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From: quasi on 11 Aug 2010 03:44
William Elliot recently posted this question:

If a commutative ring is such that all elements except 1 are zero
divisors, must it be a Boolean ring?

Remark:

Actually, William Elliot asked the same question several years ago,
and soon after that, I posted a construction showing that such a ring
need not be Boolean. In other words, I specified a construction for a
commutative ring such that not all elements are idempotent, but for
which every element except 1 is a zero divisor. Was my construction
correct? Beats me -- I hardly remember the details. It was years ago,
but as I recall, no one made objections at the time. On the other
hand, no on affirmed it either.

Below is a repost of my construction.

First, some terminology ...

For this discussion, the term ring will mean commutative ring with 1,
and with 1 != 0.

Call a ring unit-trivial if 1 is the only unit.

Note that Z is not unit-trivial since -1 is a unit distinct from 1.

It's easy to show that a unit-trivial ring must have characteristic 2.

Some simple examples of unit-trivial rings:

the field Z_2

the polynomial ring Z_2[x]

However it's not true that every ring of characteristic 2 is
unit-trivial. For example the field of rational functions Z_2(x) has
characteristic 2 but is clearly not unit-trivial.

By an embedding R->R' we will mean an injective homomorphism of the
ring R into the ring R'.

It's automatic that an embedding R->R' takes idempotents to
indempotents, non-idempotents to non-idempotents, units to units, and
zero divisors to zero divisors.

However it's not necessarily true that an embedding R->R' takes
nonunits to nonunits or non-zero-divisors to non-zero-divisors. In
fact, part of our strategy is to define embeddings which force
non-zero-divisors to become zero divisors while simultaneously not
creating any new units.

The construction ...

Let R be any unit-trivial ring.

Let X={x_M | M is a maximal ideal of R} be a set of independent
indeterminates. Consider the polynomial ring R[X]. Note -- by R[X] we
mean the ring of multivariable polynomials in the set of
indeterminates from X with coefficients in R.

Let J be the ideal of R[X] generated by the union of the sets M*x_M
for all maximal ideals M of R together with the union of the sets
{x_M1*x_M2} for all pairs of distinct maximal ideals M1,M2 of R.

Let R'=R[X]/J.

I'll make 3 claims about R', deferring the proofs until the end so as
not to distract from the main argument.

(1): The composite map R->R[X]->R[X]/J gives an embedding of R->R'.

(2): All elements of R except 1 become zero divisors in R'.

(3) R' is unit-trivial.

Thus, assuming the validity of the 3 claims, we get the following
theorem ...

Theorem 1:

Any ring unit-trivial ring R can be embedded in a unit-trivial ring R'
such that all elements of R except 1 become zero divisors in R'.

We can use direct limits to strengthen theorem 1 ...

Theorem 2:

Any unit-trivial ring R can be embedded in a ring T such that all
elements of T except 1 are zero divisors.

proof of theorem 2:

Starting with R_0=R, and applying theorem 1, we can construct an
infinite chain of embeddings R_0->R_1->R_2->... such that for all n,
R_n is unit-trivial and such that all elements of R_n except 1 are
zero divisors in R_(n+1).

Let T be the direct limit of the system R_0->R_1->R_2->...

Any element of T can be obtained from an element of R_n for some n,
but since R_(n+1) also embeds in T, it follows that all elements of T
except 1 are zero divisors.

Next, by applying theorem 2 to an appropriate ring R, we can get the
example we seek ...

Let R be any unit-trivial ring such that not every element of R is
idempotent.

For example, we could let R be the univariate polynomial ring Z_2[x].
Applying theorem 2, we can embed R into a ring T such that all
elements of T except 1 are zero divisors.

But R has non-idempotent elements hence, since R->T is an embedding,
any non-idempotent element of R maps to a non-idempotent element of T.

Therefore the ring T is a commutative ring such that all elements
except 1 are zero divisors, but not every element is idempotent.

This completes the construction.

It remains to prove the 3 claims.

Be warned -- the proofs are routine but somewhat tedious.

We need the following preliminary results.

claim (A): If p is in J, each monomial of p is in J.

This is automatic since J is generated by monomials.

claim (B): Any univariate monomial in J has the form r*(x_M)^k where r
is in M.

Suppose r*(x_M)^k is in J.

Thus, suppose r*(x_M)^k is a weighted combination of the generators of
J with weights from R[X]. Then we have an equation of the form

equation (1): r*(x_M)^k = weighted combination of the generators
of J with weights from R[X]

In equation 1, substitute 0 for all variables x_N except x_M to get

equation (2): r*(x_M)^k = weighted combination of generators of J
of the form a*x_M, with weights from R[x_M], and where a is in M.

Hence the polynomial on the right side of equation 2, viewed as an
element of R[x_M], has all coefficients in M, and on the left hand
side, the coefficient of (x_M)^k is r. It follows that r is in M, as
claimed.

claim (1) : The composite map R->R[X]->R[X]/J gives an embedding of
R->R'.

That the composite map is an embedding follows from the fact that the
ideal J in R[x], when contracted back to R, yields the 0 ideal. To see
that the contraction of J back to R is 0, simply note that all
elements of J have 0 constant term.

claim (2): All elements of R except 1 become zero divisors in R'.

