Monday, January 05, 2004
Decomposition in algebraic integers
Thanks to a post by Rick Decker, a professor at Hamilton College, I can talk about the problem with conventional thinking on algebraic integers using someone else's example.
In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his a's are roots of
a^2 - (x - 1)a + 7(x^2 + x).
Decker wanted to cast doubt on my reliance on using the constant terms to see how 7 divides both sides, by using an example where you can check at x=1, to see what the factors are, as then you have both a_1(1) and a_2(1) equal to sqrt(14).
However, Decker may have naively thought he was refuting my argument, when he didn't follow through to the logical conclusion from his own analysis, which supports it.
You see, the conventional thinking is that you can divide 7 from both sides and still be in the ring of algebraic integers, because algebraic integer are infinitely decomposable, so from a common sense perspective you might think that you can always find two algebraic integer factors of 7 to divide the two algebraic integer factors (5a_1(x) + 7) and (5a_2(x) + 7) on the left hand side.
So assume there exists algebraic integer functions w_1(x) and w_2(x), such that
w_1(x) w_2(x) = 7, and
(5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2.
The assumption is that you're still in the ring of algebraic integers with an algebraic integer x, so consider algebraic integer functions f_1(x), and f_2(x), such that
f_1(x) f_2(x) = 25x^2 + 30x + 2,
and
f_1(x) = (5a_1(x) + 7)/w_1(x)
and
f_2(x) = (5a_2(x) + 7)/w_2(x).
Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2.
Now then, replacing them with
f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have
(g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2
and letting
g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have
(5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2.
Pushed to reply further on his original post by me, Decker actually went about calculating b_1(x), and his result is
2b_1(x)^2 - x b_1(x) + x^2 + x = 0.
See:
Message-ID: <3FF8C21F.4090409@hamilton.edu>
>Date: Sun, 04 Jan 2004 20:47:11 -0500
>From: Rick Decker
Newsgroups: sci.math
>Subject: Re: JSH: Rick Decker's example
But
2b_1(x)^2 - x b_1(x) + x^2 + x = 0,
is a non-monic, and not generally reducible over Q, with an integer x, proving that b_1(x) is not in general an algebraic integer.
Now then, what conventional wisdom has clearly fallen?
Well consider that the factorization I have looks something like
AB = 7C
where A, B, C, and of course, 7 are algebraic integers.
But divide both sides by 7, and because of that non-monic, you must have cases where
DE = C, where D and E cannot be algebraic integers!!!
Dividing 7 from both sides results in factors that are NOT in general algebraic integers as shown by Decker's own result, proving that there does NOT always exist a decomposition of 7 in the ring of algebraic integers that will do the job, since the quadratic he found isn't always reducible over Q, for an algebraic integer x.
It is my hope that my use of Decker's own example, and his own analysis to get that non-monic quadratic might in some small way break through the logjam created by various posters who never back down no matter how often their positions are proven wrong.
Want more? See my blog archives:
http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html
If you wish to interpret Decker's result some different way, then you are free to try, but the gist of it is that you're pushed out of the ring of algebraic integers by dividing both sides by 7, defying conventional thinking that algebraic integer factors of 7 would always exist that can be divided from both of the algebraic integer factors on the left hand side.
[A reply to David C. Ullrich.]
You stupid shit head!!! What the fuck is wrong with you Ullrich?
No matter how many fucking times I tell you to fuck off, you keep replying to me!!!
What the fuck is your problem you shithead?
You Ullrich are a stupid piece of dumb shit who refuses to get the message when someone does NOT want to talk to you, you stupid fucking shitty asshole.
You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass stupid self somewhere to GET A FUCKING CLUE and QUIT FUCKING REPLYING TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!
FUCK OFF!!!!
Can't you get it through your stupid head?
FUCK OFF!!!!!!!!!!!!!!!!!
In his post Decker claimed to mirror my argument using a quadratic instead of a cubic, where he has
(5a_1(x) + 7)(5a_2(x) + 7) = 7(25x^2 + 30x + 2)
where his a's are roots of
a^2 - (x - 1)a + 7(x^2 + x).
Decker wanted to cast doubt on my reliance on using the constant terms to see how 7 divides both sides, by using an example where you can check at x=1, to see what the factors are, as then you have both a_1(1) and a_2(1) equal to sqrt(14).
