I never learnt ring theory and am having trouble understanding the terms "valuation, kernel, residue, ideal" as applied to groups, fields and rings. Let me ask the question this way.
1) Let [imath]xt[/imath] be the product of two fields. [imath]x[/imath] is real [imath]x\in R[/imath] and [imath]t[/imath] is a member of another ordered field [imath]t\in K[/imath] that has no intersection with the reals other than the number 0. Suppose that there is a mapping [imath]v[/imath] that maps [imath]xt[/imath] to [imath]x[/imath]. Could [imath]v[/imath] be called a "valuation"? If not, what is a valuation? If so, what is its kernel, residue and ideal?
2) Let [imath]f=\sum_{g\in G} x_g t_g[/imath] where [imath]G[/imath] is a wellordered abelian group. [imath]x_g\in R[/imath] and [imath]t_g\in K[/imath]. As before, the only element that [imath]R[/imath] and [imath]K[/imath] have in common is the number 0. Multiplication and addition are exactly as for polynomials. This time let [imath]v[/imath] be a mapping from [imath]f[/imath] to the real numbers, in this case the real number that multiplies the largest [imath]t_g[/imath]. Could [imath]v[/imath] be called a "valuation"? If not, what is a valuation? If so, what is its kernel, residue and ideal?
Is there any good reference book on this?
Ring, valuation, kernel, residue, ideal?
Moderators: gmalivuk, Moderators General, Prelates

 Posts: 26
 Joined: Wed Sep 14, 2011 3:13 am UTC
 Yakk
 Poster with most posts but no title.
 Posts: 11045
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: Ring, valuation, kernel, residue, ideal?
For (1) I think you mean subset instead of element of? Which product of two fields are you using? Your notation seems rough.
Is this what you mean by valuation?
https://en.wikipedia.org/wiki/Valuation_%28algebra%29
If so, a valuation must map 0 (the additive inverse) to infinity, among other things.
The kernel of a map is the set of elements that it maps to zero. Usually the map is structure preserving. This induces restrictions on what kind of sets the kernel can be.
Ring ideals are multiplicative absorbing sets. So for x in the ideal X, for all a in the ring, xa is also in the ideal. In some cases there are right and left ideals (ax vs xa in the ideal).
In a handwavey sense (and heck, possibly in a categorytheoretical sense), ideals are sort of like vector subspaces of the ring over itself. Now, there are other mathematical terms that use the term ideal  many (most? all?) of them end up having a similar structure, but I've seen one or two for which the connection isn't blatant.
Residue refers to the equivalence classes when you "mod out" by an ideal. So if you take the integers modulo 2, the residue classes are [0] = {...,4,2,0,+2,+4,+6,...} and [1]={all odd integers}.
Is this what you mean by valuation?
https://en.wikipedia.org/wiki/Valuation_%28algebra%29
If so, a valuation must map 0 (the additive inverse) to infinity, among other things.
The kernel of a map is the set of elements that it maps to zero. Usually the map is structure preserving. This induces restrictions on what kind of sets the kernel can be.
Ring ideals are multiplicative absorbing sets. So for x in the ideal X, for all a in the ring, xa is also in the ideal. In some cases there are right and left ideals (ax vs xa in the ideal).
In a handwavey sense (and heck, possibly in a categorytheoretical sense), ideals are sort of like vector subspaces of the ring over itself. Now, there are other mathematical terms that use the term ideal  many (most? all?) of them end up having a similar structure, but I've seen one or two for which the connection isn't blatant.
Residue refers to the equivalence classes when you "mod out" by an ideal. So if you take the integers modulo 2, the residue classes are [0] = {...,4,2,0,+2,+4,+6,...} and [1]={all odd integers}.
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

 Posts: 26
 Joined: Wed Sep 14, 2011 3:13 am UTC
Re: Ring, valuation, kernel, residue, ideal?
I'm asking because I ran into trouble understanding Todorov (2006) HAHN FIELD REPRESENTATION OF A. ROBINSON’S ASYMPTOTIC NUMBERS, http://arxiv.org/pdf/math/0601722v1.pdf
It talks about "valuation field", "valuation ring", "residue class field", "valuation ideal", "kernel" etc.
Alling (1987) "Foundations of analysis over surreal number fields" uses similar terms.
But I thought that the Ideal of a field F could only be 0 or F which makes the residue trivial. Have tried to read through wikipedia on this and have only got more confused. Also tried reading Herstein, "Topics in Algebra" but it doesn't mention valuations.
