Sunday, November 09, 2003

 

Factorization example, end of argument

Here's an example that shows the problem with the ring of algebraic integers.

Consider

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

where the c's should be algebraic integers. Notice that only two of the c's can have 7 as a factor.

Now then, you have as a zero of the factorization x = -7/c_1, so let x= -7/c_1, so you have

49(-7^3/c_1^3 + 5(49)/c_1^2 - 21/c_1 + 2) = 0

which is

2c_1^3 - 21 c_1^2 + 245 c_1 - 343 = 0.

But that is a non-monic primitive irreducible over Q, so c_1 and by symmetry c_2 cannot be algebraic integers. However they must be algebraic numbers, and it can be shown that any algebraic number can be written as the ratio of algebraic integers.

So then there must exist f, such that fc_1 is an algebraic integer, and letting g = fc_1 and multiplying both sides by f, I have

(gx + 7f)(c_2 x + 7)( c_3 x + 2) = 49f(x^3 + 5x^2 + 3x + 2)

so a zero is now x = -7f/g, which gives

49f(-7^3f^3/g^3 + 5(49)f^2/g^2 - 21f/g + 2) = 0

which is

2g^3 - 21f g^2 + 245f^2 g - 343 f^3 = 0

which proves that f must have 2 itself as a factor for g to be an algebraic integer.

But looking back again at

(c_1 x + 7)(c_2 x + 7)( c_3 x + 2) = 49(x^3 + 5x^2 + 3x + 2)

that would mean that c_3 has a factor that is 2, which can distribute to c_1 x + 7, but that leaves c_2 x + 7, which also needs a factor of 2 by symmetry.

But you see, you only have just that one 2.

Now then, are mathematicians and math groupies rational, or are you cranks?

Will you give up when you see the truth, or will you keep fighting as if mathematics is just a joke to you, as if you never really cared whether or not it was true, as long as everyone you cared about agreed with you?

What's important to you sci.math newsgroup?





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