However, Decker may have naively thought he was refuting my argument, when he didn't follow through to the logical conclusion from his own analysis, which supports it.
You see, the conventional thinking is that you can divide 7 from both sides and still be in the ring of algebraic integers, because algebraic integer are infinitely decomposable, so from a common sense perspective you might think that you can always find two algebraic integer factors of 7 to divide the two algebraic integer factors (5a_1(x) + 7) and (5a_2(x) + 7) on the left hand side.
So assume there exists algebraic integer functions w_1(x) and w_2(x), such that
w_1(x) w_2(x) = 7, and
(5a_1(x)+ 7)/w_1(x) (5a_2(x) + 7) /w_2(x) = 25x^2 + 30 x + 2.
The assumption is that you're still in the ring of algebraic integers with an algebraic integer x, so consider algebraic integer functions f_1(x), and f_2(x), such that
f_1(x) f_2(x) = 25x^2 + 30x + 2,
and
f_1(x) = (5a_1(x) + 7)/w_1(x)
and
f_2(x) = (5a_2(x) + 7)/w_2(x).
Checking at x=0, gives that f_1(0) = 1, and f_2(0) = 2.
Now then, replacing them with
f_1(0) = g_1(x) + 1, and f_2(0) = g_2(x) + 2, I have
(g_1(x) + 1)(g_2(x) + 2) = 25x^2 + 30x + 2
and letting
g_1(x) = 5 b_1(x), and g_2(x) = 5b_2(x), I have
(5 b_1(x) + 1)(5 b_2(x) + 2) = 25x^2 + 30x + 2.
Pushed to reply further on his original post by me, Decker actually went about calculating b_1(x), and his result is
2b_1(x)^2 - x b_1(x) + x^2 + x = 0.
See:
Message-ID: <3FF8C21F.4090409@hamilton.edu>
>Date: Sun, 04 Jan 2004 20:47:11 -0500
>From: Rick Decker
Newsgroups: sci.math
>Subject: Re: JSH: Rick Decker's example
But
2b_1(x)^2 - x b_1(x) + x^2 + x = 0,
is a non-monic, and not generally reducible over Q, with an integer x, proving that b_1(x) is not in general an algebraic integer.
Now then, what conventional wisdom has clearly fallen?
Well consider that the factorization I have looks something like
AB = 7C
where A, B, C, and of course, 7 are algebraic integers.
But divide both sides by 7, and because of that non-monic, you must have cases where
DE = C, where D and E cannot be algebraic integers!!!
Dividing 7 from both sides results in factors that are NOT in general algebraic integers as shown by Decker's own result, proving that there does NOT always exist a decomposition of 7 in the ring of algebraic integers that will do the job, since the quadratic he found isn't always reducible over Q, for an algebraic integer x.
It is my hope that my use of Decker's own example, and his own analysis to get that non-monic quadratic might in some small way break through the logjam created by various posters who never back down no matter how often their positions are proven wrong.
Want more? See my blog archives:
http://mathforprofit.blogspot.com/2003_10_01_mathforprofit_archive.html
If you wish to interpret Decker's result some different way, then you are free to try, but the gist of it is that you're pushed out of the ring of algebraic integers by dividing both sides by 7, defying conventional thinking that algebraic integer factors of 7 would always exist that can be divided from both of the algebraic integer factors on the left hand side.
[A reply to David C. Ullrich.]
You stupid shit head!!! What the fuck is wrong with you Ullrich?
No matter how many fucking times I tell you to fuck off, you keep replying to me!!!
What the fuck is your problem you shithead?
You Ullrich are a stupid piece of dumb shit who refuses to get the message when someone does NOT want to talk to you, you stupid fucking shitty asshole.
You are an ASSHOLE Ullrich!!! Now why don't you take your dumb ass stupid self somewhere to GET A FUCKING CLUE and QUIT FUCKING REPLYING TO ME AS IF I EVER WANT TO TALK TO YOU!!!!!!!!!!!!!
FUCK OFF!!!!
Can't you get it through your stupid head?
FUCK OFF!!!!!!!!!!!!!!!!!
# posted by James Harris @ 10:08 
