But there is a one character fix: require that the subset contains $\,{-}1\,$ vs. $\,1.\,$ Alternatively, require it to also be closed under negation (additive inverses). Perhaps they meant $\rm\,S\,$ is closed under subtraction (vs. addition), since then the subgroup test implies $\rm\,S\,$ is a subgoup of the additive group of $\rm\,R.\,$ Or, explicitly

$$\rm r,s\in S\ \Rightarrow\ r-r\ =\ 0\in S\ \Rightarrow\ 0-r\ =\: -r\in S\ \Rightarrow\ s-(-r)\: =\ s+r\in R $$

Remark $ $ A similar error appears in Greuel and Pfister: A singular introduction to commutative algebra, p. $1$, where they define a subring as a subset containing $1$ that is closed under the induced ring operations. But negation is not explicitly listed as a ring operation in their definition of a ring. Rather, similar to Atiyah and MacDonald, they say that "R, together with addition, is an abelian group". Perhaps "addition" is meant to denote the full additive group structure, so that the ring is supposed to inherit the negation operation (or equivalent axioms for inverses) from the abelian group structure, when it inherits the addition operation.

  • $\begingroup$ Thanks for the response. This shows that closure under subtraction and multiplication gives us a subring. My question was about the first statement. Is closure under addition and multiplication enough? $\endgroup$
    – Florian J
    May 22, 2011 at 15:54
  • $\begingroup$ @Florian You've already given a counterexample to show that the first statement is not sufficient (that's why I thought you were asking more). Don't be surprised by errors in popular textbooks. There are many. Google "Pertti Lounesto". $\endgroup$ May 22, 2011 at 16:49
  • $\begingroup$ OK. Thanks for the clarification. $\endgroup$
    – Florian J
    May 22, 2011 at 16:53
  • $\begingroup$ Link to the book given in this answer seems to be broken - however, the book can probably be found in many other places and maybe also a preview given in the Google Books could help. $\endgroup$ Jan 9, 2022 at 15:12
  • $\begingroup$ @Martin Thanks for the notifcation. Here is a snapshot of said page, $\endgroup$ Jan 9, 2022 at 15:42

It seems that the problem lies in what it means to be closed under addition. My interpretation of being closed under addition is that if you restrict the binary operation of addition to the subset that you want to study, then you get a well defined function.

Say, if you have a ring $(R, +, \cdot)$, then if we have a subset $S \subseteq R$, I would interpret $S$ being closed under the addition inherited from $R$ as meaning that if $a, b \in S$ then $a + b \in S$, or that the image of the map

$$+ : S \times S \longrightarrow R$$

that results from restricting the addition to the elements of $S$ lies in $S$, that is, that $+(S \times S) \subseteq S$ (however weird that notation may seem). So if this is what is what it means for $S$ to be closed under addition, then certainly $\mathbb{R}_{+}$ would satisfy the requirements in the first formulation of the "subring test" that you give, but it will not be a subring of $\mathbb{R}$ since it will not contain the additive inverses.

The same thing would happen when considering $\mathbb{N} \subseteq \mathbb{Z}$.

I just checked my copy of Atiyah-Macdonald and indeed they define a subring in this way, so maybe it is just a misunderstanding or maybe a lack of care when defining a subring.

  • $\begingroup$ Thanks for correctly understanding my question and for the response. Indeed, the first thing I checked was the definition of "closure under a binary operation" and I was and still am under the impression that you write above. I was mainly confused because while defining ideals, Atiyah Macdonald define it to be an additive subgroup closed under multiplication by elements of the ring. They could have used the phrase "additive subgroup" for defining subrings as well, but they didn't. $\endgroup$
    – Florian J
    May 22, 2011 at 16:04
  • $\begingroup$ @Florian J In fact that's exactly what other books do when defining a subring, for example I'm looking at Nathan Jacobson's Basic Algebra I and he defines a subring by saying that it is a subgroup of the additive group and a submonoid of the multiplicative monoid (he treats rings with 1 so that's why he says monoid). $\endgroup$ May 22, 2011 at 16:08
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    $\begingroup$ @Florian J: Dear Florian, Amusingly, in the case of ideals it would be okay to just require closure under addition (as well as under multiplication by arbitrary elements of the ring), because then whenever $x$ is in the ideal, we would find that $(-1)\cdot x$, i.e. $-x$, is also in the ideal. In other words, if one could switch the two statements about the additive structure in the definition of subring and in the definition of ideal, everything would be fine! Regards, $\endgroup$
    – Matt E
    May 22, 2011 at 18:22
  • $\begingroup$ @Matt E: Good observation :) $\endgroup$
    – Florian J
    May 23, 2011 at 0:38

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