Re: Understanding the quotient ring nomenclature

On Jun 19, 1:53 pm, Leland McInnes <leland.mcin...@xxxxxxxxx> wrote:
On Jun 19, 9:18 am, "Tim BandTech.com" <tttppp...@xxxxxxxxx> wrote:

On Jun 19, 3:39 am, Fishcake <kiwisqu...@xxxxxxxxxxxxxxx> wrote:

I do understand the form of the existing construction. I guess if
you'd like to quiz me on what you
think I misunderstand then I could prove this.

Okay, let's go to a question you've already been asked and have
already answered ...

<SNIP>

Let's see how absurd this statement is. Let A = R.

You are claiming that R = R[X]. That implies that these two things are equal as sets.

Yes. This would be the consistent approach.

Bzzt! Wrong.

But let's move on. Here's another question for you: Given that we are
discussing the ring A[X], is X an element of A? Now this shouldn't be
too hard, because I told you the answer just yesterday, re-quoted it,
and underlined it. ... Wait, I see you've already answered the
question today:

I believe that it is true that X is in
A.

Bzzt! Wrong again.

You see, the ring R[X] is not equal as a set to the ring R, because
the underlying set of R[X] contains elements that are not in the
underlying set of R. For starters X is an element of the underlying
set of R[X]. X is not an element of R. I'm pretty sure you've been
told this several times now.

When you say "I do understand the form of the existing construction"
you are contradicted by what you say below, which shows little or no
understanding of the construction being contemplated.

Let's try a really simple example. I am going to define a ring (B,
+,*). The underlying set of B will be the set {"red", "blue"}. The
operations are going to be defined by:
"red" + "blue" = "blue"
"red" + "red" = "red"
"blue" + "blue" = "red"
"red" * "blue" = "red"
"red" * "red" = "red"
"blue" * "blue" = "blue"
where +, and * are to be assumed to be commutative, associative, and
distributive. Please verify that this indeed forms a ring by meeting
all the ring axioms.

Now, let's define a new ring (B[X],&,#). The underlying set of B[X]
will be the set {"red", "blue", X, "blue"&X, X^2, "blue"&X&X^2,
"blue"&X^2, X&X^2, X^3, ... }. It should be obvious as to how to fill

Err... Isn't "blue"#X & "red"#XX a fine instance? Quite literally any
combination of products and sums within your notation are acceptable
constructions, whether they are in B or in B[X]. Likewise we are still
free to specify b(n) in B and just make
b2 + b4 # X^3
and still be consistent. I don't see much point in belaboring the
color instances so heavily.

out the rest of this set, but if it isn't obvious to you please ask
instead of assuming and getting it wrong). The operations are going to
be defined by:

"red" & t = t (for any element t of B[X] -- thus "red" & ("blue"&X^2)
= "blue"&X^2 for example).

"red" # t = "red" (for any element of B[X] -- thus "red" #
("blue"&X&X^2) = "red" for example).

"blue" & "blue" = "red"

"blue" # t = t (for any element of B[X] -- thus "blue" #
("blue"&X&X^2) = "blue"&X&X^2 for example).

X^n # X^m = X^(n+m) (for any natural numbers m and n)

and the rest should be clear once we assume that & and # are
commutative, associative and distributive: we can use commutativity,
associativity and distributivity to always arrive at something in the
underlying set of B[X]; for example:

("red" & X) # (blue&X^3) = X # (blue&X^3) [by "red" & t = t]
X # (blue&X^3) = (X#"blue") & (X#X^3) [by distributivity]
(X#"blue") & (X#X^3) = ("blue"#X) & (X#X^3) [by commutativity]
("blue"#X) & (X#X^3) = X & (X#X^3) [by "blue" # t = t]
X & (X#X^3) = X & X^4 [ by X^n#X^m = X^(n+m) where X = X^1 ]

and we have an element of the underlying set of B[X]. Try it yourself
on a few examples. Now note that the set given for B[X], along with
the operations given, will form a ring.

Now, a question: is X an element of the underlying set of B?

Let X = red. X^2 = red. This seems to work fine. Do you have any
instances not in B which X satisfies? We know that X is in B[X]. We
know that B is in B[X]. Therefore we know that X can be in B. What
else can X be? What else is in B[X]? Really if your so insistent then
why don't you simply create a single instance of what you are
attempting? I will do so beneath here and it will be conflicted via
the ring definition.

I am confident that the correct answer to your question above is yes:

Elements in B are in B[X]. Let's pick some elements
b(n)
in B and one more X in B[X]. Now I have enough elements to compose a
polynomial
b0 + b1 X + b2 XX + ...
which is in B[X].

We know that B[X] is at least as large as B, but we have no instances
which make it any larger than B. The usage of X only indicates that X
is in B[X]. We have no conflict with choosing X in B since B is in B
[X].

Any element in B[X] which is not in B will require additional product
rules to exist within B[X] which have not been specified in B. If
those rules existed they would not match the ring style of definition
and so B[X] would not be a ring. Thus B[X] is at least as large as B
but is no larger than B and so X must be in B. Furthermore the worry
over the product of type
b X
is alleviated since both elements are in B which covers this product
definition. If X were not in B then the ring specification in B would
not satisfy this product.

Suppose for instance B are the integers modulo 3. If we were to try B
[X] as the reals then we would claim that we could compose in B[X]
( 2 )( 5.3 )
where the only product defined was on the modulo 3 integers in B.
Somehow B[X] is supposed to address this. There is no content for the
ring definition other than back in B so this instance is conflicted.
The very construction of the product was invalid since there is no
room in B[X] to define such products especially on top of their
inherent ring B.

- Tim
.

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