Newbie Q : Integral Domains and Characteristics
junoexpress
I am very confused by the definition of "characteristic" as it applies
to an integral domain.
In Herstein, Section 3.2, pg 126, he defines an integral domain as a
commutative ring, with no zero divisors.
later (section 3.2, pg 129), he says:
"An integral domain D is said to be of finite characteristic if there
exists some positive number m s.t. ma=0 for all a in D"
My problem is that an integral divisor *by definition* has no zero
divisors, which *by definition* means that ma = o iff m=0 or a=0. So
my question is:
How can an integral domain ever have a characteritic greater than
zero?
I have a feeling the point where I am getting confused is that the
"ma" Hertsein is talking about with his definition of characteristic
is dealing with the operation of group addition, so ma = a + ....+ a
(m times), while the "ma" in the definition of zero divisor has to do
with the group operation of multiplication, but I can't quite see how
that helps me answer my paradox.
Thanks for any insight/help you can provide,
Matt
Dan Cass
The operation on the left of ma=0 is not ring multiplication.
The expression ma here is short for a+a+...+a with m copies of a being added.
Dan Cass
I read too quickly. I see you do know that ma means repeated addition of a to itself.
The "no zero divisors" is for the equation xy=0 where both x and y are in the integral domain, and xy means ring multiplication.
In short the equation ma=0 is not a special case of the above xy=0 for zero divisors, since
* The m here is a positive integer not being used as a ring element
* The "multipication" in the expression ma is not really the ring multiplication, but only shorthand for repeated addition. It's the same as the notation a^m for repeated multiplication.
fishfry
<fc992236-5048-4ba3...@f20g2000vbl.googlegroups.com>,
junoexpress <mtbre...@gmail.com> wrote:
It's a subtle point. Consider the ring of integers mod 5. In this ring,
1 + 1 + 1 + 1 + 1 = 5 = 0.
But there are no two nonzero elements of the ring whose product is 0.
If you were to say, well, 1 + 1 + 1 + 1 + 1 = 0, that's true ... but if
you write 5 + 1 = 0, note that 5 = 0 in this ring. So you have not found
two NONZERO elements of the ring whose product is 0.
Also, as others pointed out, in the notation 1 + 1 + 1 ... (n times),
the 'n' is an element of the natural numbers, not necessarily an element
of the ring itself.
Compare with the ring of integers mod 6, in which 2 * 3 = 0. That's not
an integral domain. But Z_5 is.
Bill Dubuque
>junoexpress <mtbre...@gmail.com> wrote:
>>
>> I am very confused by the definition of "characteristic" as it
>> applies to an integral domain. In Herstein, Section 3.2, pg 126,
>> he defines an integral domain as a commutative ring, with no zero
>> divisors. later (section 3.2, pg 129), he says: "An integral domain D
>> is said to be of finite characteristic if there exists some positive
>>
>> My problem is that an integral divisor by definition has no zero
>> divisors, which by definition means that ma = o iff m=0 or a=0. So my
>> question is: How can an integral domain ever have a characteritic
>> greater than zero?
>>
>> I have a feeling the point where I am getting confused is that the
>> "ma" Hertsein is talking about with his definition of characteristic
>> is dealing with the operation of group addition, so ma = a + ....+ a
>> (m times), while the "ma" in the definition of zero divisor has to do
>> with the group operation of multiplication, but I can't quite see how
>> that helps me answer my paradox.
>
> expression ma here is short for a+a+...+a with m copies of a being added.
> I read too quickly. I see you do know that ma means repeated addition of
> a to itself. The "no zero divisors" is for the equation xy=0 where both
> x and y are in the integral domain, and xy means ring multiplication.
>
> In short the equation ma=0 is not a special case of the above xy=0 for
> zero divisors, since
> * The m here is a positive integer not being used as a ring element
> * The "multipication" in the expression ma is not really the ring
> multiplication, but only shorthand for repeated addition. It's the
> same as the notation a^m for repeated multiplication.
When you later learn about algebras over a ring you will better understand
the source of this scalar/coefficient multiplication, and the underlying
vector-like (module) structure. In the case at hand it simply means that
every ring R is a Z-algebra via (n,r) -> nr = r + r + ...+ r (n times)
Here n -> n1 yields a homomorphic image Z/mZ of Z in R, and this image
is characterized by the kernel mZ, so we call m the characteristic of R
(as a Z-algebra). In fact an R-algebra A is simply a ring, possibly
noncommutative, that contains a central image R/I of R in A, i.e. every
elt of the image of R commutes with every elt of A, so the elts of R
act lie coef's or scalars in the sense that any identity from the any
poly ing R[x,y,z...] over R can be interpreted in A via an eval hom.
