MHB Rotman - Theorem 2.60 - Unique Factorization

[Math Amateur]

I am reading Chapter 2: Commutative Rings in Joseph Rotman's book, Advanced Modern Algebra (Second Edition).

I am currently focussed on Theorem 2.60 (Unique Factorization) [pages 111 - 112].

I need help to understand Rotman's use of induction (or his induction strategy) in the proof of Theorem 2.60.

Theorem 2.60 and its proof read as follows:

View attachment 2688
View attachment 2689As shown above Rotman says he is going to prove the existence of a factorization for a polynomial f(x) by induction on $$ deg(f) \ge 1$$.

I presume from reading the proof that Rotman intends to show the theorem is true for deg (f) = 1 and then assume the theorem is true for deg (f) = n and then prove that given it is true for deg (f) = n show it is then true for deg (f) = n + 1. But he does not seem to do this?

Rotman demonstrates the theorem is true for deg (f) = 1, for sure ... but then ... ?

Essentially he points out that if f(x) is irreducible we are done - fine!

Then he takes the case of f(x) not irreducible pointing out that in this case f(x) = g(x)h(x) where deg (g) < deg (f) and deg (h) < deg (f) - OK fine!

BUT ... then he states:

"By the inductive hypothesis, there are factorizations

$$ g(x) = b p_1(x) ... \ ... p_m(x) $$

and

$$ h(x) = c q_1(x) ... \ ... q_n(x) $$

where $$ b, c \in k $$ and the p's and q's are monic irreducibles."

But what exactly is the "induction hypothesis"? How exactly is Rotman following the principle of induction?

Can someone please clarify this for me?

Peter

 

[Deveno]

Rotman is using a form of induction called variously, "complete induction" or "strong induction".

Normal induction argument:

$P(n_0)$

If $P(n)$, then $P(n+1)$.

Therefore, for all $n \geq n_0 \in \Bbb N,\ P(n)$.

Strong form of induction:

$P(n_0)$.

If for all $n_0 \leq k < n,\ P(k)$, then $P(n)$.

Therefore, for all $n \geq n_0 \in \Bbb N,\ P(n)$.

In other words, we assume that for any polynomial of degree less than $\text{deg}(f)$, we have unique factorization.

If $f$ is already irreducible, the factorization is complete. If not, say $f(x) = g(x)h(x)$, where neither $g$ nor $h$ is a unit, then they are necessarily of lower degree, and we apply the induction hypothesis, since:

$\text{deg}(g),\text{deg}(h) < \text{deg}(f)$.

Note this only shows "a" factorization of $f$ into irreducible factors. However, uniqueness then follows since in an integral domain, irreducibles are prime. More generally in a UFD (and what we are doing is proving $k[x]$ IS a UFD):

Irreducible = prime.

In fact, $k[x]$ is EVEN BETTER than a UFD, it is a Euclidean domain (the degree function works as our "$d$ function").

************

The arithmetic we use in our Arabic numeral system is BASED on the homomorphism:

$\Bbb Z[x] \to \Bbb Z$ given by: $f(x) \mapsto f(10)$, where the image of $x^n$ is determined by POSITION (or "placeholder 0's" for powers of $x$ with 0 coefficients).

This is really no different than regarding an integral polynomial as a sequence of integers:

$1 + 2x + x^2 \mapsto (1,2,1,0,0,0,\dots)$

The only "unusual" aspect of this is the way we "collect like terms" (which essentially defines the multiplication and ensures the distributive law is satisfied) when we multiply.

I point this out, to highlight the DEEP connections between polynomials and integers.

There are basically two minimal ways to create a field:

1. Start with 1.
2a. If $n\cdot 1 = 0$, for some $n$, we get the prime field ${}GF_p$.

2b. If not, we get an infinite cyclic group, isomorphic to $\Bbb Z$, and thus whose field of fractions: $(\Bbb Z^{\ast})^{-1}\Bbb Z$ is isomorphic to $\Bbb Q$.

So we wind up with "basic" polynomial rings of either $\Bbb Z_p[x]$ or $\Bbb Q[x]$, and the latter turns out to be decomposable into:

$\Bbb Q \times \Bbb Z[x]$

(this is essentially what Gauss' lemma says).

Polynomials in $\Bbb Z[x]$ are like "integers in base $x$", where $x$ is an "infinitely large integer". Their arithmetic is the same laborious methods we learned in elementary school. As with integers, to understand arbitrary polynomials, we first factor them into primes.

There is a parallel process:

$\Bbb Z \to \Bbb Z/(p)$ and:

$\Bbb k[x] \to k[x]/(p(x))$

whereby we create a field out of a Euclidean domain, by modding out a prime (and thus maximal) ideal. This idea is CENTRAL to large swaths of mathematics.

 

[Math Amateur]

Rotman is using a form of induction called variously, "complete induction" or "strong induction".

