I was struggling for days with this nice problem:

Let $A$ be a finite commutative ring such that every element of $A$ can be written as product of two elements of $A$. Show that $A$ has a multiplicative unit element.

I need a hint for this problem, thank you very much.

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    @Theo Buehler : What do you mean ? , I have checked Matt E's and André Nicolas's answers and I don't think that they solved my problem – mathfan Aug 09 '11 at 02:59
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    @mathfan, you've asked four questions here previously, and not accepted the answers on any of them. If you didn't like those answers, why do you ask here again? Or if you're not familiar with the concept of "accepting answers", may I suggest you have a look at the faq? – Gerry Myerson Aug 09 '11 at 03:25
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    I'm sorry I made a slip of basic logic. – t.b. Aug 09 '11 at 08:15
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    @Theo: Welcome to the club! :) – gary Aug 09 '11 at 23:03
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    It suffices that $A$ be finitely generated. (See Jonas’s answer.) – Pierre-Yves Gaillard Aug 10 '11 at 07:53

3 Answers3


Let $a$ be an element of $A$. Then $a$ can be expressed as product of two elements, each of which can be expressed as a product of two elements, and so on forever. By the finiteness of $A$, among these expressions for $a$, some $y\in A$ appears to arbitrarily high powers.

Again by finiteness, we have $y^i=y^j$ for some positive integers $i<j$. Take a representation of $a$ as a product that uses a power of $y$ which is $i$ or greater. Then $i$ of these $y$'s can be replaced by $j$ $y$'s. This procedure does not change $a$, but it multiplies $a$ by $y^{j-i}$. Thus $a=y^{j-i}a$, and therefore there is an identity element for $a$.

We conclude that there is an identity element $1_a$ for every $a\in A$.

Now apply repeatedly the following easy to verify lemma:

Lemma: If $1_u$ is an identity element for $u$, and $1_v$ is an identity element for $v$, then $1_u+1_v-1_u1_v$ is an identity element for both $u$ and $v$. (By $s-t$ we mean $s$ plus the additive inverse of $t$.)

Comment: An equivalent way to finish the argument is to let $M$ be a maximal subset of $A$ for which there is a $u\in A$ such that $um=m$ for all $m\in M$. If $M$ is not all of $A$, we can use the lemma to extend $M$.

Or else we can obtain an explicit and even symmetric expression for a unit in terms of all the $1_a$.

André Nicolas
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    Wow! How does one find such proofs? – Pierre-Yves Gaillard Aug 09 '11 at 11:29
  • @Pierre-Yves Gaillard: A little luck. The $1_a$ was natural. To put them together I wanted something that works like orthogonal idempotents, and fooling around stumbled into something that works. – André Nicolas Aug 09 '11 at 11:39
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    @Pierre: there is a geometric intuition. Thinking of $A$ as the ring of functions on $\text{Spec } A$, to say that $1_a$ is an identity for $a$ is to say that $1_a$ is equal to $1$ on the support of $a$. Now it is not hard to visualize that $1_u + 1_v - 1_u 1_v$ is equal to $1$ on the support of both $a$ and $b$; this is just inclusion-exclusion of characteristic functions. – Qiaochu Yuan Aug 09 '11 at 14:34
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    (And what Qiaochu said is basically the geometric intuition behind orthogonal idempotents, as well.) – Dan Petersen Aug 09 '11 at 14:39
  • André: Shouldn’t your $R$ be an $A$? @Qiaochu: I think I more or less see the idea, which looks very nice. But something’s puzzling me: Are using Jonas’s notation, with $A$ possibly without 1 and $R$ with 1? (To be able to speak of the spectrum.) – Pierre-Yves Gaillard Aug 09 '11 at 14:58
  • @Pierre: no, I was using André's notation. You can still speak of the spectrum without an identity. – Qiaochu Yuan Aug 09 '11 at 15:02
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    @Pierre-Yves Gaillard: Thank you. For some reason I thought it was an English ring. Changed all (I hope) $R$ to $A$. – André Nicolas Aug 09 '11 at 15:08
  • André: You’re welcome. (I thought it would be better to avoid any confusion with Jonas’s notation.) @Qiaochu: I didn’t know. Would you give me a reference (or a brief explanation)? – Pierre-Yves Gaillard Aug 09 '11 at 15:49
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    @Pierre: I guess the subtlety is that various definitions that are equivalent for rings with identity may fail to be equivalent for rings without identity. You can take the spectrum of the unitization of $A$, then get rid of the extra point corresponding to quotienting by $A$. Again there is a geometric intuition: $A$ is analogous to the ring of functions vanishing at infinity on a locally compact non-compact Hausdorff space and its unitization is analogous to the ring of functions on the one-point compactification. – Qiaochu Yuan Aug 09 '11 at 15:59
  • @Qiaochu: OK, I see... By the way I now think the first question I asked you was silly: your consideration of the spectrum was not designed to prove the statement, but to explain the idea behind André’s proof. [If I understand correctly what you say, in the functional analysis setting one wants to unitize only algebras which definitely don’t have an identity (that’s why you speak of locally compact **non-compact** Hausdorff space).] – Pierre-Yves Gaillard Aug 09 '11 at 16:22
  • Actually we can use induction on the cardinality of the ring : for n=1,2 it is clear , now suppose it is true for all k \leq n and we will show it is true for n+1 , take R a ring with cardinality n+1 , let a be an element , if the ideal aR={0} , it is an easy case , other wise the quotient ring R/aR has cardinality – mathfan Aug 09 '11 at 17:26
  • @mathfan: Yes, that would also work nicely, and bypass the construction of my $1_a$. – André Nicolas Aug 09 '11 at 18:08
  • @mathfan: How do you handle the easy case? – Pierre-Yves Gaillard Aug 10 '11 at 05:13
  • @Pierre-Yves Gaillard : well if aR={0} we can find x such that : (a+x)R <> {0} so by case 2 there is e such that ea+ex=a+x that is a=ex-x (*) now xR<> {0} otherwise a+x=0 hence we can find e' such that : e'x=x , multipliying (*) by e' we get : 0= e'a=ee'x-e'x=ex-x=a (remember aR={0}) – mathfan Aug 30 '11 at 19:06

