Polynomial Rings

(a) Coefficients of $f+g$ above $\max(\deg f,\deg g)$ are sums of zero coefficients from both $f$ and $g$, hence zero. So $f+g$ has no nonzero coefficient beyond that maximum.

(b) Writing $f=\sum a_ix^i$ ($\deg f=m$) and $g=\sum b_jx^j$ ($\deg g=n$), every coefficient of $fg$ at a power $x^k$ with $k>m+n$ is a sum of terms $a_ib_j$ with $i+j=k>m+n$; since $i\le m$ forces $j=k-i > n$, giving $b_j=0$ (or symmetric), every such term vanishes.

Let $a_m,b_n$ be the leading coefficients of $f,g$ (both nonzero). The coefficient of $x^{m+n}$ in $fg$ is exactly $a_mb_n$ (every other pairing $i+j=m+n$ with $i\neq m$ would need $j>n$, giving $b_j=0$; similarly for $j\neq n$). Since $R$ has no zero divisors and $a_m,b_n\neq0$, $a_mb_n\neq0$. So $fg$ has a nonzero coefficient at exactly degree $m+n$, and by Theorem 14.1(b) no nonzero coefficient beyond it: $\deg(fg)=m+n=\deg f+\deg g$.

($\Leftarrow$) Suppose $R$ is a domain. If $f,g\in R[x]$ are both nonzero, Theorem 14.2 gives $\deg(fg)=\deg f+\deg g$, a finite nonnegative number — so $fg\neq0$ (which would need $\deg(fg)=-\infty$). No zero divisors: $R[x]$ is a domain (commutative and unital, inherited directly from $R$).

($\Rightarrow$) Suppose $R[x]$ is a domain. The constant polynomials form a subring of $R[x]$ isomorphic to $R$ (an easy check: constants add and multiply exactly as their coefficients do). If $R$ had a zero divisor $ab=0$ with $a,b\neq0$, the corresponding constant polynomials would give $ab=0$ in $R[x]$ too, with both factors nonzero — contradicting $R[x]$ being a domain. So $R$ has no zero divisors: a domain.

If $u\in R^\times$, the constant polynomial $u$ has inverse the constant polynomial $u^{-1}$ in $R[x]$, so $u\in R[x]^\times$.

Conversely, suppose $fg=1$ in $R[x]$. By Theorem 14.2 (domain, both $f,g$ nonzero since their product is nonzero), $\deg f+\deg g = \deg(1) = 0$. Since degrees are nonnegative integers, $\deg f=\deg g=0$: both $f,g$ are constants, say $f=a,g=b\in R$ with $ab=1$. So $a\in R^\times$, and $f=a$ is (the constant polynomial corresponding to) a unit of $R$.

Existence, by strong induction on $\deg f$ (treating $f=0$ or $\deg f<\deg g$ as base cases with $q=0,r=f$). If $\deg f\geq\deg g$, write $f=a_mx^m+\cdots$, $g=b_nx^n+\cdots$ with $m\geq n$, $a_m,b_n\neq0$. Since $F$ is a field, $b_n^{-1}$ exists; form:

The $x^m$ terms cancel by construction, so $\deg f_1 < \deg f$. By the induction hypothesis, $f_1 = q_1g+r$ with $r=0$ or $\deg r<\deg g$. Then $f = f_1 + \frac{a_m}{b_n}x^{m-n}g = \big(q_1+\frac{a_m}{b_n}x^{m-n}\big)g + r$, giving the required $q$.

Uniqueness. Suppose $f=q_1g+r_1=q_2g+r_2$ with each $r_i=0$ or $\deg r_i<\deg g$. Then $(q_1-q_2)g = r_2-r_1$. If $q_1\neq q_2$, the left side has degree $\deg(q_1-q_2)+\deg g \geq \deg g$ (Theorem 14.2, valid since $F$ is a domain), while the right side has degree $<\deg g$ (Theorem 14.1(a), since both $r_i$ do) — a contradiction. So $q_1=q_2$, forcing $r_1=r_2$ too.

By Theorem 14.5 (division algorithm with $g=x-a$, degree 1): $f = q(x-a)+r$ with $r=0$ or $\deg r<1$, i.e. $r$ is a constant $c\in F$. Evaluating at $x=a$: $f(a) = q(a)(a-a)+c = c$. So $r=f(a)$ always. Then $(x-a)\mid f \iff r=0 \iff f(a)=0$.

By induction on $n$. If $n=0$, $f$ is a nonzero constant, with no roots — consistent with "at most 0."

Suppose $\deg f=n\geq1$ and $f$ has a root $a\in F$. By Theorem 14.6, $f=(x-a)h$ for some $h\in F[x]$, and by Theorem 14.2 (domain), $\deg h = n-1$. If $b\neq a$ is another root of $f$: $0=f(b)=(b-a)h(b)$. Since $F$ is a field (no zero divisors) and $b-a\neq0$, $h(b)=0$: every root of $f$ other than $a$ is a root of $h$. By the induction hypothesis, $h$ has at most $n-1$ distinct roots, so $f$ has at most $1+(n-1)=n$ distinct roots (counting $a$ itself). If $f$ has no roots at all, the bound "at most $n$" holds trivially.