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Is the set of all function vanishing at both $0$ and $1$ is maximal ideal ? Yes/no

My attempt : i know that it will not prime ideal take $f(x) = x-1 $ and $ g(x) = x$. where $f, g \in C[0,1]$

But here im confused about maximal ideal.?

I thinks yes

jasmine
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    A maximal ideal of what ring? Which functions exactly (domain, codomain...)? – Qi Zhu Aug 21 '19 at 18:56
  • If your ring is $C[0,1]$, then your attempt already answers the question. It's not prime, hence not maximal. – Qi Zhu Aug 21 '19 at 18:59
  • @QiZhu domain is $ [0,1]$ – jasmine Aug 21 '19 at 18:59
  • @QiZhu every prime ideal need not be maximal !!!! – jasmine Aug 21 '19 at 19:00
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    Read my whole comment and answer all of it. Yes, prime ideals need not be maximal - however maximal ideals are prime. – Qi Zhu Aug 21 '19 at 19:01
  • @QiZhu $C[0,1]$ is a infinite ring so every prime ideal need not be maximal.....but if Ring is finite then we can say every prime is maxim al – jasmine Aug 21 '19 at 19:02
  • Please read my comments precisely before answering... Prime implying maximality and maximality implying prime are two entirely different things. – Qi Zhu Aug 21 '19 at 19:12
  • @QiZhu ya i have read ur comment but im not getting what exactly u want to say pliz elaborate more.....or u can write ur answer in answer box – jasmine Aug 21 '19 at 19:15
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    Then please next time ask questions instead of immediately bashing what other people are writing. I have tried to explain my comment as an answer below. – Qi Zhu Aug 21 '19 at 19:21

2 Answers2

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I'll suppose the ring in question is $C[0,1]$. Let the proposed ideal be $I$. Now, $I$ is not prime as you have noted (since $fg \in I$ but $f \not \in I, g \not \in I$), but then surely it cannot be maximal.

(There seems to be some confusion about maximality and primality. One very important fact is that every maximal ideal is prime. You can read this up in every elementary abstract algebra book. So if $I$ were maximal, then it has to be prime. But that isn't the case as you have shown.)

Qi Zhu
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Hint: Consider the set of all functions vanishing at $0$. It is an ideal. How does it compare to the one you're asking the question about.

Scientifica
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  • u mean at $0$ both $f$ and $g$ is not vanish – jasmine Aug 21 '19 at 19:04
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    @jasmine Nope; forget about any particular $f$ and $g$ as the ones you defined. An ideal $I$ is maximal if there is no bigger ideal other than the whole ring. You're asking the question if $I$, the set of functions vanishing at $0$ and $1$, is a maximal ideal or not. I propose to consider $J$ the set of functions vanishing at $0$ (but not necessarily at $1$). Then $I\subsetneq J\neq C[0,1]$, and $J$ is an ideal. – Scientifica Aug 21 '19 at 19:51
  • ya i got its @Scientifica thanks u.. – jasmine Aug 21 '19 at 19:55
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    @jasmine: But note that to establish $I\subsetneq J$ you should exhibit a function that is in $J$ but in in $I$. In this case, $x\mapsto x^2$ would do it. – Arturo Magidin Aug 21 '19 at 21:07
  • @ArturoMagidin thanks u sir – jasmine Aug 21 '19 at 21:09