E..g the binomial theorem (x+y)^n = .... maps into any Z-algebra = ring,
the freshman's dream (x+y)^p = x^p + y^p maps into any Z/p-algebra;
i.e. R[x,y] is the *universal* R-algebra on two-generators, etc.
As above, the kernel ideal I "characterizes" the type of the R-algebra
--Bill Dubuque
Arturo Magidin
> Hi,
>
> I am very confused by the definition of "characteristic" as it applies
> to an integral domain.
>
> In Herstein, Section 3.2, pg 126, he defines an integral domain as a
> commutative ring, with no zero divisors.
> later (section 3.2, pg 129), he says:
> "An integral domain D is said to be of finite characteristic if there
> exists some positive number m s.t. ma=0 for all a in D"
>
> My problem is that an integral divisor *by definition* has no zero
> divisors, which *by definition* means that ma = o iff m=0 or a=0.
The "m" is not an element of D; it is a nonnegative integer. It is
really shorthand for "add a to itself m times". Remember: given *any*
ring R, and any element a of R, we define:
0a = 0;
(n+1)a = na + a for any nonnegative integer n;
(-n)a = -(na) for any positive integer n.
The "n" is not in the ring, it's notation. Likewise, above, the a is
in D, but them "m" is not (usually) in D, it is part of the shorthand
notation.
>So
> my question is:
> How can an integral domain ever have a characteritic greater than
> zero?
>
> I have a feeling the point where I am getting confused is that the
> "ma" Hertsein is talking about with his definition of characteristic
> is dealing with the operation of group addition, so ma = a + ....+ a
> (m times), while the "ma" in the definition of zero divisor has to do
> with the group operation of multiplication, but I can't quite see how
> that helps me answer my paradox.
It answers it *precisely*, because the "ma" there is not a product in
the ring, it is a shorthand for a sum. There is, in fact, no
"multiplication" of elements of D going on in the equation "ma = 0".
--
Arturo Magidin
junoexpress
> In article
> <fc992236-5048-4ba3-98d8-b0b80367e...@f20g2000vbl.googlegroups.com>,
>
>
>
> junoexpress <mtbrenne...@gmail.com> wrote:
> > Hi,
>
> > I am very confused by the definition of "characteristic" as it applies
> > to an integral domain.
>
> > In Herstein, Section 3.2, pg 126, he defines an integral domain as a
> > commutative ring, with no zero divisors.
> > later (section 3.2, pg 129), he says:
> > "An integral domain D is said to be of finite characteristic if there
> > exists some positive number m s.t. ma=0 for all a in D"
>
> > My problem is that an integral divisor *by definition* has no zero
> > divisors, which *by definition* means that ma = o iff m=0 or a=0. So
> > my question is:
> > How can an integral domain ever have a characteritic greater than
> > zero?
>
>
> > Thanks for any insight/help you can provide,
> > Matt
>
> It's a subtle point. Consider the ring of integers mod 5. In this ring,
> 1 + 1 + 1 + 1 + 1 = 5 = 0.
Yes, it was for me, but thanks to the responses, I understand now.
BTW, I don't know if this helps any of you who teach abstract algebra,
but I can tell you what my particular problem was.
Let R be a ring and consider the expression: a + a, where a is in R.
The "shorthand notation" as pointed out is 2a, where a is in R, 2 is
in Z, and the implicit operation (of group addition) is not the ring
multiplication operation
My problem was that when I factored out a (let's say from the RHS) of
this expression I obtained:
a + a = (1 + 1)*a
Now *all* of the operations on the RHS of this expression are the ring
operations: 1 + 1 is the addition operation in the ring and the * is
the ring multiplication operation.
Now my problem becomes pretty obvious since the obvious temptation is
to let 1 + 1 = 2 (where now 2 is in Z), which IIUC, is false. Rather 1
+ 1 is just that: it is the unit element added to itself, and that's
all it is. It could be viewed, I guess, as a generator for adding any
element to itself twice, but you can't go beyond that and equate it
with a natural number (rather it denotes a natural number of
operations).
Anyhow, thanks again for the help,
Matt
Bill Dubuque
>On Nov 4, 10:01 am, fishfry <BLOCKSPAMfish...@your-mailbox.com> wrote:
>>>
>>> I am very confused by the definition of "characteristic" as it applies
>>> to an integral domain.
>>>
>>> In Herstein, Section 3.2, pg 126, he defines an integral domain as a
>>> commutative ring, with no zero divisors.
>>> later (section 3.2, pg 129), he says:
>>> "An integral domain D is said to be of finite characteristic if there
>>> exists some positive number m s.t. ma=0 for all a in D"
>>
>>> My problem is that an integral divisor *by definition* has no zero
>>> divisors, which *by definition* means that ma = o iff m=0 or a=0. So
>>> my question is:
>>> How can an integral domain ever have a characteritic greater than
>>> zero?