Normal induction argument:

$P(n_0)$

If $P(n)$, then $P(n+1)$.

Therefore, for all $n \geq n_0 \in \Bbb N,\ P(n)$.

Strong form of induction:

$P(n_0)$.

If for all $n_0 \leq k < n,\ P(k)$, then $P(n)$.

Therefore, for all $n \geq n_0 \in \Bbb N,\ P(n)$.

In other words, we assume that for any polynomial of degree less than $\text{deg}(f)$, we have unique factorization.

If $f$ is already irreducible, the factorization is complete. If not, say $f(x) = g(x)h(x)$, where neither $g$ nor $h$ is a unit, then they are necessarily of lower degree, and we apply the induction hypothesis, since:

$\text{deg}(g),\text{deg}(h) < \text{deg}(f)$.

Note this only shows "a" factorization of $f$ into irreducible factors. However, uniqueness then follows since in an integral domain, irreducibles are prime. More generally in a UFD (and what we are doing is proving $k[x]$ IS a UFD):

Irreducible = prime.

In fact, $k[x]$ is EVEN BETTER than a UFD, it is a Euclidean domain (the degree function works as our "$d$ function").

************

The arithmetic we use in our Arabic numeral system is BASED on the homomorphism:

$\Bbb Z[x] \to \Bbb Z$ given by: $f(x) \mapsto f(10)$, where the image of $x^n$ is determined by POSITION (or "placeholder 0's" for powers of $x$ with 0 coefficients).

This is really no different than regarding an integral polynomial as a sequence of integers:

$1 + 2x + x^2 \mapsto (1,2,1,0,0,0,\dots)$

The only "unusual" aspect of this is the way we "collect like terms" (which essentially defines the multiplication and ensures the distributive law is satisfied) when we multiply.

I point this out, to highlight the DEEP connections between polynomials and integers.

There are basically two minimal ways to create a field:

1. Start with 1.
2a. If $n\cdot 1 = 0$, for some $n$, we get the prime field ${}GF_p$.

2b. If not, we get an infinite cyclic group, isomorphic to $\Bbb Z$, and thus whose field of fractions: $(\Bbb Z^{\ast})^{-1}\Bbb Z$ is isomorphic to $\Bbb Q$.

So we wind up with "basic" polynomial rings of either $\Bbb Z_p[x]$ or $\Bbb Q[x]$, and the latter turns out to be decomposable into:

$\Bbb Q \times \Bbb Z[x]$

(this is essentially what Gauss' lemma says).

Polynomials in $\Bbb Z[x]$ are like "integers in base $x$", where $x$ is an "infinitely large integer". Their arithmetic is the same laborious methods we learned in elementary school. As with integers, to understand arbitrary polynomials, we first factor them into primes.

There is a parallel process:

$\Bbb Z \to \Bbb Z/(p)$ and:

$\Bbb k[x] \to k[x]/(p(x))$

whereby we create a field out of a Euclidean domain, by modding out a prime (and thus maximal) ideal. This idea is CENTRAL to large swaths of mathematics.

Hi Deveno,

You write:

"If $f$ is already irreducible, the factorization is complete. If not, say $f(x) = g(x)h(x)$, where neither $g$ nor $h$ is a unit, then they are necessarily of lower degree, and we apply the induction hypothesis, since:

$\text{deg}(g),\text{deg}(h) < \text{deg}(f)$."

I had reflected on my post and come to this conclusion after reading on the Principle of Induction - specifically Hungerford (Abstract Algebra: An Introduction) Appendix C where he calls the strong form of induction, The Principle of Complete Induction (page 522) ... but you beat me to it ...

BUT ... ... you have also provided me with some further ideas which I am now reflecting on ... ... most helpful ...

Thank you for your support and help ... ...

Peter

 

[Math Amateur]

Hi Deveno,

You write:

"If $f$ is already irreducible, the factorization is complete. If not, say $f(x) = g(x)h(x)$, where neither $g$ nor $h$ is a unit, then they are necessarily of lower degree, and we apply the induction hypothesis, since:

$\text{deg}(g),\text{deg}(h) < \text{deg}(f)$."

I had reflected on my post and come to this conclusion after reading on the Principle of Induction - specifically Hungerford (Abstract Algebra: An Introduction) Appendix C where he calls the strong form of induction, The Principle of Complete Induction (page 522) ... but you beat me to it ...

BUT ... ... you have also provided me with some further ideas which I am now reflecting on ... ... most helpful ...

Thank you for your support and help ... ...

Peter

I now have a further two questions regarding the second part of the proof!


Question 1

Rotman says his proof is based on induction on $$M = max \{ m,n \} \ge 1$$.

Rotman shows that the 'base step" $$M = 1$$ holds i.e, we have uniqueness i.e. $$ p_1(x) = q_1(x) $$.

Now, assuming he is using strong induction, we now assume uniqueness holds for $$ 1 \leq M \lt t $$ and need to show that uniqueness holds for $$M = t$$.