It talks about "valuation field", "valuation ring", "residue class field", "valuation ideal", "kernel" etc.
Alling (1987) "Foundations of analysis over surreal number fields" uses similar terms.
But I thought that the Ideal of a field F could only be 0 or F which makes the residue trivial. Have tried to read through wikipedia on this and have only got more confused. Also tried reading Herstein, "Topics in Algebra" but it doesn't mention valuations.
Re: Ring, valuation, kernel, residue, ideal?
Just to make sure, you have noticed that there is an explanation of these terms on page 4 of the paper you linked? In that case, could you be more specific about what it is you would like us to explain?
Regarding your most recent question, the valuation ideal is not an ideal in the valuation field (as you say, if it were, it would be trivial), but in the valuation ring, which is the subset of the valuation field, containing the elements of nonnegative value. It turns out this ideal is maximal in the valuation ring, so the quotient ring is a field, called the residue class field.
Regarding your most recent question, the valuation ideal is not an ideal in the valuation field (as you say, if it were, it would be trivial), but in the valuation ring, which is the subset of the valuation field, containing the elements of nonnegative value. It turns out this ideal is maximal in the valuation ring, so the quotient ring is a field, called the residue class field.

 Posts: 26
 Joined: Wed Sep 14, 2011 3:13 am UTC
Re: Ring, valuation, kernel, residue, ideal?
I was beginning to despair of ever understanding the term "valuation", because each source I looked at seemed to have a different definition (eg. K to R\0, to R union infinity, to R^{+}, to nonnegative R, to Z, and to Z^{+}), when I stumbled across the book "Valued Fields" (2005) by Engler and Prestel.
The explanation given there is very simple. The valuation is the negative logarithm of the absolute value. I have no idea why anyone would want to do that, but it does simplify my understanding enormously. I was confusing a valuation with an absolute value, which was understandable because even the application of an absolute value to a nonArchimedean field is not quite trivial.
So taking a simple example, the complex numbers C, a valuation maps it to R union infinity. What then is the valuation ring, the ideal of the valuation ring, the residue class field and the kernel of that valuation?
After understanding that, it will be up to me to figure out how an absolute value applies on fields generated by Hahn series, hyperreals and the surreal numbers.
The explanation given there is very simple. The valuation is the negative logarithm of the absolute value. I have no idea why anyone would want to do that, but it does simplify my understanding enormously. I was confusing a valuation with an absolute value, which was understandable because even the application of an absolute value to a nonArchimedean field is not quite trivial.
So taking a simple example, the complex numbers C, a valuation maps it to R union infinity. What then is the valuation ring, the ideal of the valuation ring, the residue class field and the kernel of that valuation?
After understanding that, it will be up to me to figure out how an absolute value applies on fields generated by Hahn series, hyperreals and the surreal numbers.
 Yakk
 Poster with most posts but no title.
 Posts: 11045
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: Ring, valuation, kernel, residue, ideal?
A valuation is the negative logarithm of the absolute value. There is no the valuation  a valuation is a type of operation.
I can think of an obvious reason why you'd want to do that  it transforms multiplication into addition. And if you have ever used a slide rule, you'd think that was useful sometimes.
So, valuation is a form of abstraction and generalization of the concept of logarithm to general fields.
So for the complex numbers C, there is a valuation v that maps it to R union infinity given by v(z) = ln(z) if z!=0, and infinity if z=0. There are other valuations from C, and some of them map to R.
Now, what set of elements are greater than or equal to 0 when mapped through v? Ie, v^{1}(R^{+} U {0, infinity})?
This is the valuation ring in C with respect to v.
Next, the prime ideal/maximal ideal/valuation ideal of v is the same, with 0 removed  v^{1}(R^{+} U { infinity}), which also equals the valuation ring, minus v[sup]1[/sub]({0}). These are elements of the valuation ring of C with respect to v.
Finally, we take the valuation ring of C with respect to v, then mod out by the valuation ideal. And this gives us the residue field of v.
Now, this is a simple example  but a simpler example might be just using R instead of C as the space we are working over. Admittedly, the residue field ends up being pretty boring in this case!
I can think of an obvious reason why you'd want to do that  it transforms multiplication into addition. And if you have ever used a slide rule, you'd think that was useful sometimes.
So, valuation is a form of abstraction and generalization of the concept of logarithm to general fields.
So for the complex numbers C, there is a valuation v that maps it to R union infinity given by v(z) = ln(z) if z!=0, and infinity if z=0. There are other valuations from C, and some of them map to R.
