Polynomial Rings _ Bland - Theorem 6.3.17

[Math Amateur]

I am reading Paul E. Bland's book, "The Basics of Abstract Algebra".

I am currently focused on Chapter 6: Polynomial Rings.

I need help with an aspect of Theorem 6.3.17.

Theorem 6.3.17 requires awareness of the notation of Definition 6.3.15 which reads as follows:
https://www.physicsforums.com/attachments/4690Theorem 6.3.17 reads as follows:https://www.physicsforums.com/attachments/4691In the above Theorem we read the following:" ... ... and it is not difficult to show that

\(\displaystyle [f]_p (x) = [g]_p (x) [h]_p (x)\) ... ... "Although it seems plausible that \(\displaystyle [f]_p (x) = [g]_p (x) [h]_p (x)\), I am unable to rigorously demonstrate that this is the case.

Can someone help me to prove that \(\displaystyle [f]_p (x) = [g]_p (x) [h]_p (x)\)?

Hope someone can help ... ...

Peter

 

[Taryn1]

Wow. What kind of math IS this? I'm only barely in pre-calc.

 

[Deveno]

Note that $a \mapsto [a]_p$ is a ring-homomorphism $\Bbb Z \to \Bbb Z/p\Bbb Z$.

Given $g(x) = \sum\limits_{i = 0}^m a_ix^i \in \Bbb Z[x]$, and $h(x) = \sum\limits_{i = 0}^n b_ix^i \in \Bbb Z[x]$

we have:

$f(x) = g(x)h(x) = \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} a_ib_jx^{i+j}\right)$

thus:

$[g]_p(x)[h]_p(x) = \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} [a_i]_p[b_j]_px^{i+j}\right)$

$= \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} [a_ib_j]_px^{i+j}\right)$

$= \sum\limits_{k = 0}^{m+n}\left(\left[\sum\limits_{i+j=k} a_ib_j\right]_px^k\right) = [f]_p(x)$

For example, we have in $\Bbb Z[x]$:

$x^3 - 12x^2 + 26x + 45 = (x - 5)(x^2 - 7x - 9)$.

Reducing the cubic mod $3$, we obtain:

$x^3 + 2x$, and reducing the factors mod $3$, we have $x + 1$ and $x^2 + 2x$, and:

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

In other words, polynomials "mod $p$" and polynomials over the integers share many of the same algebraic rules.

 

[Taryn1]

Wow.

 

[Deveno]

Wow. What kind of math IS this? I'm only barely in pre-calc.

Surely you have heard of polynomials, even in pre-calc, yes?

One of the main goals of working with polynomials is to try to "factor" them. This is sort of like factoring positive integers into primes. Ideally, one tries to totally "split" the polynomial into linear (degree 1) factors, because that tells you the roots.

For example, take $x^2 - 5x + 6 = (x - 2)(x - 3)$.

The factorization on the right tells us if:

$x^2 - 5x + 6 = 0$, then either $x - 2 = 0 \iff x = 2$ or $x - 3 = 0 \iff x = 3$.

However, sometimes you can't do this, for example, any attempt to try to factor (in the integers):

$x^2 + x + 1$ will fail (it turns out the roots aren't even real numbers, much less integers).

Such an "unfactorable" polynomial is the polynomial equivalent (for INTEGER polynomials) of a prime number, and is called "irreducible" (over the integers-if we allow more "complicated kinds of numbers", it might be factorable over those things).

In the middle ages, "solving" (factoring) polynomials was all the rage, and mathematicians actually held sorts of "mathematical duels" in the form of "I bet you can't factor THIS!". In the mid 19th century, a bright young upstart with the wrong politics, and a habit of getting himself into trouble with the authorities, Evariste Galois, studied how "shuffling the roots" of a polynomial affected the "number structure" the roots lived in. This was one of the beginnings of what we now called "group theory", and which now forms part of the "study of algebraic structures" known as "abstract algebra".

Galois' work laid the groundwork for what is known as the Abel-Ruffini Theorem: factoring polynomials of degree 5 and higher, is by and large, a lost cause (we can factor SOME, but there is no analogue of the "quadratic formula" for polynomials of degree 5 and up. The formula for a degree 4 polynomial is ugly, and chock-full of square and cube roots).

 

[Math Amateur]

Wow. What kind of math IS this? I'm only barely in pre-calc.

Hi Taryn, it is Abstract Algebra, a truly beautiful subject ... very rigorous and pure ...

My question is very definitely undergraduate ... not postgraduate ... I have to say ...

Peter

- - - Updated - - -

Note that $a \mapsto [a]_p$ is a ring-homomorphism $\Bbb Z \to \Bbb Z/p\Bbb Z$.

Given $g(x) = \sum\limits_{i = 0}^m a_ix^i \in \Bbb Z[x]$, and $h(x) = \sum\limits_{i = 0}^n b_ix^i \in \Bbb Z[x]$

we have:

$f(x) = g(x)h(x) = \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} a_ib_jx^{i+j}\right)$

thus:

$[g]_p(x)[h]_p(x) = \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} [a_i]_p[b_j]_px^{i+j}\right)$

$= \sum\limits_{k = 0}^{m+n}\left(\sum\limits_{i+j = k} [a_ib_j]_px^{i+j}\right)$

$= \sum\limits_{k = 0}^{m+n}\left(\left[\sum\limits_{i+j=k} a_ib_j\right]_px^k\right) = [f]_p(x)$

For example, we have in $\Bbb Z[x]$:

$x^3 - 12x^2 + 26x + 45 = (x - 5)(x^2 - 7x - 9)$.

Reducing the cubic mod $3$, we obtain:

$x^3 + 2x$, and reducing the factors mod $3$, we have $x + 1$ and $x^2 + 2x$, and:

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

In other words, polynomials "mod $p$" and polynomials over the integers share many of the same algebraic rules.

Thanks for the help Deveno ...

Just reflecting on what you have written ...

Your support is invaluable ...

Thanks again,

Peter[PS you definitely should be a university teacher ... ]

 

[Taryn1]

Thanks Devono! And yeah, I use polynomials. Just not on that level!

 

[Deveno]

Thanks Devono! And yeah, I use polynomials. Just not on that level!

Well, it's not so different from solving "regular algebra".

For example, to solve:

$2x + 5 = 13$, one does the following:$2x + 5 + (-5) = 13 + (-5)$
$2x + (5 + (-5)) = 13 + (-5)$
$2x + 0 = 13 - 5$
$2x = 8$
$\frac{1}{2}(2x) = \frac{1}{2}(8)$
$(\frac{1}{2}(2))x = 4$
$1x = 4$
$x = 4$.

A more "polynomial approach" would be:

$2x +5 = 13 \implies 2x - 8 = 0$.

$2x - 8 = 2(x - 4) = 0 \implies 2 = 0$ or $x - 4 = 0$.

Since $2 \neq 0$, it must be that $x - 4 = 0 \implies x = 4$.

In "abstract" algebra, one forgets about the "numbers" part (the $2,5$ and $13$ in my example), and just focuses on the "rules" part. It turns out there's a lot you can say just from "knowing the rules" even when you don't know what "things" you have.

 

[Taryn1]

Cool, thanks. It makes a little more sense now. (Whew)