I have an Artinian local ring $(R,\mathfrak{m})$ and I know that such a ring is complete in its $\mathfrak{m}$-adic topology. An Artinian ring also has the property that regular elements (non zero divisors) are units, so, in the local case, the length of any regular sequence in $R$ is $0.$ If I'm not mistaken, this means that the residue field $k = R/\mathfrak{m}$ embeds in $R,$ so that $R$ is isomorphic to $k[[x_1, \ldots, x_n]]/I,$ with $n$ being the number of generators of $\mathfrak{m},$ and $I$ an $\mathfrak{m}$-primary ideal (as mentioned here).
My question is motivated by the fact that since the ideal $I$ contains a power of $\mathfrak{m},$ say $\mathfrak{m}^t,$ all monomials of degree $t$ or higher are annihilated in taking the quotient by $I$ of $k[[x_1, \ldots, x_n]].$ So, could we not replace $k[[x_1, \ldots, x_n]]$ by the polynomial ring $k[x_1, \ldots, x_n]$? If this is not possible, an explanation would be appreciated, since I am not particularly well-versed in completions or Cohen structure theory.