Is this a maximal ideal of the ring of formal power series

abstract-algebracommutative-algebraformal-power-seriesmaximal-and-prime-ideals

Let $ k $ be an algebraically closed field, and $ k[[T]] $ be the ring of formal power series in variables $ (T_{1},\dots,T_{n}) = T. $ Let $ \mathfrak{m}^{l} $ be the ideal of $ k[[T]] $ consisting of formal power series with no terms of degree less than $ l.$ Is this ideal maximal in $ k[[T]]$?

I think that it is not possible to embed $ \mathfrak{m}^{l} $ into some $ \mathfrak{m}^{j} $ with $ j > l, $ but I don't think this is sufficient since it may be that the maximal ideals of $ k[[T]] $ are not strictly of the form $ \mathfrak{m}^{l}$ (I don't know either way).

It is also not clear to me that $ k[[T]]/\mathfrak{m}^{l} \cong k. $

EDIT: It is true that $ \mathfrak{m}^{j} \subset \mathfrak{m}^{l} $ where $j>l$. So we have a chain $$ k[[T]] \supset \mathfrak{m}^{1} \supset \mathfrak{m}^{2} \supset \mathfrak{m}^3 \supset \dots $$

I guess this means that $ \mathfrak{m}^{1} $ might be maximal. Is that right? It looks like $ k[[T]]/\mathfrak{m}^{1} \cong k. $

Best Answer

It's not a maximal ideal unless $\ell = 1$, since $\mathfrak{m}^\ell \subset \mathfrak{m}^1$, which, concretely, is the ideal of power series with no constant term. When $\ell = 1$, we have $$ \frac{k[[T]]}{\mathfrak{m}} \stackrel{\sim}{\longrightarrow} k $$ by the rule $f \mapsto f(0)$.