>>
>> 1 + 1 + 1 + 1 + 1 = 5 = 0.
>
> BTW, I don't know if this helps any of you who teach abstract algebra,
> but I can tell you what my particular problem was.
> Let R be a ring and consider the expression: a + a, where a is in R.
> The "shorthand notation" as pointed out is 2a, where a is in R, 2 is
> in Z, and the implicit operation (of group addition) is not the ring
> multiplication operation
> My problem was that when I factored out a (let's say from the RHS) of
> this expression I obtained:
> a + a = (1 + 1)*a
> Now *all* of the operations on the RHS of this expression are the ring
> operations: 1 + 1 is the addition operation in the ring and the * is
> the ring multiplication operation.
> Now my problem becomes pretty obvious since the obvious temptation is
> to let 1 + 1 = 2 (where now 2 is in Z), which IIUC, is false. Rather 1
> + 1 is just that: it is the unit element added to itself, and that's
> all it is. It could be viewed, I guess, as a generator for adding any
> element to itself twice, but you can't go beyond that and equate it
> with a natural number (rather it denotes a natural number of
> operations).
Every ring R contains an image of Z via the hom 1 -> 1_R in R,
so 2 -> 1_R + 1_R, 3 -> 1_R + 1_R + 1_R, etc. The yields an
image of Z in R, namely Z/mZ where mZ in the kernel of the hom
i.e. m is the least integer that maps to 0 via the homomorphism.
Hence m "characterizes" the image of Z in R. This allows us,
by abuse of notation, to view integers as elts of every ring R,
with the implicit understanding that they must be interpreted
modulo m. This is useful because it allows us to transfer all
known integer ring identities from Z into any ring. For example
we can transfer polynomial identities such as the binomial
theorem (x+y)^n = ... from Z[x,y] to R[x,y] for any ring R
simply by interpreting the binomial coefs (mod m), where
m is the characteristic of R. So, e.g. when m = p is prime
we deduce that the Freshman's Dream holds (x+y)^p = x^p + y^p
over any ring of characteristic p. Similarly we can view
polynomial factorizations in any ring which contains an
image of the coefficient ring, e.g.
x^4 + 3 x^2 + 1 = (x^2 - i x + 1) (x^2 + i x + 1)
is interpretable in any Gaussian integer algebra, i.e. any ring
containing an image of the Gaussian integers Z[i], e.g. in Z/5Z
with i = 2 (so i^2 = -1 in Z/5). Later you will learn that this
is a prototypical example of a *universal* object, specifically
the polynomial ring R[x,y,z...] is the universal R-algebra on
the generators x,y,z... Thus every identity true in R[x,y,z...]
can be mapped into any R-algebra, i.e. every ring containing
a central image of the coefficient ring R (central because
commutativity is assumed when multiplying polys, e.g. rx = xr
in R[x]; so poly evaluation is a hom iff coefs are central)
--Bill Dubuque
Arturo Magidin
Here is what you are seeing: if R is a ring with 1, then there is
always a "canonical" ring homomorphism from Z to R, that maps the 1 of
Z to the 1 of R.
Call this map phi. When you write "a+a = 2a", what you really have is
that a+a = phi(2)a, and now the operations are all happening in R, no
problem. It is by abuse of notation that we write 2a, 3a, etc. The
same way that we work in Z/nZ (integers modulo n) writing things like
5 instead of (5+nZ) or [5], which we techically should.
You can define the characteristic in the following way: given a ring R
with 1, let phi:Z-->R be the canonical ring homomorphism. Let nZ be
the kernel of phi, with n>=0. Then we define the characteristic of R
to be n. Since the image of phi is isomorphic to Z/ker(phi) = Z/nZ, it
follows that if R is a domain, then Z/nZ cannot have any zero
divisors, so n must be 0 or a prime, so the characteristic of an
integral domain must be 0 or a prime.
With this definition, the fact that nx = 0 for all x in R becomes a
theorem.
You can extend the ideas to rings without 1 (in which case you would
not be able to factor a+a as (1+1)a anyway); then for every a in R you
get a *group* homomorphism phi:Z-->R by mapping 1 to a. You can still
talk about its kernel.
--
Arturo Magidin
Arturo Magidin
As my ring theory colleagues here keep telling people...
"Every ring R *WITH UNITY*..."
More importantly of course is that in the category of rings-with-one
(and homomorphisms that map 1 to 1), the map is canonical (Z is an
initial object). Otherwise, any idempotent in R will do for "an image
of Z" inside of R.
--
Arturo Magidin
Bill Dubuque
>Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>>
>> Every ring R contains an image of Z via the hom 1 -> 1_R in R,
>
> As my ring theory colleagues here keep telling people...
