[proofplan] Write $f$ and $g$ as finite sums of monomials with nonzero coefficients, and use homogeneity to identify the total degrees of all monomials appearing in each polynomial. Distributivity shows that every monomial appearing before cancellation in $fg$ has total degree $d+e$, so once we know $fg \neq 0$, the product is homogeneous of degree $d+e$. To prove nonvanishing without appealing to an external theorem, choose lexicographically maximal exponent vectors in the supports of $f$ and $g$ and show that their sum contributes a unique nonzero coefficient to the product. [/proofplan] [step:Expand both polynomials over their homogeneous supports] For a multi-index $\alpha = (\alpha_1,\ldots,\alpha_n) \in (\mathbb{N} \cup \{0\})^n$, define \begin{align*} |\alpha| = \alpha_1 + \cdots + \alpha_n \end{align*} and \begin{align*} x^\alpha = x_1^{\alpha_1}\cdots x_n^{\alpha_n}. \end{align*} Since $f$ and $g$ are nonzero polynomials, there exist finite nonempty subsets $A,B \subset (\mathbb{N} \cup \{0\})^n$ and coefficients $c_\alpha \in k^\times$ for $\alpha \in A$ and $a_\beta \in k^\times$ for $\beta \in B$ such that \begin{align*} f = \sum_{\alpha \in A} c_\alpha x^\alpha \end{align*} and \begin{align*} g = \sum_{\beta \in B} a_\beta x^\beta. \end{align*} Because $f$ is homogeneous of degree $d$, every exponent vector in $A$ satisfies $|\alpha|=d$. Because $g$ is homogeneous of degree $e$, every exponent vector in $B$ satisfies $|\beta|=e$. [/step] [step:Show every product monomial has total degree $d+e$] By distributivity in the [polynomial ring](/page/Polynomial%20Ring) $k[x_1,\ldots,x_n]$, \begin{align*} fg = \sum_{\alpha \in A} \sum_{\beta \in B} c_\alpha a_\beta x^{\alpha+\beta}. \end{align*} For each pair $(\alpha,\beta) \in A \times B$, the total degree of the monomial $x^{\alpha+\beta}$ is \begin{align*} |\alpha+\beta| = |\alpha| + |\beta| = d+e. \end{align*} Thus every monomial appearing in this expansion of $fg$ has total degree $d+e$. Therefore, if $fg$ is nonzero, then after collecting like terms every remaining monomial still has total degree $d+e$, so $fg$ is homogeneous of degree $d+e$. [/step] [step:Choose lexicographically maximal monomials to prove the product is nonzero] Put the lexicographic order on $(\mathbb{N} \cup \{0\})^n$: for distinct multi-indices $\mu$ and $\nu$, write $\mu <_{\mathrm{lex}} \nu$ if, at the first index $j$ for which $\mu_j \neq \nu_j$, one has $\mu_j < \nu_j$. Since $A$ and $B$ are finite nonempty sets, there are lexicographically maximal elements $\alpha_0 \in A$ and $\beta_0 \in B$. Define \begin{align*} \gamma_0 = \alpha_0 + \beta_0. \end{align*} We claim that the only pair $(\alpha,\beta) \in A \times B$ satisfying $\alpha+\beta=\gamma_0$ is $(\alpha_0,\beta_0)$. Indeed, suppose $\alpha+\beta=\gamma_0$. If $\alpha \neq \alpha_0$, then maximality of $\alpha_0$ gives $\alpha <_{\mathrm{lex}} \alpha_0$. Let $j$ be the first index for which $\alpha_j \neq (\alpha_0)_j$. Then $\alpha_i=(\alpha_0)_i$ for all $i<j$ and $\alpha_j<(\alpha_0)_j$. From $\alpha+\beta=\alpha_0+\beta_0$, it follows that $\beta_i=(\beta_0)_i$ for all $i<j$ and $\beta_j>(\beta_0)_j$. Hence $\beta_0 <_{\mathrm{lex}} \beta$, contradicting the maximality of $\beta_0$ in $B$. Therefore $\alpha=\alpha_0$, and then $\beta=\beta_0$. The coefficient of $x^{\gamma_0}$ in $fg$ is therefore exactly $c_{\alpha_0}a_{\beta_0}$. Since $c_{\alpha_0},a_{\beta_0} \in k^\times$ and $k$ is a field, their product is nonzero. Hence $fg \neq 0$. [guided] The only possible obstruction to proving $fg \neq 0$ is cancellation among different products of monomials. To rule this out, we choose a monomial in $f$ and a monomial in $g$ whose product cannot be produced in any other way. Define the lexicographic order on $(\mathbb{N} \cup \{0\})^n$ as follows: for distinct multi-indices $\mu$ and $\nu$, write $\mu <_{\mathrm{lex}} \nu$ if the first coordinate where they differ is smaller for $\mu$ than for $\nu$. Since the support sets $A$ and $B$ are finite and nonempty, each has a lexicographically maximal element. Let $\alpha_0 \in A$ be maximal in $A$, and let $\beta_0 \in B$ be maximal in $B$. Define \begin{align*} \gamma_0 = \alpha_0 + \beta_0. \end{align*} We now prove that $x^{\gamma_0}$ has only one source in the product expansion. Suppose $(\alpha,\beta) \in A \times B$ satisfies \begin{align*} \alpha+\beta=\gamma_0. \end{align*} If $\alpha \neq \alpha_0$, then, because $\alpha_0$ is lexicographically maximal in $A$, we must have $\alpha <_{\mathrm{lex}} \alpha_0$. Let $j$ be the first index at which $\alpha$ and $\alpha_0$ differ. Then $\alpha_i=(\alpha_0)_i$ for all $i<j$, while $\alpha_j<(\alpha_0)_j$. Comparing the equality $\alpha+\beta=\alpha_0+\beta_0$ coordinate by coordinate gives $\beta_i=(\beta_0)_i$ for all $i<j$ and $\beta_j>(\beta_0)_j$. Thus $\beta_0 <_{\mathrm{lex}} \beta$, which contradicts the maximality of $\beta_0$ in $B$. Therefore $\alpha=\alpha_0$. Substituting this into $\alpha+\beta=\alpha_0+\beta_0$ gives $\beta=\beta_0$. Hence the exponent $\gamma_0$ occurs from exactly one pair, namely $(\alpha_0,\beta_0)$. The coefficient of $x^{\gamma_0}$ in the product is consequently \begin{align*} c_{\alpha_0}a_{\beta_0}. \end{align*} Both factors are nonzero elements of the field $k$, and [fields have no zero divisors](/theorems/8301). Hence $c_{\alpha_0}a_{\beta_0} \neq 0$. So the coefficient of $x^{\gamma_0}$ in $fg$ is nonzero, and therefore $fg \neq 0$. [/guided] [/step] [step:Conclude homogeneity of the nonzero product] The preceding step proves that $fg$ is nonzero. The earlier degree computation showed that every monomial appearing in $fg$ after collecting like terms has total degree $d+e$. Hence $fg$ is a nonzero [homogeneous polynomial](/page/Homogeneous%20Polynomial) of degree $d+e$, as required. [/step]