Ring $C([0,1])$

Question

Consider the interval $I=[0,1]$ and let $R$ be the ring of continuous function $I \to \mathbb{R}$ on the interval $I$. Let $M \subset R$ be the subset of functions which are zero at $1/2$, in other words: $$M:=\{f \in R| f(1/2)=0\}$$ Is the short exact sequence of $R$-module $$0 \to M \to R \to R/M \to 0$$ split?

I have showed that $M$ is maximal ideal in $R$ and $R/M \cong \mathbb{R}$. However, I dont't know how to show whether the sequence is split or not. I'm trying to claim that $\mathbb{R}$ is a projective $R$ module but have no idea.

Answer 1

It depends.

If you consider this sequence as an exact sequence of $\mathbb{R}$-modules (vector spaces), then it obviously splits (every sequence of vector spaces splits - it is a general fact) and one such splitting is explicitly given in the answer by Tsemo Aristide, who interpreted your question in this manner.

On the other hand if you ask if this sequence is a split exact sequence of $R$-modules for $R= C([0,1])$ (which seems to be your real intention), then the answer is no. Indeed, suppose that $s:R/M\rightarrow R$ is a splitting of the projection $p:R\rightarrow R/M$ in the category of $R$-modules. Pick $f = s(1)\in R$. Then $f(\frac{1}{2}) = 1$ and since $f$ is continuous, we derive that there exists $x_0\in [0,1]\setminus \{\frac{1}{2}\}$ such that $f(x_0)\neq 0$. There exists a function $g\in R$ such that $$g\left(\frac{1}{2}\right) = 1,\,g\left(x_0\right) = 2$$ For instance pick $g$ to be piecewise affine function that satisfy these constraints. Then $g\cdot f \neq f$ because $$(g\cdot f)(x_0) = g(x_0)\cdot f(x_0) = 2f(x_0) \neq f(x_0)$$ On the other hand $g \cdot f,f \in s\left(R/M\right)$, since $s$ is a morphism of $R$-modules, and also $$p(g\cdot f) = 1 = p(f)$$ Since $s$ is a splitting of $p$, we derive that $g\cdot f = f$. This is a contradiction.

Answer 2

The quotient map $p:R\rightarrow R/M$ of the exact sequence $0\rightarrow R\rightarrow M/R\rightarrow 0$, is the evaluation $p(f)=f({1\over 2})$, in this case $g:\mathbb{R}\rightarrow R$ defined by $g(a)=f_a$ where $f_a$ is the constant function defined by $f_a(x)=a$ is a splitting.

Answer 3

Another way to see that it does not split as a sequence of $R$ modules:

If it were to split, then that would mean $M$ is a direct summand of $R$.

But given an element $f$ that isn't in $M$, you can always multiply by a continuous, piecewise-linear "V" function $g$ which is zero at $1/2$ to get a nonzero element of $M$. This amounts to $M$ being an essential submodule of $R$.

Of course, a maximal right ideal $M$ cannot be both a summand and essential right submodule of $R$ at the same time.