### Thursday, September 16, 2010

## JSH: Coverage issue of algebraic integers

Unique issues arise with the ring of algebraic integers as a result of the very definition that gives the set, where the rule that algebraic integers be roots of monic polynomials with integer coefficients can be shown to force a dramatic coverage issue, where key numbers are left out.

The problem is similar to what happens if you consider only evens, so 2 and 6 do NOT then have a factor in common because 3 is odd, so it has been left out.

Seeing the more complex problem with the ring of algebraic integers requires only elementary methods using some basic algebra.

I'll need only one proof:

The equation (g_1(x) + 1)(g_2(x) + 2) = P(x), when P(x) is a quadratic with integer coefficients, and g_1(0) = g_2(0) = 0, where the g's are arbitrary functions of x, with a specific case not required for the proof itself, may for certain highly specific cases not exist in the ring of algebraic integers.

Proof:

Proving that statement requires use of a constant factor, so I'll multiply by 9, and now consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

Then by the distributive property, I have:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x).

Let f_1(x) and f_2(x) be non-rational roots of the same monic quadratic with integer coefficients—so they are being required to be algebraic integers—then f_1(x) = 9g_1(x), requires one of the f's to have 9 as a factor, while the other must be coprime to 9, since

f_2(x) = g_2(x) - 7

and g_1(0) = g_2(0) = 0.

But it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

The result has important implications for the coverage of algebraic integers, as consider the following carefully peculiar way to generate functions.

With P(x) = 441x^2 - 35x + 2, now moving to the ring of algebraic integers, multiplying by 9, and re-grouping terms gives:

9*P(x) = (81x^2 - 18x)7^2 + (9x-1)(9)(7) + 9^2

which allows me now to get a non-polynomial factorization with

9*P(x) = (7a_1(x) + 9)(7a_2(x)+ 9)

where the a's are roots of

a^2 - (9x-1)a + (81x^2 - 18x) = 0

and with f_1(x) = 7a_1(x) and f_2(x) = 7a_2(x), I have functions that fulfill the conditions above.

The very purpose of the odd construction here is to get functions where the f's are roots of the same monic polynomial with integer coefficients, where just one of the f's equals 0 at x=0, and that has been achieved.

Now notice that at x=1, I have the f's are roots of:

a^2 - 8a + 63 = 0

where it is not allowed in the ring of algebraic integers for either of the roots to have 7 as a factor.

Now consider again the expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

by the proof above the part 9(g_1(x) + 1)(g_2(x) + 2) is blocked from the ring of algebraic integers at x=1, while:

(f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

is STILL in the ring! And notice that you are not able to in general divide off the 9, in the ring of algebraic integers.

But that is not a problem for the field of complex numbers!

The sudden move to a field may be a surprise, but notice I've hit an odd limitation: the ring of algebraic integers provably will not allow me to generally divide off the 9, but I'm curious! I want to do it anyway! So how? Go to the field of complex numbers and there is no block. Dividing off the 9 there gives:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9)/9 = P(x)

where I haven't guessed at how the 9 divides through the middle but of course there are an

And notice that (g_1(x) + 1)(g_2(x) + 2) = P(x) exists in the field of complex numbers!!!

So then, what was the ring of algebraic integers hiding? One might presume that the g's are forced in some way to be fractions but it's not clear why that would be the case! And worse, what might you presume to be in the denominator for any type of fraction?

Oddly enough, if you check against the solution for the f's—you can solve for the f's using the quadratic formula—you would be forced to say, factors in common with 9.

But there is no 9. And I could just as soon use 7, if I wished, and did so for years with such demonstrations.

So the ring of algebraic integers is hiding, what? And how?

The obvious answer is, the ring of algebraic integers is using unit factors, and holding the 9 hostage, one might say:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/9 + u_1)(f_2(x)*u_2 + 9*u_2) = P(x)

where I'm dividing through as I did because f_1(x) = 9*g_1(x), from above.

