Let $k$ be a field. Is the morphism $k[x_1,\ldots]\to k[[x_1,\ldots]]$ (countably infinite number of variables) flat? For a Noetherian ring $R$, the map $R\to \hat{R}$ to its completion is flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

Let $k$ be a field. For $R=k[x_1,\ldots]$ with countably infinite number of variables, [due to the discussion in the comments] we have to make the following distinction between $k[[x_1,\ldots]]$ and the completion $\hat{R}$ of $R$ at the ideal $(x_1,\ldots)$: note that the former admits elements which can have infinitely many monomials of the same degree whereas the latter can not (e.g. $\sum_i x_i\in k[[x_1,\ldots]]$ but $\notin\hat{R}$). There are two questions

1. Is the morphism $R\to\hat{R}$ flat? If $R$ were any Noetherian ring, the map $R\to \hat{R}$ to its completion is always flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

2. Is the morphism $R\to k[[x_1,\ldots]]$ flat? This is answered positively by ayanta below.

For a Noetherian ring $R$, the map $R\to \hat{R}$ to its completion is flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

Let $k$ be a field. Is the morphism $k[x_1,\ldots]\to k[[x_1,\ldots]]$ (countably infinite number of variables) flat? For a Noetherian ring $R$, the map $R\to \hat{R}$ to its completion is flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

For a Noetherian ring $R$, the map $R\to \hat{R}$ to it's completion is flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.

For a Noetherian ring $R$, the map $R\to \hat{R}$ to its completion is flat (see e.g. Atiyah-MacDonald 10.14) but my intuition breaks down for non-Noetherian rings.