Question

1. Let R be an integral domain and f(x), g(x)$\in$R[x] . Assume that the leading coefficient of g(x) is a unit in R. Verify that the Division Algorithm holds for f(x) as dividend and g(x) as divisor.

1. Give an example in$\mathrm{\mathbb{Z}}\left[x\right]$ to show that part (a) may be false if the leading coefficient of g(x) is not a unit.

Short answer

1. It is proved that the division algorithm holds for dividend f(x) and divisor g(x) .
2. It is proved that the division algorithm is invalid if the leading coefficient of g(x) is not a unit.

Step-by-step solution

Polynomial Arithmetic 

If any given function R[x] is a ring, then the commutative, associative, and distributive laws hold such that the function f(x)+g(x) exists.

Polynomial Operations

(a)

When deg f(x)$<$deg g(x), then we have:

f(x)=0g(x)+f(x)

Now, according to theorem 4.6, if deg f(x)=0$\Rightarrow$deg g(x)=0 , then we must have elements such that:

$$\begin{gathered} f\left(x\right)=a\in R \\ g\left(x\right)=b\in R \end{gathered}$$

Then, due to the leading coefficient of $g\left(x\right)$, we have b as a unit such that:

$a=\left(a{b}^{-1}\right)b+0$

Similarly, when $degf\left(x\right)>0$, we have:

$$\begin{gathered} f\left(x\right)=\sum _{i=0}^m{a}_i{x}^i \\ g\left(x\right)=\sum _{i=0}^m{b}_j{x}^j \end{gathered}$$

Then, due to the leading coefficient of g(x), we have b_n as a unit such that:

$f\left(x\right)-a_m{b}_{n}g\left(x\right)x^{m-n}$

Hence proved, division algorithm holds for dividend f(x) and g(x) divisor .

Polynomial Operations

(b)

The polynomials x⁴ and 2x² in $\mathrm{\mathbb{Z}}\left[x\right]$ could not follow division algorithm because 2x² does not divide $x^4$.

There will always be some quotient q(x) and remainder r(x) , such that:

f(x) = g(x)q(x)+r(x)

Now, in case of x⁴ and 2x² , 2 is not a unit in $\mathrm{\mathbb{Z}}$.

Hence proved, division algorithm is invalid if leading coefficient of g(x) is not a unit.