Some might ask, why unit factors?

The answer is, only unit factors can be generated spontaneously, like 9 isn't a unit factor and had to multiply times P(x) to be a factor, but once I divide 9 off, P(x) only has 1 itself as a factor, so only factors of 1 are available i.e. unit factors.

But now there is a truly odd conclusion that follows: if mathematicians try to divide off 9, from say, the roots of

a^2 - 8a + 63 = 0

what they are getting are the unit factor u_1, which is an algebraic integer but not a unit in that ring, and also 9*u_2, where u_2 is hidden as it's not a member of the ring! Like with 2 and 6 when you only consider evens, you find that 3 must remain wrapped.

So u_2 is a hidden unit, and the coverage problem is revealed.

Why does it have to be a unit again? Because only units can generate spontaneously.

But what is the ring of algebraic integers DOING??!!! Why is it doing this thing here?

The 'why' of the problem is actually fairly simple.

Answering the mystery is as simple as noting a corollary from the ring of integers that is not allowed in the ring of algebraic integers, as note that in the ring of integers, any member of that ring can be paired with a unit as roots of a quadratic, for example:

x^2 + 3x + 2 = (x + 2)(x + 1)

While that is allowed in the ring of algebraic integers, the more general:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is not allowed at all for integer b, if the u's are non-rational. That gives numbers that are properly units which are blocked from being units in that ring.

So why wasn't that noticed over a hundred years ago when mathematicians were testing out the ring of algebraic integers and LOOKING for coverage problems?

Good question.

With a hundred years of naive usage of the ring of algebraic integers a serious amount of resistance has emerged today so that the simple mathematics above is for the most part, years old. I've explained often, trying different ways to explain, as I persist in trying to get acknowledgement of a serious problem.

Some may wonder why it's such a big deal. Well consider one of the tricks mathematicians in number theory often use, which is to use the result that something is not a unit in the ring of algebraic integers to "prove" it has factors like above such a faux proof would involve showing the factors in common with 9 remain for both f's. Any conclusion they get from that wrong argument is not mathematically proven and can actually be false!

Imagine if people DID only use evens, and if you tried to explain to people that 2 can divide from 6 to give 3, and they said that 3 didn't exist, you were crazy and that 2 and 6 were coprime? How might that change our world today?

While mathematicians resist acknowledging this error they can do flawed mathematical arguments, and possibly appear to prove just about anything, which is an odd thing about serious math errors.

That could be enticing for some.

After all, mathematics is VERY hard as a discipline. Can you imagine being a "pure math" mathematician and starting over?

This time to do mathematics that is actually correct?

What if instead you could just play pretend? After all, your entire career has been smoke and mirrors anyway, if it is "pure math" only in areas where this error holds sway.

This post is yet another attempt to help chip away at the resistance. And you know it does amaze me the newer math students who must not wish to achieve greatness in the field, as knowing there is this error, which they can work through themselves as its elementary methods and mostly rather simple algebra, they must know that there is no great proof to be found in actuality if you rely on the error!

So they will never achieve true greatness while under its shadow.

To the extent that they learn in error anyway, they are accepting that they will never make a great discovery—or maybe any real discovery at all—and that could be part of the problem! Possibly deep down many of today's math students don't feel capable of ever making a great discovery anyway.

At least with the error, they can get paid, raise a family maybe. Pay a mortgage.

The problem is similar to what happens if you consider only evens, so 2 and 6 do NOT then have a factor in common because 3 is odd, so it has been left out.

Seeing the more complex problem with the ring of algebraic integers requires only elementary methods using some basic algebra.

I'll need only one proof:

The equation (g_1(x) + 1)(g_2(x) + 2) = P(x), when P(x) is a quadratic with integer coefficients, and g_1(0) = g_2(0) = 0, where the g's are arbitrary functions of x, with a specific case not required for the proof itself, may for certain highly specific cases not exist in the ring of algebraic integers.