>
> "Every ring R *WITH UNITY*..."
>
> More importantly of course is that in the category of rings-with-one
> (and homomorphisms that map 1 to 1), the map is canonical (Z is an
> initial object).
That's implicit in what I wrote above. Namely, 1_R in R denotes the
unit elt of R, and, by definition, the hom maps 1 in Z to 1_R in R.
> Otherwise, any idempotent in R will do for "an image of Z" inside of R.
No, that wouldn't yield a ring (with 1) containment, i.e. a subring
since it would have a different unit elt. Recall our prior discussion
on this very issue [1] where I gave a counterexample to a related claim
you made. I've appended it below since it may prove instructive here:
Arturo Magidin <mag...@math.berkeley.edu> wrote on Nov 7 2003:
>Bill Dubuque <w...@nestle.ai.mit.edu> wrote in [1]:
>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>>>Bill Dubuque <w...@nestle.ai.mit.edu> wrote:
>>>>>>Arturo Magidin <mag...@math.berkeley.edu> wrote:
>>>>>>>
>>>>>>> So in ANY ring that contains the integers, 7 and 22 are coprime
>>>>>>> (under either definition).
>>>>>>
>>>>>> That's true in any ring R since R contains a homomorphic image of Z,
>>>>>> Z -> Z*1_R, i.e. any ring is a Z-algebra. But any ring homomorphism
>>>>>> must preserve the relation 22 - 3(7) = 1.
>>>>>
>>>>> Hmmm... Only if you assume that ring morphisms map 1 to 1, which
>>>>> is not necessarily a given either. Even assuming rings have a 1,
>>>>> the zero map is usually considered a valid homomorphism,
>>>>> and your conclusion would be incorrect there.
>>>>
>>>> For Rings (with 1, as I assume above) ring morphisms must preserve 1,
>>>> so the zero map is not a morphism of Rings with 1.
>>>
>>> Granted; like I said, "if you assume that ring morphisms map 1 to 1."
>>
>> That's implicit in my statement, but I agree it's worth highlighting.
>>
>>> On the other hand, you lose some things by assuming that: you
>>> don't get isomorphic copies of the rings in direct products. Some
>>> conventions are better than others, depending on the situation.
>>
>> Indeed. But I think your critique is better targeted at your own:
>>
>>> LEMMA. If S is a ring, and x and y in S are coprime in the sense that
>>> there exist a and b in S such that a x + b y = 1, then for any ring
>>> R that contains S, x and y are also coprime (in the same sense).
>>
>> COUNTEREXAMPLE Let S = Z x {0} < R = Z x Z, a = (3,0), b = (2,0)
>>
>> aS + bS = S = aR + bR < R, so a,b are coprime in S but not in R
>>
>> Your lemma needs S < R is a sub(ring with 1), so 1_S = 1_R.
>> "R contains S" is not sufficient, as the counterexample shows.
>>
>> However, my statement remains true here: 3 - 2 = 1 in Z
>>
>> maps to (3 - 2 = 1) 1_R -> (3,3) - (2,2) = (1,1) in R
>
> Ouch. Your absolutely right (except, I wasn't critising, just
> commenting...) I had a rather nasty double standard there, where on
> the one hand I left open the possibility that ring morphisms do not
> preserve 1, but on the other I was tacitly assuming that subrings did.
> Which of course, makes little sense, as then the image of a morphism
> would not necessarily be a subring. I shall be more careful in the
> future...
[1] http://groups.google.com/group/sci.math/msg/685de5892a0d09c0
http://google.com/groups?selm=y8zu15g...@nestle.ai.mit.edu
Arturo Magidin
> Arturo Magidin <magi...@member.ams.org> wrote:
> >Bill Dubuque <w...@nestle.csail.mit.edu> wrote:
>
> >> Every ring R contains an image of Z via the hom 1 -> 1_R in R,
>
> > As my ring theory colleagues here keep telling people...
>
> > "Every ring R *WITH UNITY*..."
>
> > More importantly of course is that in the category of rings-with-one
> > (and homomorphisms that map 1 to 1), the map is canonical (Z is an
> > initial object).
>
> That's implicit in what I wrote above. Namely, 1_R in R denotes the
> unit elt of R, and, by definition, the hom maps 1 in Z to 1_R in R.
>
> > Otherwise, any idempotent in R will do for "an image of Z" inside of R.
>
> No, that wouldn't yield a ring (with 1) containment, i.e. a subring
> since it would have a different unit elt.
That's implicit in the "otherwise": that is, when discussing rings
which may not have a 1, and discussing homomorphism that need not
preserve it even when it exists.
--
Arturo Magidin