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Let $R$ be the ring of all continuous functions from the interval $[0,1]$ to $\Bbb R$. For each $a\in[0,1]$ let $$A_a=\{f\in R \mid f(a)=0\}$$

Now firstly, this is part of an assignment problem, please give me hints/insight into the problem, and not a solution.

I want to show that $A_a$ is a maximal ideal - but I am not sure how to think of my set. I have dealt with polynomial rings, perhaps some analogy?

Now, how to know this is a subring? Well if $f_1(a)=0$ and $f_2(a)=0$ clearly this is closed under subtraction and multiplication. Showing it is maximal is harder and I am not sure where to start.

user26857
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ContinuityOfFailure
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2 Answers2

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You're supposed to see that $A_a$ is an ideal. That's stronger than being a subring. Anyway...

but I am not sure how to think of my set.

Hint: it's a kernel.

anon
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  • And if you require your subrings to have their own multiplicative identity (let alone requiring them to have the original ring's identity) then $A_a$ is in fact not a subring. – anon May 29 '15 at 10:26
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Consider the application $$\phi_a:R\to\Bbb R$$ defined by $\phi_a(f)=f(a)$.

Is $\phi_a$ a morphism? Is it surjective? What is its kernel? Remember that given a ring $S$ and an ideal $\mathfrak m$, then $\mathfrak m$ is maximal if and only if the quotient ring $S/\mathfrak m$ is a field.

ajotatxe
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  • Secondary dumb question: In terms of polynomials I can see exactly what a quotient ring is when taking an ideal. But here I cannot. What does it mean to mod out a set of functions? – ContinuityOfFailure May 29 '15 at 10:45
  • @ContinuityOfFailure It means you identify any two functions as being "the same" (in your quotient) if they evaluate on $a$ to the same value. Of course, this partitions all of the functions according to what value they take on $a$, and these values are the elements of $\Bbb R$... – anon May 29 '15 at 10:48
  • @Anon So in $\frac{\Bbb c([0,1],\Bbb R)}{A_a}$ if we have $f_1(a)=f_2(a)$ then $f_1=f_2$? – ContinuityOfFailure May 29 '15 at 11:55
  • Yes, since $(f_1-f_2)(a)=0$ means $f_1-f_2\in A_a\implies f_1\equiv f_2$ in the quotient. – anon May 29 '15 at 12:00
  • @Anon Oh I didn't see this before opening that new question sorry!(you didn't tag me) – ContinuityOfFailure May 29 '15 at 12:13