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Saturday, February 28, 2015

Blocking algebra with algebraic integers

In the ring of algebraic integers the expression:

P(x) = (g1(x) + 1)(g2(x) + 2)

is blocked if P(x) is a primitive quadratic irreducible over Q, and g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Proof:

Let P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a quadratic irreducible over Q, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

Introduce new functions f1(x), and f2(x), and non-zero integers c1 and c2, where I'll use the second one first, as let:

g2(x) = f2(x) + c2

where c1 = c2 + 2

Now make the substitution:

P(x) = (g1(x) + 1)(f2(x) + c2 +  2) = (g1(x) + 1)(f2(x) + c1)

Now multiply both sides by c1,

c1*P(x) =  (c1g1(x) + c1)(f2(x) + c1)

And now, let g1(x) = f1(x)/c1, and make that substitution to get:

c1*P(x) =  (f1(x) + c1)(f2(x) + c1)

Indistinguishable, the f's must be roots of the same polynomial function, as symmetry has been forced.

But now the requirement that g1(x) = f1(x)/c1, means it is blocked in the ring of algebraic integers unless f1(x) is itself a polynomial function. But if P(x) is irreducible over Q, then that is not possible; therefore, g1(x) cannot be an algebraic integer function.

Proof complete.

Here's an example of the blocking with actual equations that fulfill all of the restrictions.

Let c2 = 5, f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

so the f's are algebraic integer functions, since the polynomial expression is monic, as long as x is an algebraic integer. That then requires that

P(x) = 175x2 - 15x + 2

which is not reducible over Q.

Notice you can see the consequences if you let x = 1, then a2 - 6a + 35 = 0, which would imply that just one of the roots has 7 as a factor, which is NOT possible in the ring of algebraic integers, as in that ring neither can have 7 as a factor.

So it is proven that algebra that is allowed for polynomial factorizations is blocked within the ring of algebraic integers for non-polynomial factorizations.


James Harris

Wednesday, February 25, 2015

Algebra and some manipulations

You can do some interesting things with algebra just by starting things in a special way, like consider the following.

Consider:

P(x) = (g1(x) + 1)(g2(x) + 2)

where P(x) is a primitive quadratic irreducible over Q, g1(0) = g2(0) = 0, but g1(x) does not equal 0 for all x.

If you think that is without problems, watch as I do some simple algebra. So at this point you just have a polynomial which is a product of two functions of x, which are normalized, how hard can that be?

Let's play.

Introduce new functions f1(x), and f2(x), where I'll use the second one first, as let

g2(x) = f2(x) + 5.

No objections there, right? If there are, go away, you don't know algebra.

Now make the substitution:

P(x) = (g1(x) + 1)(f2(x) + 5 +  2) = (g1(x) + 1)(f2(x) + 7)

Wow, boring. But let's multiply both sides by 7, why? Just follow along...

7*P(x) =  (7g1(x) + 7)(f2(x) + 7)

If you're wondering why I multiplied by 7 the way I did, you're not thinking symmetry!

As now of course, let g1(x) = f1(x)/7, and make that substitution to get:

7*P(x) =  (f1(x) + 7)(f2(x) + 7)

And we're done.

It turns out that if P(x) is an irreducible quadratic polynomial, then g1(x) does NOT exist in the ring of algebraic integers.

Neat!

Here's an example of the blocking with actual equations that fulfill all of the restrictions.

Let f1(x) = 5a1(x), and f2(x) = 5a2(x), where the a's are roots of

a2 - (7x-1)a + (49x2 - 14x) = 0

so the f's are algebraic integer functions, since the polynomial expression is monic, as long as x is an algebraic integer. That then requires that

P(x) = 175x2 - 15x + 2

which is not reducible over Q.

What's wrong with that? Well it implies that one of the f's has 7 as a factor, which it cannot in the ring of algebraic integers, except for at x = 0.

Why do you have a problem? Goes back to the asymmetry in the beginning:

P(x) = (g1(x) + 1)(g2(x) + 2)

Turns out you can't have the 1 and 2 with normalized functions of x, and be in the ring of algebraic integers if P(x) is irreducible.

Of course if P(x) IS reducible, then you can, like have g1(x) = x, and g2(x) = x, as simple examples that ARE possible. Then, of course P(x) = x2 + 3x + 2.

But if P(x) is irreducible, then P(x) = (g1(x) + 1)(g2(x) + 2) cannot exist in the ring of algebraic integers. It's what I call a non-polynomial factorization.

So notice, algebraically we can write the expression, but regardless if P(x) is irreducible it's not allowed to exist within the ring of algebraic integers.

The ring of algebraic integers actually blocks algebra in that situation.

However, the expression does exist in the field of complex numbers!

Oh yeah, notice I chose to multiply by 7 above, but could have used 5, or 11 or any other nonzero integer, so why exactly is the ring of algebraic integers blocking?

Inquiring minds would want to know!!!


James Harris

Tuesday, February 17, 2015

Questions of interest

One of the great things about having your own blog is access to information that others just should not have, as of course it is important to keep up with how much interest the blog is generating.

That information is important for lots of reasons, which is why I think it is kept much more close in modern blogs than in the past, when for instance you could often see visitor counts. These days for most that kind of information tells too much.

It's a highly competitive world. And on the web you are competing with the world.

There are other clues though, like I deliberately named this blog something I thought of as rather generic. That approach is risky, as you may never go high in web search, but if that is possible, then it is a great way to see whether or not interest is being drawn.

So, was wondering if it would take over search results.

In Google, it did.

And it can be hard to get a feel for what that means unless you ask yourself a question:

Wouldn't you like to be able to tell people that you found some math?

Where you could direct them, not with a paper handed to them, or a link, but with simple instructions: just go to Google and search--some math

If you think that's easy to accomplish, why don't you do it?

That's global. Have fun checking me on that one.

Or even better, take it over. If you find some math that is better, shouldn't that #1 be you?

It's weird to consider, but highly improbable that even a leading university, let alone any individual human being, could take that search position from me now, as you're looking at a global effect.

It would be cool though if someone did! As it would require some math, use of that phrase, and that math would have to have global impact, so that enough people would do whatever it is they do that drives things up in search! Google knows of course, as I guess do the other search engines, while right now Google is the best at it.

But web technology is still rather new. More than likely in the near future the fullness of that reality will be clear, while for now most people, including me, still struggle to understand what it takes.

Oh yeah, I'm assuming a bit here about how that search went for you!

This blog should be in the top 10 with that search. If not, oh well. Lots less impressive.

And don't I look silly then?

You see, I can't control that search--other people do. Their interest is what's reflected.

And advanced technological companies like Google have built their business around that interest. So it is as objective as you can get.

There is no plausible reason for why Google would bend any of its rules to suit my wants.

Which means an unbelievably powerful validation comes from the best of modern technology saying there are people interested! Cool.

If that interest goes away, then I lose search position.

And I can sit here and type what I want about what's true now, but it's a dynamic reality so it's a constant competition, every minute, of every day.

That competition is unending.

So yeah, if you think it worth challenging that position, which I think is a great thing for others to do, understand you fight to hold it from then on, without let-up, 24 hours a day, 365 days a year, against anyone else anywhere in the world who can take it from you.

Competition can be good for us all. Humanity benefits, remember. Pushing the limits of human knowledge is not a bad thing, and it should be recognized.

And web search is an objective measure that lets you know: some people are interested.


James Harris