MHB Polynomials Over a Field - Rotman, Lemma 3.87

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.6 Unique Factorization ...

I need help with an aspect of the proof of Lemma 3.87 ...

The relevant text from Rotman's book is as follows:
View attachment 4653
In the proof of Lemma 3.87 above, we read the following:

" ... ... By induction, there are $$d_j(x) \in k[x]$$ with each $$d_j(x) = 0$$ or $$deg(d_j) \lt deg(b)$$, such that

$$g(x) = d_m b^{m-1} + \ ... \ + d_2 b = d_1$$ ... ... "Can someone help me with the proof of this statement ... that is, how to set up for induction in this case, and work through the proof by induction ...

Help with this matter will be much appreciated ...

Peter

 

[Deveno]

Suppose $f(x) = a$, a constant. Then we may take $d_j(x) = 0$ for every $j > 0$, and $d_0(x) = a = f(x)$.

Since $\text{deg}(d_0) = 0$ and $\text{deg}(b) \geq 1$, the theorem's hypotheses are satisfied, in this case. This is our "base" case, inducting on the degree of $f$.

With our inductive hypothesis, we assume the theorem holds for any polynomial $g$ with $0 \leq \text{deg}(g) < \text{deg}(f)$.

If $\text{deg}(b) < \text{deg}(f)$, dividing $f$ by $b$ yields such a $g$. Note if $\text{deg}(b) >\text{deg}(f)$, we can simply take $d_j(x) = 0$ for $j > 0$, and $d_0(x) = f(x)$, as in the constant case.

The case $\text{deg}(b) = \text{deg}(f)$ is slightly more interesting. Note that since $\deg{b} \geq 1$, this is true (in this case of equal degrees of $b$ and $f$) for $f$. In this case, we take $d_j(x) = 0$, for all $j > 1$.

If the leading coefficient of $b$ is, say $a_0$, and the leading coefficient of $f$ is $c_0$, we may take $d_1(x) = \dfrac{c_0}{a_0}$, which has degree 0, which is less than that of $b$.

Then $f(x) - d_1(x)b(x)$ is either the 0-polynomial, or has degree < the degree of $f$ = the degree of $b$, and we may take $d_0(x) = f(x) - d_1(x)b(x)$, which gives:

$f(x) = d_1(x)b(x) + d_0(x)$.

Otherwise, we are down to the case where $\text{deg}(b) < \text{deg}(f)$.

In this case, we have $f = gb + r$, where $\text{deg}(r) < \text{deg}(b)$, or $r = 0$. I leave the $r = 0$ case to you (hint: we can do it in one term).

So, by our inductive hypothesis, we have:

$g(x) = d'_{m-1}(x)b(x)^{m-1} +\cdots + d'_1(x)b(x) + d'_0$.

(the degree of $g$ is at most one less than the degree of $f$, hence the "$m-1$"-the coefficients may actually stop earlier, we may have $d_j = 0$ for all $j > k$ for some $0 \leq k \leq m-1$).

Let $d_j(x) = d'_{j-1}(x)$, and $d_0(x) = r(x)$.

Then $f(x) = g(x)b(x) + r(x) = (d'_{m-1}(x)b(x)^{m-1} + \cdots + d'_1(x)b(x) + d'_0)b(x) + r(x)$

$= d'_{m-1}b(x)^m + \cdots + d'_1(x)b(x)^2 + d'_0b(x) + r(x)$

$= d_m(x)b(x)^m + \cdots + d_2(x)b(x)^2 + d_1(x)b(x) + d_0(x)$.

So if the theorem holds for any polynomial of degree less than $f$, it holds for $f$ as well, and thus for ALL $f$.

 

[Math Amateur]

Suppose $f(x) = a$, a constant. Then we may take $d_j(x) = 0$ for every $j > 0$, and $d_0(x) = a = f(x)$.

Since $\text{deg}(d_0) = 0$ and $\text{deg}(b) \geq 1$, the theorem's hypotheses are satisfied, in this case. This is our "base" case, inducting on the degree of $f$.

With our inductive hypothesis, we assume the theorem holds for any polynomial $g$ with $0 \leq \text{deg}(g) < \text{deg}(f)$.

If $\text{deg}(b) < \text{deg}(f)$, dividing $f$ by $b$ yields such a $g$. Note if $\text{deg}(b) >\text{deg}(f)$, we can simply take $d_j(x) = 0$ for $j > 0$, and $d_0(x) = f(x)$, as in the constant case.

The case $\text{deg}(b) = \text{deg}(f)$ is slightly more interesting. Note that since $\deg{b} \geq 1$, this is true (in this case of equal degrees of $b$ and $f$) for $f$. In this case, we take $d_j(x) = 0$, for all $j > 1$.

If the leading coefficient of $b$ is, say $a_0$, and the leading coefficient of $f$ is $c_0$, we may take $d_1(x) = \dfrac{c_0}{a_0}$, which has degree 0, which is less than that of $b$.

Then $f(x) - d_1(x)b(x)$ is either the 0-polynomial, or has degree < the degree of $f$ = the degree of $b$, and we may take $d_0(x) = f(x) - d_1(x)b(x)$, which gives:

$f(x) = d_1(x)b(x) + d_0(x)$.

Otherwise, we are down to the case where $\text{deg}(b) < \text{deg}(f)$.

In this case, we have $f = gb + r$, where $\text{deg}(r) < \text{deg}(b)$, or $r = 0$. I leave the $r = 0$ case to you (hint: we can do it in one term).

So, by our inductive hypothesis, we have:

$g(x) = d'_{m-1}(x)b(x)^{m-1} +\cdots + d'_1(x)b(x) + d'_0$.

(the degree of $g$ is at most one less than the degree of $f$, hence the "$m-1$"-the coefficients may actually stop earlier, we may have $d_j = 0$ for all $j > k$ for some $0 \leq k \leq m-1$).

Let $d_j(x) = d'_{j-1}(x)$, and $d_0(x) = r(x)$.

Then $f(x) = g(x)b(x) + r(x) = (d'_{m-1}(x)b(x)^{m-1} + \cdots + d'_1(x)b(x) + d'_0)b(x) + r(x)$

$= d'_{m-1}b(x)^m + \cdots + d'_1(x)b(x)^2 + d'_0b(x) + r(x)$

$= d_m(x)b(x)^m + \cdots + d_2(x)b(x)^2 + d_1(x)b(x) + d_0(x)$.

So if the theorem holds for any polynomial of degree less than $f$, it holds for $f$ as well, and thus for ALL $f$.

Hi Deveno,

Thanks for a really clear post ... extremely helpful!

Another example of induction for me as well ... a case of complete or strong induction, I think ...

Thanks again for your support ...

Peter

 

[Deveno]

Yes, it would be strong induction, because unless $b$ is of degree $1$, we go more than "one notch down".