Question

(Degree Rule) LetDbe an integral domain andf(x), g(x)∈D[x]. Prove that deg(f(x)·g(x)) = degf(x) + degg(x). Show, by example, that for commutative ringRit is possible that degf(x)g(x)

Answer 1

Complete Question

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Answer:

From the question we are told that

Let D be an integral domain  and f(x),g(x)∈ D[x]

We are to prove that

if f(x),g(x)∈ D[x] then deg(f(x)g(x)) = deg f(x) + deg  g (x)

to do this we need to show that the commutative ring R is possible. That  f(x) and g(x) are non-zero elements in R[x], deg f(x)g(x) < g=deg f(x) + deg g(x)  

\

Let say that g(x) = $b_qx^q +b_{q-1}^{q-1}+...+b_1x +b_0$ with deg(g(x)) = q

f(x) = $a_nx^n +a_{n-1}^{n-1} +...+a_1x +a_0$ with its deg(f(x)) =n

What this means is that $a_n$ and $b_q$ are not non-zero coefficients

That to say  $a_n \neq 0$ and $b_q \neq 0$

Looking at the product of the two function

f(x)g(x) =$(a_nx^n+a_{n-1}x^{n-1}+...+a_1x +a_0)(b_qx^q+b_{q-1}x^{q-1}+...+b_1x+b_0)$

           =$a_nb_qx^{n+q}+...+a_0b_0$

Looking at the above equation in terms degree

  de(f(x)g(x)) = $deg(a_nb_qx^{n+q}+...+a_0b_0)$

                    = n+q

                   = $deg(f(x))+deg(g(x))$

Looking at the above equation we have proven that

                      deg(f(x)g(x)) = deg f(x) + deg  g (x)

Now considering this example

 f(x) = $3x^2$  ∈ $R_2[x]$  and g(x) = $3x^2$ ∈  $R_2[x]$  

Notice that f(x) = g(x) $\neq$ 0

Let take a closer look at their product

   f(x) g(x) = ($4x^2$ ).($4x^2$ )

                = $16x^4$

    To obtain their degree we input mod (power in this case 4)

               = $0x^4$

               = 0

So the degree of the polynomial is 0

Since the both polynomial are the same then

      deg(f(x)) =  deg(g(x))

                     = deg ($2x^2$)

                     = 2

and we know from our calculation that the

       deg(f(x)g(x)) = deg (0)

                           = 0

     and looking at this we can see that it is not equal to the individual degrees of the polynomial added together i.e 2+2 = 4

     Thus  deg(f(x)g(x))  < degf(x) + deg g(x)

Note   This only true when the ring is a cumulative ring

Step-by-step explanation:

In order to get a better understanding of the solution above Let explain so terms

RING

  In mathematics we can defined a  ring a a collection R of item that has the ability to perform binary operations that define generally addition and multiplication

Now when talk about commutative ring it mean that the operation that the collection is equipped to perform is commutative in nature