Exercise A0.III.1.14

Proposition A

Let $R$ be a commutative ring and let $f, g \in R[x].$ Then

$$\deg(f + g) \leq \max(\deg(f),\deg(g)).$$

Proof

We will prove this by induction on $\max(\deg(f),\deg(g)).$ Suppose both polynomials are constant, so their degrees are $0.$ In this case, their sum is also a constant, so the base case holds.

Suppose $\max(\deg(f),\deg(g)) = n.$ Let $c_1, c_2 \in R$ be such that $f = c_1x^n +f',$ $g = c_2 x^n + g'$ with degrees of $f'$ and $g'$ strictly below $n.$ In this case, $\deg(f' + g') < n$ (by inductive hypothesis).

We know that $\deg(f + g) = \deg((c_1 + c_2) x^n + f' + g').$

Suppose $c_1 + c_2 = 0.$ In this case $\deg(f + g) = \deg(f' + g') \leq n.$ Otherwise, if $c_1 + c_2 \neq 0$ then $\deg(f + g) = \deg((c_1 + c_2) x^n) = n,$ which is also less or equal to $n.$

$\blacksquare$

Proposition B

Let $R$ be an integral domain and let $f, g \in R[x]$ be non-zero. Then

$$\deg(fg) = \deg(f) + \deg(g).$$

Proof

Let $n = \deg{f}$ and $m = \deg(g).$ Let $c_1x^n$ be the leading term of $f$ and let $c_2x^m$ be the leading term of $g.$ Note that $c_1 c_2$ is non-zero, since both $c_1$ and $c_2$ are non-zero and $R$ is an integral domain. Hence, the leading term of $fg$ is $c_1c_2x^{m + n},$ so the degree of $fg$ is at least $m + n.$

We also note that $\deg(fg)$ cannot be more than $m + n,$ since

$$fg = (c_1x^n + f')(c_2 x^m + g') = c_1 c_2 x^{n + m} + \text{lower degree terms}.$$

$\blacksquare$