Just a few short lines will do.
Next of course we really ought to consider how minor abuse can lead to major abuse.
Post by Mike Terry
Post by Mike Terry Post by Tim Golden BandTech.com Post by Mike Terry Post by Tim Golden BandTech.com
Mon 31 Aug 2020 08:47:53 AM EDT
Abstract Algebra Broken
So this issue I can only repeat myself on so many times. Still I'll try another variation.
Possibly eventually you will be able to at least regurgitate my own position to yourself since my position is completely unchanged.
[snip rest of ridiculous rant]
X + ...
Post by Mike Terry Post by Tim Golden BandTech.com Post by Mike Terry
This all reads like you are serious, but you clearly have never studied
abstract algebra, and know next to nothing about it - certainly not
enough to criticise even a basic construction which you don't
understand. Or is it some kind of subtle joke? (If so I miss the point.)
is not a binary operation. agree?
Assuming a1 is an element of the underlying ring...
a) the polynomial a1 X
[i.e. whose X coefficient is a1, and all other coefficients zero]
b) the product of the polynomial a1
[i.e. whose constant coefficient is a1, and all others zero]
and the polynomial X
This above (b) is it; just as I wrote in the O.P. except I would not call this 'the polynomial a1'
OK, so you are saying that you mean the product of a1 and X, so clearly
the product operation involved is a binary operation. [It involves two
There are now two possibilities: you consider a1 to be a polynomial, in
which case we have regular polynomial multiplication, which is properly
defined, or you consider a1 to be a member of the underlying ring.
It seems your issue is with the second approach. Lets focus on
polynomials over R, which is what you discuss below.
I'll assume you're ok with the definition of a polynomial over R. There
are several equivalent approaches to how these are defined, and I could
expand on this, but they all lead the same way, defining addition and
multiplication /of polynomials/.
But now we have something else: a binary operation taking a real number
and a real polynomial. Of course such an operation needs to be properly
defined to have a meaning, as it's not covered by the definition of
polynomial multiplication. Let's write the op as a binary function sm,
so sm: (R x R[X]) ---> R[x]. [R is the set of real numbers, and R[x]
the set of polynomials over R.] I choose the notation "sm" for "scalar
And here is the definition for sm: if a is in R and p is a polynomial
over R, represented as
p = p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
sm(a,p) = (a p_0) + (a p_1) x + (a p_2) x^2 + ... + (a p_n) x^n
There is an important point here. When I wrote p = p_0 + p_1 x + p_2
x^2 + ... + p_n x^n above, I am not writing a long sequence of sums and
products on the right hand side (rhs) of the equals! If I were, you
could correctly claim the definition is circular! I am writing a
polynomial specified in whatever notation polynomials have been
previously defined, except admittedly I've just assumed for convenience
they've been introduced as expressions of this form. (Even though there
are other technical ways of introducing them, authors would typically
make a point that they can be represented in this notation, at least for
So sm is well defined, no problem with this approach either.
Effectively, this is considering R[x] to be a module over its base ring
R, and sm is just like multiplying vectors in a vector space by a scalar
in its base field. [A module is akin to a vector space, except it has a
base /ring/ rather than a base /field/.]
And of course, now that we've defined sm, we can go on and prove basic
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n =
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n)
(Here, the + signs on lhs of equals are just part of the notation for a
polynomial p, not an operation sign, while the + signs on rhs ARE the
binary operation acting on R[x].)
And just to add to notation confusion, it is typical tradition to write
scalar multiplication of sm(a, p) simply as a p, as you did initially,
in which case we can also write
sm(p_0, x^0) + sm(p_1, x^1) + ... + sm(p_n, x^n) =
p_0 + p_1 x + p_2 x^2 + ... + p_n x^n
but the meaning of rhs here is quite different from meaning of lhs
above: the rhs here IS a long sequence of (polynomial) sums and
Does it seem terrible to you that the notation is ambiguous? The point
is that we realise it is ambiguous, but provably no harm is done, as
we've proved that all the different interpretation lead to the same
result, so no harm is done. This is common practice in mathematics,
balancing simplicity of notation against formal syntactical correctness.