Now at this point Rotman writes:

"For the inductive step, the given equation shows that $$ p_m(x) \ | \ q_1(x) ... \ ... \ q_n(x) $$.

My question is as follows:

How does this follow? That is how/why can we conclude that $$ p_m(x) \ | \ q_1(x) ... \ ... \ q_n(x) $$?



Question 2

On page 112 after the proof of Theorem 2.60 Rotman writes:

"Thereom 2.60 shows that if $$ deg(f) \geq 1 $$ then f has factorizations; moreover, if all the exponents $$ e_i \gt 0 $$, then the factors in this prime factorization are unique."

My question is as follows: why does the uniqueness of the factors depend on the exponents of the$$ p_i(x)$$ being greater than zero - it does not seem to make sense - if the exponent was zero then the polynomial would not be a factor?

Can someone please clarify this issue for me?Help would be very much appreciated.

Peter***EDIT regarding Question 2*** I should have noted that Rotman writes further to my quote in Question 2:

"The statement of Proposition 2.61 below illustrates the convenience of allowing some $$e_i = 0$$"

I am now investigating Proposition 2.61!

 

[Deveno]

I think his writing style is a bit unclear here.

If we have an equation:

$ap_1(x)\cdots p_m(x) = bq_1(x)\cdots q_n(x)$

then $p_m(x)$ divides the left-hand side, so clearly divides the right.

Recall that a prime element of a ring is $p$ such that:

$p|ab$ implies $p|a$ or $p|b$.

Applying that to this theorem we have:

$p_m(x)|b$ or $p_m(x)|q_1(x)\cdots q_n(x)$.

The first condition is impossible, since $p_m(x)$ has degree $\geq$ 1, while $b$ has degree 0.

Hence $p_m$ divides the product of $q$'s, and thus must divide some single $q_i(x)$, by repeatedly applying the definition of prime (possibly up to $n-1$ times).

It really doesn't matter which $q_i$ that $p_m$ divides, we wind up with an equation:

$ap_1\cdots p_m(x) = (bk)q_1(x)\cdots q_{i-1}(x)q_{i+1}(x)\cdots q_n(x)p_m(x)$

and we have a new equation (by cancelling the factor $p_m(x)$ from both sides) where we have the maximum degree of both sides is less than $M$. NOW we apply our induction hypothesis.

Note that since all the $p$'s and $q$'s are monic, the factor $k$ must be 1.

A lot of details in this proof, come from the fact that units "gum up the picture": for any polynomial $f(x)$ we have the distinct factorizations, for any unit $u$:

$f(x) = 1f(x)$

$f(x) = u(u^{-1}f(x))$

Insisting the factors be monic, is akin to insisting the prime integer factors of an integer be greater than 1. This avoids such ambiguities such as:

$6 = -2 \ast (-3)$
$6 = (-1) \ast (-1) \ast (2) \ast (3)$

which are, clearly, two distinct factorizations of 6.

 

[Math Amateur]

I think his writing style is a bit unclear here.

If we have an equation:

$ap_1(x)\cdots p_m(x) = bq_1(x)\cdots q_n(x)$

then $p_m(x)$ divides the left-hand side, so clearly divides the right.

Recall that a prime element of a ring is $p$ such that:

$p|ab$ implies $p|a$ or $p|b$.

Applying that to this theorem we have:

$p_m(x)|b$ or $p_m(x)|q_1(x)\cdots q_n(x)$.

The first condition is impossible, since $p_m(x)$ has degree $\geq$ 1, while $b$ has degree 0.

Hence $p_m$ divides the product of $q$'s, and thus must divide some single $q_i(x)$, by repeatedly applying the definition of prime (possibly up to $n-1$ times).

It really doesn't matter which $q_i$ that $p_m$ divides, we wind up with an equation:

$ap_1\cdots p_m(x) = (bk)q_1(x)\cdots q_{i-1}(x)q_{i+1}(x)\cdots q_n(x)p_m(x)$

and we have a new equation (by cancelling the factor $p_m(x)$ from both sides) where we have the maximum degree of both sides is less than $M$. NOW we apply our induction hypothesis.

Note that since all the $p$'s and $q$'s are monic, the factor $k$ must be 1.

A lot of details in this proof, come from the fact that units "gum up the picture": for any polynomial $f(x)$ we have the distinct factorizations, for any unit $u$:

$f(x) = 1f(x)$

$f(x) = u(u^{-1}f(x))$

Insisting the factors be monic, is akin to insisting the prime integer factors of an integer be greater than 1. This avoids such ambiguities such as:

$6 = -2 \ast (-3)$
$6 = (-1) \ast (-1) \ast (2) \ast (3)$

which are, clearly, two distinct factorizations of 6.

Thanks so much Deveno ... Most supportive and helpful post ... Enables me to go forward in my revision of ring theory ...

Thanks again,

Peter