[proofplan] We prove each part by direct computation with leading coefficients. Parts (i) and (ii) follow from examining the coefficients of $f + g$ and $fg$ at each degree and observing that no term of degree exceeding the claimed bound can appear. For part (iii), the key observation is that in an integral domain the product of the two leading coefficients is nonzero, which pins down the degree of $fg$ exactly. The "in particular" statement then follows immediately: a product of nonzero polynomials has a well-defined degree, so it cannot be zero. [/proofplan] [step:Write $f$ and $g$ in terms of their coefficients and leading terms] Let $n = \deg f$ and $m = \deg g$. Write \begin{align*} f &= a_n x^n + a_{n-1} x^{n-1} + \dots + a_0, \\ g &= b_m x^m + b_{m-1} x^{m-1} + \dots + b_0, \end{align*} where $a_i, b_j \in R$ for all $i, j$, with $a_n \neq 0_R$ and $b_m \neq 0_R$ (since $f$ and $g$ are nonzero with the stated degrees). We adopt the convention that $a_k = 0_R$ for $k > n$ and $b_k = 0_R$ for $k > m$. [guided] Before proving any of the three parts, we set up notation. Let $n = \deg f$ and $m = \deg g$. Since $f$ is a nonzero polynomial of degree $n$, we can write \begin{align*} f &= a_n x^n + a_{n-1} x^{n-1} + \dots + a_0, \end{align*} where $a_i \in R$ for $0 \le i \le n$ and $a_n \neq 0_R$. The coefficient $a_n$ is the **leading coefficient** of $f$. Similarly, \begin{align*} g &= b_m x^m + b_{m-1} x^{m-1} + \dots + b_0, \end{align*} where $b_j \in R$ for $0 \le j \le m$ and $b_m \neq 0_R$. We set $a_k = 0_R$ for $k > n$ and $b_k = 0_R$ for $k > m$, so that addition and multiplication formulas can be written uniformly without case splits on the indices. [/guided] [/step] [step:Prove that $\deg(f + g) \le \max(\deg f, \deg g)$] Set $d = \max(n, m)$. The coefficient of $x^k$ in $f + g$ is $a_k + b_k$. For $k > d$, both $a_k = 0_R$ and $b_k = 0_R$, so $a_k + b_k = 0_R$. Therefore $f + g$ has no nonzero coefficient in degree exceeding $d$, which gives $\deg(f + g) \le d = \max(\deg f, \deg g)$. [guided] The idea is straightforward: addition of polynomials is performed coefficient-by-coefficient, and we check that no coefficient beyond degree $d = \max(n, m)$ can be nonzero. The coefficient of $x^k$ in $f + g$ is $a_k + b_k$. For any $k > d$, both $k > n$ and $k > m$ hold, so our convention gives $a_k = 0_R$ and $b_k = 0_R$. Therefore $a_k + b_k = 0_R + 0_R = 0_R$ for all $k > d$. Since every coefficient of $f + g$ in degree strictly above $d$ vanishes, the polynomial $f + g$ has degree at most $d$. That is, $\deg(f + g) \le d = \max(\deg f, \deg g)$. Note that the inequality can be strict: if $n = m$ and $a_n + b_n = 0_R$ (i.e., the leading terms cancel), then $\deg(f + g) < \max(\deg f, \deg g)$. For example, take $f = x^2 + 1$ and $g = -x^2 + x$ in $\mathbb{Z}[x]$; then $\max(\deg f, \deg g) = 2$ but $f + g = x + 1$ has degree $1$. This is why part (i) is stated as an inequality rather than an equality — cancellation of leading terms can strictly lower the degree. [/guided] [/step] [step:Prove that $\deg(fg) \le \deg f + \deg g$] By the definition of polynomial multiplication, the coefficient of $x^k$ in $fg$ is $c_k = \sum_{i+j=k} a_i b_j$. For $k > n + m$, every pair $(i, j)$ with $i + j = k$ satisfies either $i > n$ or $j > m$ (since $i \le n$ and $j \le m$ would give $i + j \le n + m < k$). Therefore $a_i b_j = 0_R$ for each such pair, giving $c_k = 0_R$. This means $fg$ has no nonzero coefficient in degree exceeding $n + m$, so $\deg(fg) \le n + m = \deg f + \deg g$. [guided] By the definition of polynomial multiplication in $R[x]$, the coefficient of $x^k$ in $fg$ is \begin{align*} c_k = \sum_{\substack{i + j = k \\ i, j \ge 0}} a_i b_j. \end{align*} We must show that $c_k = 0_R$ for all $k > n + m$. Take any $k > n + m$ and consider a pair $(i, j)$ with $i + j = k$ and $i, j \ge 0$. Suppose for contradiction that $i \le n$ and $j \le m$. Then $i + j \le n + m < k = i + j$, a contradiction. So for every summand in $c_k$, either $i > n$ (in which case $a_i = 0_R$) or $j > m$ (in which case $b_j = 0_R$). In both cases, $a_i b_j = 0_R$. Since every term in the sum vanishes, $c_k = 0_R$. Therefore $fg$ has no nonzero term of degree exceeding $n + m$, and we conclude $\deg(fg) \le \deg f + \deg g$. Why is this only an inequality? The coefficient of $x^{n+m}$ in $fg$ is $c_{n+m} = a_n b_m$, and in a general commutative ring this product can be zero even though $a_n \neq 0_R$ and $b_m \neq 0_R$ (when $R$ has zero divisors). Part (iii) addresses exactly this point. [/guided] [/step] [step:Prove that $\deg(fg) = \deg f + \deg g$ when $R$ is an integral domain] Assume $R$ is an integral domain. From the previous step, $\deg(fg) \le n + m$. The coefficient of $x^{n+m}$ in $fg$ is \begin{align*} c_{n+m} = \sum_{\substack{i + j = n + m \\ i, j \ge 0}} a_i b_j = a_n b_m, \end{align*} since for any other pair $(i, j)$ with $i + j = n + m$ and $(i, j) \neq (n, m)$, either $i > n$ (giving $a_i = 0_R$) or $i < n$ which forces $j = n + m - i > m$ (giving $b_j = 0_R$). Since $R$ is an integral domain and $a_n \neq 0_R$, $b_m \neq 0_R$, we have $a_n b_m \neq 0_R$. Therefore $c_{n+m} \neq 0_R$, so $\deg(fg) = n + m = \deg f + \deg g$. [guided] Now assume $R$ is an integral domain. We already established in the previous step that $\deg(fg) \le n + m$. To obtain the reverse inequality, we show that the coefficient of $x^{n+m}$ in $fg$ is nonzero. The coefficient of $x^{n+m}$ in $fg$ is \begin{align*} c_{n+m} = \sum_{\substack{i + j = n+m \\ i, j \ge 0}} a_i b_j. \end{align*} We claim that the only contributing term is $a_n b_m$. Consider a pair $(i, j)$ with $i + j = n + m$, $i \ge 0$, $j \ge 0$, and $(i, j) \neq (n, m)$. There are two cases: - If $i > n$, then $a_i = 0_R$, so $a_i b_j = 0_R$. - If $i < n$, then $j = n + m - i > m$, so $b_j = 0_R$, giving $a_i b_j = 0_R$. In either case the term vanishes. Therefore $c_{n+m} = a_n b_m$. This is where the integral domain hypothesis is consumed. Since $R$ is an integral domain, $R$ has no zero divisors: if $a_n \neq 0_R$ and $b_m \neq 0_R$, then $a_n b_m \neq 0_R$. We have $a_n \neq 0_R$ (since $\deg f = n$) and $b_m \neq 0_R$ (since $\deg g = m$), so $c_{n+m} = a_n b_m \neq 0_R$. Therefore $fg$ has a nonzero coefficient in degree $n + m$, giving $\deg(fg) \ge n + m$. Combined with the upper bound from part (ii), we conclude $\deg(fg) = n + m = \deg f + \deg g$. [/guided] [/step] [step:Deduce that $R[x]$ is an integral domain when $R$ is] Assume $R$ is an integral domain. Let $f, g \in R[x]$ be nonzero polynomials. By part (iii), $\deg(fg) = \deg f + \deg g$. Since $\deg f \ge 0$ and $\deg g \ge 0$, the product $fg$ has a well-defined non-negative degree, so in particular $fg \neq 0$. Therefore $R[x]$ has no zero divisors. Since $R$ is a commutative ring with $1_R \neq 0_R$ and $1_R \in R[x]$ serves as the multiplicative identity of $R[x]$, we conclude that $R[x]$ is an integral domain. [guided] Finally, we show that if $R$ is an integral domain, then so is $R[x]$. Recall that an integral domain is a commutative ring with $1 \neq 0$ and no zero divisors. We must verify all three properties for $R[x]$. First, $R[x]$ is a commutative ring: polynomial addition and multiplication are commutative (since the coefficient ring $R$ is commutative), and the ring axioms are inherited from $R$. The multiplicative identity is the constant polynomial $1_R$. Since $R$ is an integral domain, $1_R \neq 0_R$, so the identity in $R[x]$ is nonzero. It remains to show $R[x]$ has no zero divisors. Let $f, g \in R[x]$ with $f \neq 0$ and $g \neq 0$. Since $f$ and $g$ are nonzero polynomials, they each have a well-defined degree with $\deg f \ge 0$ and $\deg g \ge 0$. By part (iii), $\deg(fg) = \deg f + \deg g \ge 0$. A polynomial with a well-defined non-negative degree is nonzero (its leading coefficient is nonzero by definition of degree), so $fg \neq 0$. This shows that the product of any two nonzero elements of $R[x]$ is nonzero, i.e., $R[x]$ has no zero divisors. Combined with commutativity and $1_{R[x]} \neq 0_{R[x]}$, we conclude that $R[x]$ is an integral domain. [/guided] [/step]