If \(F\) is a field, show that the product of two non-zero elements of \(F^{\infty}\) is non-zero.
Solution
Let \[ f=(f_0,f_1,\ldots),\qquad g=(g_0,g_1,\ldots) \] be nonzero elements of \(F^\infty\). Let \(m\) be the least index with \(f_m\ne0\) and \(n\) the least index with \(g_n\ne0\). By the convolution definition of multiplication in \(F^\infty\), \[ (fg)_{m+n}=\sum_{i=0}^{m+n}f_i g_{m+n-i}. \] If \(i<m\), then \(f_i=0\); if \(i>m\), then \(m+n-i<n\), so \(g_{m+n-i}=0\). Thus only the term \(i=m\) survives: \[ (fg)_{m+n}=f_mg_n\ne0. \] Hence \(fg\ne0\).