Let r be an element of R, r=/=1. Then r is in some maximal ideal M of
R. But then in R', r*x_M=0. But by claim (B), x_M is not in J, hence
x_M is not 0 in R', and therefore r is a zero-divisor, as claimed.

claim (3): R' is unit-trivial.

Suppose instead there exists f',g' in R' such that f'*g'=1 and
f',g'=/=1.

Choose f,g in R[X], reduced mod J, such that f maps to f' and g maps
to g'. By reduced mod J, we mean no monomial term of either f or g is
in J.

Let a_0 be the constant term of f, and b_0 be the constant term of g.
Then f*g=1 mod J implies a0*b0=1 mod J. Since J contracts to 0 in R,
it follows that a0*b0=1. Since R is unit-trivial, it follows that a0=1
and b0=1. Thus, we can write f=f1+1, g=g1+1, where f1,g1 have no
constant term. Also, since f=/=1, g=/=1, it follows that f1,g1 are
both nonzero

Since f,g are reduced mod J, each monomial in either f1, g1 is
univariate.

Let var_f = the set of indeterminates actually present in f, and let
var_g = the set of indeterminates actually present in g.

Suppose var_f intersect var_g is empty. Then since the product of any
2 distinct indeterminates is in J, f1*g1=0 mod J, hence
f*g=(f1+1)*(g1+1)=f1*g1+f1+g1+1=f1+g1+1 mod J. But then f*g=1 mod J
implies f1+g1=0 mod J. But f1 and g1 are both nonconstant and reduced
mod J. Since they have distinct indeterminates, the can be no
cancellation of terms mod J, so f1+g1=0 is impossible.

Hence f and g have some common indeterminate, x_M, say.

For ease of notation, let x=x_M.

Let a_j*x^j be the leading term of f with respect to x, and let
b_k*x^k be the leading term of g with respect to x. Since the terms
are reduced, we know that a_j and b_k are not in M.

Then, the leading term of f*g with respect to x is a_j*b_k*x^(j+k).
Since it's the only term of it's kind, no cancellation with another
term is possible. But f*g=1 mod J implies f*g-1 is in J, hence by
claim (A), all monomials of f*g-1 are in J. Since a_j and b_k are not
in M, a_j*b_k is not in M, hence, by claim (B), the term
a_j*b_k*x^(j+k) is not in J, so this term survives mod J in f*g as
well as in f*g-1, contradiction. This proves claim (3).

The proof is now complete.

quasi
From: Bill Dubuque on 11 Aug 2010 17:37
quasi <quasi(a)null.set> wrote:
> William Elliot recently posted this question:
>
> If a commutative ring is such that all elements except 1
> are zero divisors, must it be a Boolean ring?
>
> Remark:
>
> Actually, William Elliot asked the same question several years ago,
> and soon after that, I posted a construction showing that such a ring
> need not be Boolean. In other words, I specified a construction for a
> commutative ring such that not all elements are idempotent, but for
> which every element except 1 is a zero divisor. Was my construction
> correct? Beats me -- I hardly remember the details. It was years ago,
> but as I recall, no one made objections at the time. On the other
> hand, no on affirmed it either.
>
> Below is a repost of my construction. [...]

Cohn showed how to do this long ago. Namely, as a corollary to another
result, he shows that any commutative ring with a unit-element 1,
which has no invertible elements other than 1, can be embedded in
a ring in which every element unequal to 1 is a zerodivisor. See

P. M. Cohn. Rings of zero-divisors. Proc. Amer. Math. Soc. 9 (1958), 909-914
http://www.ams.org/journals/proc/1958-009-06/S0002-9939-1958-0103202-2/S0002-9939-1958-0103202-2.pdf

--Bill Dubuque
From: quasi on 11 Aug 2010 19:55
On 11 Aug 2010 17:37:38 -0400, Bill Dubuque <wgd(a)nestle.csail.mit.edu>
wrote:

>quasi <quasi(a)null.set> wrote:
>> William Elliot recently posted this question:
>>
>> If a commutative ring is such that all elements except 1
>> are zero divisors, must it be a Boolean ring?
>>
>> Remark:
>>
>> Actually, William Elliot asked the same question several years ago,
>> and soon after that, I posted a construction showing that such a ring
>> need not be Boolean. In other words, I specified a construction for a
>> commutative ring such that not all elements are idempotent, but for
>> which every element except 1 is a zero divisor. Was my construction
>> correct? Beats me -- I hardly remember the details. It was years ago,
>> but as I recall, no one made objections at the time. On the other
>> hand, no on affirmed it either.
>>
>> Below is a repost of my construction. [...]
>
>Cohn showed how to do this long ago. Namely, as a corollary to another
>result, he shows that any commutative ring with a unit-element 1,
>which has no invertible elements other than 1, can be embedded in
>a ring in which every element unequal to 1 is a zerodivisor. See
>
>P. M. Cohn. Rings of zero-divisors. Proc. Amer. Math. Soc. 9 (1958), 909-914
>http://www.ams.org/journals/proc/1958-009-06/S0002-9939-1958-0103202-2/S0002-9939-1958-0103202-2.pdf

Thanks for the link.

Interestingly, Cohn's proof looks to be essentially the same as mine,
which suggests to me that the underlying proof idea is a natural one.

quasi
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