Let $k$ be a field, $R=k[x_1,x_2,x_3...]=k[x_i]_{\mathbb{N}_0}$ and $\mathfrak{m}=(x_1,x_2,x_3,...)$.
I want to check whether the following equality is true or not:
$\varprojlim{\mathfrak{m}/\mathfrak{m}^{i+1}}= \mathfrak{m}(\varprojlim{R/\mathfrak{m}^i})$
I claim that this equality is true. And I claim
$\varprojlim{\mathfrak{m}/\mathfrak{m}^{i+1}}=L= \mathfrak{m}(\varprojlim{R/\mathfrak{m}^i})$, where $L=\{f\in k[[x_i]]_{i \in \mathbb{N}_0} | f(0)=0\}$
My try:
I define $\phi_i: L \longrightarrow \mathfrak{m}/\mathfrak{m}^{i+1}: p \mapsto p + \mathfrak{m}^{i+1}$.
Let $F:\mathbb{N}^{op} \longrightarrow \mathcal{C}$ such that $F(i)=\mathfrak{m}/\mathfrak{m}^{i+1}$ and $F(i\geq j)= \mathfrak{m}/\mathfrak{m}^{i+1} \longrightarrow \mathfrak{m}/\mathfrak{m}^{j+1}$.
We want to check that $F(i\geq j) \circ \phi_j=\phi_i$:
$F(i\geq j): \mathfrak{m}/\mathfrak{m}^{i+1} \longrightarrow \mathfrak{m}/\mathfrak{m}^{j+1}: p + \mathfrak{m}^{i+1} \mapsto p + \mathfrak{m}^{j+1}$
$F(i\geq j) \circ \phi_i: L \longrightarrow p/\mathfrak{m}^{j+1}: p \mapsto p + \mathfrak{m}^{j+1}$
$\phi_j: L \longrightarrow \mathfrak{m}/\mathfrak{m}^{j+1}$
And this is $F(i\geq j)\circ \phi_i =\phi_j$.
Now we have to see that if there is a $(L',\alpha)$ satisfying the same condition, then there exists a unique morphism $u:L\longrightarrow L'$ such that $\phi \circ u = \alpha$. But I do not know how to check that.
I know I still need to prove that the right inverse limit is equal to $L$.
But how could I prove the fact mentioned above?
Thank you.
In case someone come back to this post one day, I will answer what I found.
The equality does not hold.
$\mathfrak{m}(\varprojlim{R/\mathfrak{m}^i})\subset \varprojlim{\mathfrak{m}/\mathfrak{m}^{i+1}} $
To see it, we can check that $$g(x_1,x_2,x_3,...)=\sum_{n=1}^{\infty} x_n^n$$ is not in $\mathfrak{m}(\varprojlim{R/\mathfrak{m}^i})$. But it is in $\varprojlim{\mathfrak{m}/\mathfrak{m}^{i+1}}$.