HINT $\ $ The hypothesis implies that $\rm\:A^2 = A\:.\:$ Therefore, by invoking the simple Lemma in this answer, we deduce that $\rm\:A\:$ is principal, generated by an idempotent.

NOTE $\ $ The cited Lemma is a generalization of the proof hinted by Jonas (and Pierre). As here, this Lemma often proves handy so it is well-worth knowing.

Bill Dubuque
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This can be solved using Nakayama's lemma. The first version stated in the linked article is quoted below. The Wikipedia article includes a proof and a reference to Commutative ring theory by Matsumura.

Let $R$ be a commutative ring with identity $1$. . . . Let $I$ be an ideal in $R$, and $M$ a finitely-generated module over $R$. If $IM = M$, then there exists an $r \in R$ with $r \equiv 1$ (mod $I\ $), such that $rM = 0$.

Consider what happens when $R$ is the unitalization$^1$ of $A$, and when $I$ and $M$ are both $A$.

$^1$ The unitalization of $A$ can be defined as $R=A\times \mathbb Z$ with the operations $(a,m)+(b,n)=(a+b,m+n)$ and $(a,m)(b,n)=(ab+mb+na,mn)$.

Jonas Meyer
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  • Sorry! I didn’t read you answer carefully enough! I’ll delete my comment (and repost the reference to Pete’s notes (just as a complement to your Wikipedia reference)). – Pierre-Yves Gaillard Aug 09 '11 at 07:06
  • +1: Very nice! Thank you! (For another online proof of Nakayama’s Lemma, see Theorem 47 p. 46 in these [commutative algebra notes](http://math.uga.edu/~pete/integral.pdf) by [Pete L. Clark](http://math.stackexchange.com/users/299/pete-l-clark).) – Pierre-Yves Gaillard Aug 09 '11 at 07:32
  • Minor variation (in the wording): Embed $A$ into a commutative ring $R$ with 1 (see answer). Let $x_1,\dots,x_n$ be the distinct elements of $A$. We have $x_i=a_{ij}\,x_j$ with $1\le i,j\le n$. Write this in matrix form as $X=BX$, or $(I-B)X=0$ (here $I$ is the identity matrix). Multiplying on the left by the adjugate, we get $\det(I-B)x_i=0$ for all $i$, and the determinant is of the form $1-a$ with $a$ in $A$. – Pierre-Yves Gaillard Aug 09 '11 at 13:04
  • @Pierre and Jonas: see the Lemma cited in my answer for a more general way. – Bill Dubuque Aug 09 '11 at 14:56
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    Dear Jonas: Your argument shows that it suffices that $A$ be finitely generated. – Pierre-Yves Gaillard Aug 10 '11 at 07:52
  • In "my" lemma about stitching together two local identities, the lemma does not care whether these identities are for a single element or for a large subring. – André Nicolas Aug 10 '11 at 20:29