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.
> 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
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.
> > 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
> 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.
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.
> 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
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.
Dan Cass <dc...@sjfc.edu> wrote: >junoexpress <mtbrenne...@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 >> 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.
> 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. > 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
On Nov 4, 5:45 am, 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.
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".
> > 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).
>>> 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?
>> 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).
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)
> > > 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).
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.
> >>> 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?
> >> 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).
> 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). Otherwise, any idempotent in R will do for "an image of Z" inside of R.
>> 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 <magi...@math.berkeley.edu> wrote on Nov 7 2003:
>>>>>>> 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...
> >> 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.
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