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Let $R$ be the ring of germs of continuous functions $\mathbb R \rightarrow \mathbb R$. It is clear that this is a local ring with maximal ideal $\mathcal m$ consisting of those germs $f$ with $f(0)=0$.

What is not clear to me is why $(\mathcal m)^2 =\mathcal m$ should hold. Perhaps someone can give me a small hint how to show this.

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@MattN. : I think $(m)^2 = \{\sum f_i g_i | f_i, g_i \in m\}$. –  Joel Cohen Jul 2 '12 at 7:52
    
@JoelCohen Yes. Right. –  Matt N. Jul 2 '12 at 7:59
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This provides a counterexample for Nakayama's Lemma (without f.g. assumption). –  Martin Brandenburg Jul 2 '12 at 8:26
    
This was my motivation to consider this non-noetherian ring. –  Cyril Jul 2 '12 at 12:46

2 Answers 2

up vote 7 down vote accepted

For $f\in \mathfrak m$ we have $f=\sqrt [3]f\cdot (\sqrt [3]f)^2$ and $\sqrt [3]f, (\sqrt [3]f)^2\in \mathfrak m$, so that indeed $\mathfrak m\subset \mathfrak m^2$.
The other inclusion is trivial and we have proved that $$\mathfrak m^2 = \mathfrak m $$ Edit: warning !
The analogue of the above result is false if one replaces the ring of germs of continuous functions by the ring of germs $\mathcal C_0^k\; (1\leq k\leq \infty)$ of $k$ times continuously differentiable functions at the origin of $\mathbb R$.
Indeed if $f\in \mathfrak m^2$, we can write $f=\sum g_kh_k\; (g_k, h_k\in \mathfrak m)$ and differentiating yields $f'(0)=\sum g_k'(0)h_k(0)+\sum g_k'(0)h_k(0)=0$, so that elements $f\in \mathfrak m^2$ satisfy $f'(0)=0$ .
In particular $x\notin \mathfrak m^2$, although $x\in \mathfrak m$.
Hence $$\mathfrak m^2 \subsetneq \mathfrak m$$

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The inclusion $(m)^2 \subset m$ is obvious. To prove the other inclusion, take $f \in m$. We wish to prove that $f \in (m)^2$. Define $f_+(x) = \max(f(x),0)$ and $f_-(x) = \min(f(x),0)$. Note that we have $f_+, f_- \in m$, $f_+ \ge 0$, $f_- \le 0$ and

$$f = f_+ + f_-$$ Now define $g = \sqrt{f_+}$ and $h = \sqrt{-f_-}$. We also have $h,g \in m$, and

$$f = g^2 - h^2 \in (m)^2$$

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