Remark

Let $u$ and $v$ be elements of $A[(X_i)_i \in ,]$, and suppose that $A$ is an integral domain, $v$ is a non-zero multiple of $u$ and $v$ is homogeneous; then $u$ is also homogeneous. For let $u' \in A[(X_i)_i ,_1]$ be such that $v = uu'$; we have $u \neq 0$, $u' \neq 0$, and if

are the decompositions of $u$ and $u'$ into homogeneous components, with $u_h \neq 0$, $u_k \neq 0$, $u_{h'} \neq 0$, $u_{k'} \neq 0$, then $v = u_h u_{h'} + u_h u_{h'+1} + \ldots + u_k u_{k'}$ and $u_h u_{h'}$ is non-zero homogeneous of degree $h + h'$ while $u_k u_{k'}$ is non-zero homogeneous of degree $k + k'$ (Prop. 8). Since $v$ is homogeneous, we have $h + h' = k + k'$ whence $h = k$, $h' = k'$.