It is true that the polynomial form
0 + a1 X + 0 X X + 0 X X X + ...
is equal to
so I see how you come to name it so. These systems of umpteen variables are true across all values and so to select zeros for most of the values is a fine usage. Now we have something simple to discuss.
Further we can use real values for a1, as is the custom in abstract algebra as they often implement the 'polynomial with real coefficients'. It is particularly at this point that full breakage occurs, though the usage of an undefined X in a product, even with itself, is a fairly sore point as well.
No, there is nothing "undefined" here. You've not properly grasped the
definition of polynomials I think. Perhaps my explanation above will
help, but possibly we need to go back to the basics of how polynomials
are rigorously defined, and the basic operations on them.
I wouldn't mind doing that if you're serious, or alternatively let me
know if you still think something is undefined...
X of course is in its 'abstract' form. a1 and X are not in the same set. Yet there is a product being taken. This product ought to get rather some attention since it is another operator... and wasn't it just moments ago in this subject that the ring operators were so carefully laid out?
OK, the problem here is either that you missed the definition for the
scalar product, or possibly the author of the text you're using omitted
it for whatever reason. That does not make abstract algebra "wrong" in
any way, and the problem with your OP was that it comes across as a rant.
Well, lets say it was a rant, but for most people if they were trying to
understand a new field of knowledge and didn't follow something, they
would ASK FOR HELP IN EXPLAINING THEIR CONFUSION, rather than rant on
about how the field is a pile of crap etc.. Do you see how the latter
behaviour justifiably invites laughter and (frankly) scorn from readers?
Anyway, I gave the definitions for scalar multiplication above, so
hopefully all is clear now! :)
Where was the discussion of this new non-binary operator and their non-binary sums by the way as we study
a0 + a1 X
this sum can only further the problem, as now we definitely have a real (as chosen above) in sum with a non-real entity.
In this context, when you consider the sum a0 + a1 X, it is understood
that a0 is the /polynomial/ a0 X^0, and it is simply by tradition that
we often write the shorter a0. Similar to the situation above with sm,
there are basic (provable) results which underly this slight abuse of
notation, rendering it harmless.
Specifically, it is shown that the set of polynomials of the form (a0
x^0) together with polynomial addition and multiplication is
/isomorphic/ to the set of real numbers with real number addition and
multiplication. That is, we have the correspondence
a0 x^0 <----> a0
and we show this correspondence respects the operations of addition and
multiplication appropriate for each side of the correspondence.
Example: (3 x^0) + (5 x^0) = (8 X^0) <----> 8 = 3 + 5, and also
3 x^0 <----> 3 and 5 x^0 <----> 5. (Yes, this is as obvious is it
seems!) So algebraicly R and the set of polynomials of form a x^0
behave exactly the same, and we informally identify them together in day
to day use. (This is like we identify the real number 2 with the
natural number 2, although it can be argued they are conceptually
How such a direct contradiction in a subject that is supposedly pristine can be propagated and absorbed by so many for so long is surely a statement with broader consequences. All that most can do is to deny the breakage. Here at least a lamb has offered itself up.
I hope I've shown there is no "breakage". At worst there is some minor
abuse of notation going on, which does no harm, and is completely
understood by everybody except you.
It may not be your fault that you missed out on a fuller explanation in
your studies, but your underlying response to this (your attitude in
posting a rant) is down to you...
Thanks Mike and I hope you will pardon the rhetoric for there is actual content to discuss here. The strictness of the ring definition; it was well built. The sum and the product are sufficient without the reverse operators. This polynomial stage though; then the quotient and ideal; these things are very dirty. Should mathematicians really be taking up particle/wave duality without explicitly stating it? At least the physicists bother to explain the contradiction before they eat it en mass. The mathematicians cover it up. That is not mathematics at all, and yet the constructions stands freely and mostly unchallenged. The consequences are broad even if non-mathematical. The separation of philosophy from mathematics and from physics might just have a wee bit to do with this. These are false divisions. As the ring provides there is no need for division. We ought to do without it in the name of simplicity.
Post by Mike Terry
[i.e. whose X coefficient is 1, and all others zero]
c) something else?
[e.g. maybe X represents multiplication or something!]