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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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    $\begingroup$ @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 $\endgroup$
    – mathfan
    Aug 9, 2011 at 2:59
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    $\begingroup$ @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? $\endgroup$
    – Gerry Myerson
    Aug 9, 2011 at 3:25
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    $\begingroup$ I'm sorry I made a slip of basic logic. $\endgroup$
    – t.b.
    Aug 9, 2011 at 8:15
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    $\begingroup$ @Theo: Welcome to the club! :) $\endgroup$
    – gary
    Aug 9, 2011 at 23:03
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    $\begingroup$ It suffices that $A$ be finitely generated. (See Jonas’s answer.) $\endgroup$
    – Pierre-Yves Gaillard
    Aug 10, 2011 at 7:53

3 Answers 3

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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 $1_a$.

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    $\begingroup$ Wow! How does one find such proofs? $\endgroup$
    – Pierre-Yves Gaillard
    Aug 9, 2011 at 11:29
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    $\begingroup$ @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. $\endgroup$
    – Qiaochu Yuan
    Aug 9, 2011 at 14:34
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    $\begingroup$ (And what Qiaochu said is basically the geometric intuition behind orthogonal idempotents, as well.) $\endgroup$
    – Dan Petersen
    Aug 9, 2011 at 14:39
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    $\begingroup$ @Pierre-Yves Gaillard: Thank you. For some reason I thought it was an English ring. Changed all (I hope) $R$ to $A$. $\endgroup$
    – André Nicolas
    Aug 9, 2011 at 15:08
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    $\begingroup$ @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. $\endgroup$
    – Qiaochu Yuan
    Aug 9, 2011 at 15:59
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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.

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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 $(a,m)(b,n)=(ab+mb+na,mn)$.

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  • $\begingroup$ 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)). $\endgroup$
    – Pierre-Yves Gaillard
    Aug 9, 2011 at 7:06
  • $\begingroup$ +1: Very nice! Thank you! (For another online proof of Nakayama’s Lemma, see Theorem 47 p. 46 in these commutative algebra notes by Pete L. Clark.) $\endgroup$
    – Pierre-Yves Gaillard
    Aug 9, 2011 at 7:32
  • $\begingroup$ 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 $A$. $\endgroup$
    – Pierre-Yves Gaillard
    Aug 9, 2011 at 13:04
  • $\begingroup$ @Pierre and Jonas: see the Lemma cited in my answer for a more general way. $\endgroup$
    – Bill Dubuque
    Aug 9, 2011 at 14:56
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    $\begingroup$ Dear Jonas: Your argument shows that it suffices that $A$ be finitely generated. $\endgroup$
    – Pierre-Yves Gaillard
    Aug 10, 2011 at 7:52

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