In an integral domain deg(pq) = deg(p) + deg(q)

[Mr Davis 97]

Homework Statement

Let $R$ be an integral domain and $p_1(x),p_2(x) \in R[x]$ with neither equal to $0$. Show that the degree of $p_1(x)p_2(x)$ is the sum of the degrees of $p_1(x)$ and $p_2(x)$.

Homework Equations

The Attempt at a Solution

Here is my attempt.
Let $p_1(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ and $p_2(x) = b_m x^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0$. Evidently $\operatorname{deg} (p_1) = n$ and $\operatorname{deg} (p_2) = m$

Upon multiplication of $p_1$ and $p_2$, the highest order term will be $a_n b_m x^{n+m}$. Since $a_n, b_m \in R$, an integral domain, and $a_n \ne 0$ and $b_m \ne 0$, it must be the case that $a_n b_m \ne 0$. Hence $\operatorname{deg} (p_1 p_2) = \operatorname{deg} (p_1) + \operatorname{deg} (p_2) = m + n$.

 

[fresh_42]

Homework Statement

Let $R$ be an integral domain and $p_1(x),p_2(x) \in R[x]$ with neither equal to $0$. Show that the degree of $p_1(x)p_2(x)$ is the sum of the degrees of $p_1(x)$ and $p_2(x)$.

Homework Equations

The Attempt at a Solution

Here is my attempt.
Let $p_1(x) = a_n x^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$ and $p_2(x) = b_m x^m + b_{m-1}x^{m-1} + \dots + b_1x + b_0$. Evidently $\operatorname{deg} (p_1) = n$ and $\operatorname{deg} (p_2) = m$

Upon multiplication of $p_1$ and $p_2$, the highest order term will be $a_n b_m x^{n+m}$. Since $a_n, b_m \in R$, an integral domain, and $a_n \ne 0$ and $b_m \ne 0$, it must be the case that $a_n b_m \ne 0$. Hence $\operatorname{deg} (p_1 p_2) = \operatorname{deg} (p_1) + \operatorname{deg} (p_2) = m + n$.

I would have worded it differently ($a_n \neq 0 \neq b_m$ at the beginning as part of the definition of degree), but it is correct.

 

[Mr Davis 97]

I would have worded it differently ($a_n \neq 0 \neq b_m$ at the beginning as part of the definition of degree), but it is correct.

Also, one thing. To what extent is it rigorous enough to just say that "Upon multiplication of $p_1$ and $p_2$, the highest order term will be $a_n b_m x^{n+m}$"?

 

[fresh_42]

Also, one thing. To what extent is it rigorous enough to just say that "Upon multiplication of $p_1$ and $p_2$, the highest order term will be $a_n b_m x^{n+m}$"?

It is. You can write $p_1(x)\cdot p_2(x) = (a_nx^n + \ldots )\cdot (b_mx^m + \ldots ) = a_nb_mx^{n+m} + \ldots$ or even
$$
p_1(x)\cdot p_2(x) = \left( \sum_{i=0}^{n}a_i x^i \right) \cdot \left( \sum_{j=0}^{m}b_j x^j \right) = \sum_{k=0}^{n+m} \sum_{i+j=k}a_ib_j x^{i+j}
$$
instead, but this isn't really necessary in this case.

 

[nuuskur]

An integral domain contains no zero divisors, therefore leading term will not vanish as you correctly deduced.