Now, what set of elements are greater than or equal to 0 when mapped through v? Ie, v^{1}(R^{+} U {0, infinity})?
This is the valuation ring in C with respect to v.
Next, the prime ideal/maximal ideal/valuation ideal of v is the same, with 0 removed  v^{1}(R^{+} U { infinity}), which also equals the valuation ring, minus v[sup]1[/sub]({0}). These are elements of the valuation ring of C with respect to v.
Finally, we take the valuation ring of C with respect to v, then mod out by the valuation ideal. And this gives us the residue field of v.
Now, this is a simple example  but a simpler example might be just using R instead of C as the space we are working over. Admittedly, the residue field ends up being pretty boring in this case!
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

 Posts: 26
 Joined: Wed Sep 14, 2011 3:13 am UTC
Re: Ring, valuation, kernel, residue, ideal?
I'm still getting stuck. The use of "valuation" in this paper from 1939 doesn't look to be anything like the negative logarithm of an absolute value. Again "residue class" is used.
http://projecteuclid.org/DPubS/Reposito ... 1183502257
The proof of [MacLean, 1939] Thoerem 1 begins:
[imath]x=\sum a_\alpha t^\alpha[/imath].
In the power series field S=K{t^{G}} we introduce a valuation V by setting [imath]V(x)=\alpha_1[/imath] if [imath]a_{\alpha_1}t^{\alpha_1}[/imath] is the first nonvanishing term in the power series for x. In this valuation, G is the value group and K the field of residue classes. Furthermore, [Krull, 1931] S is maximal with respect to this valuation, in the sense that any proper extension S'>S to which the valuation V has been extended must either have a larger value group or a larger residue class field than S.
I need to understand this because by elementary methods I am coming to an opposite conclusion to Theorem 1, so there has to be something deeply buried in the highfalutin maths that is invalidating my elementary approach.
http://projecteuclid.org/DPubS/Reposito ... 1183502257
The proof of [MacLean, 1939] Thoerem 1 begins:
[imath]x=\sum a_\alpha t^\alpha[/imath].
In the power series field S=K{t^{G}} we introduce a valuation V by setting [imath]V(x)=\alpha_1[/imath] if [imath]a_{\alpha_1}t^{\alpha_1}[/imath] is the first nonvanishing term in the power series for x. In this valuation, G is the value group and K the field of residue classes. Furthermore, [Krull, 1931] S is maximal with respect to this valuation, in the sense that any proper extension S'>S to which the valuation V has been extended must either have a larger value group or a larger residue class field than S.
I need to understand this because by elementary methods I am coming to an opposite conclusion to Theorem 1, so there has to be something deeply buried in the highfalutin maths that is invalidating my elementary approach.
 Yakk
 Poster with most posts but no title.
 Posts: 11045
 Joined: Sat Jan 27, 2007 7:27 pm UTC
 Location: E pur si muove
Re: Ring, valuation, kernel, residue, ideal?
That looks sort of logarithmic to me? Did you check the axioms of valuations?
As a quick stab, if you have two power series whose first nonvanishing terms are alpha_1 and alpha_2, the first nonvanishing term of their product will be (alpha_1 + alpha_2)  that is pretty damn logarithmicesque in behavior to me. How is V not a valuation? What axiom does it violate?
Have you done the modding out of the simple/simpler case yet?
Ie, we have R and our valuation is V(x) := ln(x). Have you actually looked at the valuation ring and the residue classes?
As a quick stab, if you have two power series whose first nonvanishing terms are alpha_1 and alpha_2, the first nonvanishing term of their product will be (alpha_1 + alpha_2)  that is pretty damn logarithmicesque in behavior to me. How is V not a valuation? What axiom does it violate?
Have you done the modding out of the simple/simpler case yet?
Ie, we have R and our valuation is V(x) := ln(x). Have you actually looked at the valuation ring and the residue classes?
One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision  BR
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.
Re: Ring, valuation, kernel, residue, ideal?
Different authors use different nonequivalent concepts of "valuation".
I often saw the "surjective homomorphism of the multiplicative group of K to the additive group of R where v(a + b) >= min{v(a), v(b)} extended by v(0) = infinity" definition.
These axioms are true for the valuation you mentioned.
I often saw the "surjective homomorphism of the multiplicative group of K to the additive group of R where v(a + b) >= min{v(a), v(b)} extended by v(0) = infinity" definition.
These axioms are true for the valuation you mentioned.
Who is online
Users browsing this forum: No registered users and 6 guests