Let $A$ be the ring of continuous maps $f: [0,1] \rightarrow \mathbb{R}$. Prove that there exists a prime ideal $Q \in \operatorname{Spec} (A)$ which is not of the form $I_p = \{ f \in A \mid f(p)=0 \}$ with $p \in [0,1]$.
Hint: use localization.
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Consider the ideal $J\subset A$ of functions $f\in A$ such that the limit for $x$ tending to $\frac 12$ (but remaining $\neq \frac 12$ ) of $\frac {f(x)}{(x-\frac 12)^r}$ is $0$ for all $r=1,2,3,...$ .
Then $J$ is not prime, but is a reduced ideal (i.e. equal to its nilradical) .
So $J$ is the intersection of the prime ideals containing it.
Since the only $I_p$ containing $J$ is $I_{\frac 12}$ there and since $J\neq I_{\frac 12}$ there must exist some prime ideal $\mathfrak p\subset A$ with $\mathfrak p\neq I_p$ for all $p\in [0,1]$.
Remark: why $J$ is not prime
Consider in $A$ the functions $f(x)= \operatorname {max} (0,x-\frac 12)$ and $g(x)= \operatorname {max} (0,\frac 12 -x)$.
Then $fg=0\in J$ although $f,g\notin J$. Hence the ideal $J$ is not prime.
Georges Elencwajg
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How would you ensure that such ideal is not $I_0$ or $I_1$? – Pavel Čoupek May 06 '14 at 21:29
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@Pavel: you are right. I have corrected my answer by furnishing prime ideals different from all $I_p$'s – Georges Elencwajg May 06 '14 at 23:24
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The nilradical of $J$ consists of the $f\in A$ which have some power in $J$: $\exists N, f^N\in J$. – Georges Elencwajg May 07 '14 at 08:05
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It is a radical. There exists another radical, the intersection of the maximal ideals containing the ideal, which is called the Jacobson radical. – Georges Elencwajg May 07 '14 at 10:28
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why is $J = nil(J)$ ? – Gauss May 07 '14 at 12:02
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I think $J=0$. Is it true? – Gauss May 07 '14 at 12:54
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@gregoryduran No, $J$ contains e.g. $\exp \left(-\frac{1}{(x-\frac{1}{2})^2}\right)$, where the value at $\frac{1}{2}$ is understood to be $0$. If a power of $f$ vanishes faster than every power of $(x-\frac12)$, then $f$ itself must vanish faster than every power, therefore $J = \sqrt{J}$. – Daniel Fischer May 07 '14 at 12:59
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Thanks for the comment, @Daniel: I couldn't have said it as well as you! – Georges Elencwajg May 07 '14 at 13:23
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@GeorgesElencwajg Why $I_{1/2}$ is the only $I_p$ containing $J$? – karparvar Mar 11 '16 at 16:01
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@karparvar : Because Daniel's exponential function function in the comment above is in $J$ but in no $I_p$ except $I_{1/2}$ . – Georges Elencwajg Mar 11 '16 at 18:15
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I notice that the hint was to “use localization” and that the other answer does not seem to have utilized it. Just today I saw someone mention the probable desired trick, and it seems too ingenious not to share. Here it is.
Take $S$ to be the subset of nonzero polynomial functions restricted to $[0,1]$. It is a mutiplicative set. Therefore by basic commutative algebra, there is a prime ideal disjoint from $S$. Clearly $P$ is not of the form described, since to do so would require it to contain $x-a$ for some $a\in[0,1]$.
rschwieb
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