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This is something I've been trying to work out this evening.

Let $R$ be the ring of continuous real-valued functions on $[0,1]$ with pointwise addition and multiplication. For $t\in [0,1]$, the map $\phi_t\colon f\to f(t)$ is a ring homomorphism of $R$ to $\mathbb{R}$. I'm trying to show that every ring homomorphism of $R\to\mathbb{R}$ has this form.

Suppose otherwise, that there is some $\phi\neq\phi_t$, and thus there is some $f_t\in R$ such that $\phi(f_t)\neq \phi_t(f_t)=f_t(t)$. Define $g_t=f_t-\phi(f_t)1\in R$. Here $\phi(f_t)1$ is the constant function sending $[0,1]$ to $\phi(f_t)$. Then $g_t(t)\neq 0$. My first small question is why does $\phi(g_t)=0$? It seems only that $\phi(g_t)=\phi(f_t)-\phi(\phi(f_t)1)$.

I would like to conclude that there are only finitely many $t_i$ such that $g(x)=\sum g_{t_i}^2(x)\neq 0$ for all $x$. Then $g^{-1}=1/g(x)\in R$, but $\phi(g)=0$, contradicting the fact that homomorphisms map units to units. How can we be sure there are only finitely many $g_{t_i}$ such that the sum of their squares is never $0$? Thanks.

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Regarding your small question: $\phi(1) = 1$ (as $\phi$ is a ring morphism), so $\phi(\phi(f_t)1) = \phi(f_t)$. –  martini Jun 18 '12 at 7:09
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@martini Thanks, but how exactly do you get that? Since $\phi$ isn't a module homomorphism, I don't see why $\phi(\phi(f_t)1)=\phi(f_t)\phi(1)=\phi(f_t)$. –  Jakucha Jun 18 '12 at 7:39
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@Jakucha: You're right, this is more a subtle point than I thought. Let $f: \mathbb R \to \mathbb R$ be defined by $f(x) = \phi(x1)$. This is also a ring homomorphism since $(x1)(y1) = (xy)1$, likewise for addition. But the only endomorphism of $\mathbb R$ is the identity (first show that $f(q)=q$ for rational $q$, then use $f(c^2) \ge 0$ to fill in the rest). –  Erick Wong Jun 18 '12 at 7:58
    
@ErickWong Oh thanks, so $f$ is nondecreasing, but fixes $\mathbb{Q}$. This makes it pretty clear that $f$ is the identity, since if $q_1\leq x\leq q_2$, then $q_1\leq f(x)\leq q_2$ for any $x$ and $q_i\in\mathbb{Q}$. –  Jakucha Jun 18 '12 at 8:02
    
@Jakucha: Yep! Could not have said it better myself. –  Erick Wong Jun 18 '12 at 8:05
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1 Answer

up vote 4 down vote accepted

Use compactness. For each $t$ the set $\{x : g_t(x) \ne 0\}$ is open and contains $t$, so the union of all these sets is $[0,1]$, meaning they form an open cover.

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Thanks, my topology is rusty. –  Jakucha Jun 18 '12 at 7:45
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