Proof:

Proving that statement requires use of a constant factor, so I'll multiply by 9, and now consider the following expression in the ring of algebraic integers:

9(g_1(x) + 1)(g_2(x) + 2) = 9*P(x)

where g_1(0) = g_2(0) = 0, and P(x) is a quadratic with integer coefficients; therefore, trivially, the last coefficient of P(x) is 2.

Then by the distributive property, I have:

(9g_1(x) + 9)(g_2(x) + 2) = 9*P(x)

which must always exist then if the original expression exists. But now consider functions f_1(x), and f_2(x), such that

f_1(x) = 9g_1(x), and f_2(x) = g_2(x) - 7, so:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

where again g_1(0) = g_2(0) = 0, and notice one of the f's equals 0, at x=0, and P(x).

Let f_1(x) and f_2(x) be non-rational roots of the same monic quadratic with integer coefficients—so they are being required to be algebraic integers—then f_1(x) = 9g_1(x), requires one of the f's to have 9 as a factor, while the other must be coprime to 9, since

f_2(x) = g_2(x) - 7

and g_1(0) = g_2(0) = 0.

But it is well-known that in the ring of algebraic integers that just one root of a monic polynomial with integer coefficients irreducible over Q may not have a non-unit rational as a factor if ALL of the roots do not have it as a factor.

Therefore 9 is a factor of NEITHER of the f's in the ring of algebraic integers, and there is a contradiction.

Proof complete.

The result has important implications for the coverage of algebraic integers, as consider the following carefully peculiar way to generate functions.

With P(x) = 441x^2 - 35x + 2, now moving to the ring of algebraic integers, multiplying by 9, and re-grouping terms gives:

9*P(x) = (81x^2 - 18x)7^2 + (9x-1)(9)(7) + 9^2

which allows me now to get a non-polynomial factorization with

9*P(x) = (7a_1(x) + 9)(7a_2(x)+ 9)

where the a's are roots of

a^2 - (9x-1)a + (81x^2 - 18x) = 0

and with f_1(x) = 7a_1(x) and f_2(x) = 7a_2(x), I have functions that fulfill the conditions above.

The very purpose of the odd construction here is to get functions where the f's are roots of the same monic polynomial with integer coefficients, where just one of the f's equals 0 at x=0, and that has been achieved.

Now notice that at x=1, I have the f's are roots of:

a^2 - 8a + 63 = 0

where it is not allowed in the ring of algebraic integers for either of the roots to have 7 as a factor.

Now consider again the expression:

9(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

by the proof above the part 9(g_1(x) + 1)(g_2(x) + 2) is blocked from the ring of algebraic integers at x=1, while:

(f_1(x) + 9)(f_2(x) + 9) = 9*P(x)

is STILL in the ring! And notice that you are not able to in general divide off the 9, in the ring of algebraic integers.

But that is not a problem for the field of complex numbers!

The sudden move to a field may be a surprise, but notice I've hit an odd limitation: the ring of algebraic integers provably will not allow me to generally divide off the 9, but I'm curious! I want to do it anyway! So how? Go to the field of complex numbers and there is no block. Dividing off the 9 there gives:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x) + 9)(f_2(x) + 9)/9 = P(x)

where I haven't guessed at how the 9 divides through the middle but of course there are an

**infinity**of ways to do it in the field.And notice that (g_1(x) + 1)(g_2(x) + 2) = P(x) exists in the field of complex numbers!!!

So then, what was the ring of algebraic integers hiding? One might presume that the g's are forced in some way to be fractions but it's not clear why that would be the case! And worse, what might you presume to be in the denominator for any type of fraction?

Oddly enough, if you check against the solution for the f's—you can solve for the f's using the quadratic formula—you would be forced to say, factors in common with 9.

But there is no 9. And I could just as soon use 7, if I wished, and did so for years with such demonstrations.

So the ring of algebraic integers is hiding, what? And how?

The obvious answer is, the ring of algebraic integers is using unit factors, and holding the 9 hostage, one might say:

(g_1(x) + 1)(g_2(x) + 2) = (f_1(x)*u_1/9 + u_1)(f_2(x)*u_2 + 9*u_2) = P(x)

where I'm dividing through as I did because f_1(x) = 9*g_1(x), from above.

Some might ask, why unit factors?

The answer is, only unit factors can be generated spontaneously, like 9 isn't a unit factor and had to multiply times P(x) to be a factor, but once I divide 9 off, P(x) only has 1 itself as a factor, so only factors of 1 are available i.e. unit factors.

But now there is a truly odd conclusion that follows: if mathematicians try to divide off 9, from say, the roots of

a^2 - 8a + 63 = 0

what they are getting are the unit factor u_1, which is an algebraic integer but not a unit in that ring, and also 9*u_2, where u_2 is hidden as it's not a member of the ring! Like with 2 and 6 when you only consider evens, you find that 3 must remain wrapped.

So u_2 is a hidden unit, and the coverage problem is revealed.

Why does it have to be a unit again? Because only units can generate spontaneously.

But what is the ring of algebraic integers DOING??!!! Why is it doing this thing here?

The 'why' of the problem is actually fairly simple.

Answering the mystery is as simple as noting a corollary from the ring of integers that is not allowed in the ring of algebraic integers, as note that in the ring of integers, any member of that ring can be paired with a unit as roots of a quadratic, for example:

x^2 + 3x + 2 = (x + 2)(x + 1)

While that is allowed in the ring of algebraic integers, the more general:

x^2 + bx + 2 = (x + 2u_1)(x + u_2)

is not allowed at all for integer b, if the u's are non-rational. That gives numbers that are properly units which are blocked from being units in that ring.

So why wasn't that noticed over a hundred years ago when mathematicians were testing out the ring of algebraic integers and LOOKING for coverage problems?

Good question.

With a hundred years of naive usage of the ring of algebraic integers a serious amount of resistance has emerged today so that the simple mathematics above is for the most part, years old. I've explained often, trying different ways to explain, as I persist in trying to get acknowledgement of a serious problem.

Some may wonder why it's such a big deal. Well consider one of the tricks mathematicians in number theory often use, which is to use the result that something is not a unit in the ring of algebraic integers to "prove" it has factors like above such a faux proof would involve showing the factors in common with 9 remain for both f's. Any conclusion they get from that wrong argument is not mathematically proven and can actually be false!

Imagine if people DID only use evens, and if you tried to explain to people that 2 can divide from 6 to give 3, and they said that 3 didn't exist, you were crazy and that 2 and 6 were coprime? How might that change our world today?

While mathematicians resist acknowledging this error they can do flawed mathematical arguments, and possibly appear to prove just about anything, which is an odd thing about serious math errors.

That could be enticing for some.

After all, mathematics is VERY hard as a discipline. Can you imagine being a "pure math" mathematician and starting over?

This time to do mathematics that is actually correct?

What if instead you could just play pretend? After all, your entire career has been smoke and mirrors anyway, if it is "pure math" only in areas where this error holds sway.

This post is yet another attempt to help chip away at the resistance. And you know it does amaze me the newer math students who must not wish to achieve greatness in the field, as knowing there is this error, which they can work through themselves as its elementary methods and mostly rather simple algebra, they must know that there is no great proof to be found in actuality if you rely on the error!

So they will never achieve true greatness while under its shadow.

To the extent that they learn in error anyway, they are accepting that they will never make a great discovery—or maybe any real discovery at all—and that could be part of the problem! Possibly deep down many of today's math students don't feel capable of ever making a great discovery anyway.

At least with the error, they can get paid, raise a family maybe. Pay a mortgage.