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Here is the question that I am trying to solve:

Let $R$ be the ring of continuous functions from $\mathbb R$ to $\mathbb R.$ Let $I \subset R$ be the ideal of functions with compact support(i.e., $f(x) = 0$ for $|x|$ sufficiently large).\ $(a)$ Show that $I$ is not a prime ideal of $R.$\ $(b)$ Is there an ideal $J$ of $R$ that contains $I$ such that $J$ is prime? Why or why not?

My thoughts:

For part$(a).$

I know that this ring does not have an identity but how this will help me in showing that this $I$ is not prime, could someone explain this to me please or at least show me the proof.

For part$(b).$

I do not know how to even think about it, any help will be greatly appreciated.

Emptymind
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    The ring does have an identity (the constant-1 function), assuming multiplication is pointwise. For (a): $I$ is prime iff whenever the product $fg$ of two functions is in $I$, then either $f$ or $g$ is in $I$. Can you find two functions with non-compact support that multiply to a function with compact support? – Kenanski Bowspleefi Nov 01 '23 at 15:57
  • No there is nothing on the top of my head @KenanskiBowspleefi – Emptymind Nov 01 '23 at 17:43
  • Maybe consider the function $$f(x) = \begin{cases} \sin(x),& \sin(x)>0\ 0,&\sin(x) \leq 0.\end{cases}$$ This function doesn’t have compact support. Try shifting $f(x)$ horizontally to get a function $g(x)$ that is $0$ when $f(x)$ is nonzero (and that also doesn’t have compact support). So then $fg\in I$ because $f(x)g(x)=0$ for all $x$, but neither $f$ nor $g$ is in $I$. – Kenanski Bowspleefi Nov 01 '23 at 20:14
  • @KenanskiBowspleefi could you please read my last comment below the answer given below? – Emptymind Nov 02 '23 at 11:19

1 Answers1

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For b: by Zorn's lemma, there exists a maximal ideal containing $I$. In particular, this maximal ideal is prime.

Nilav
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  • I am not allowed to use Zorn's Lemma – Emptymind Nov 01 '23 at 15:48
  • @Emptymind Frankly, it's a bit bogus to spring a random requirement like that now. For one thing, you should have said it up front. For another thing, it is perfectly reasonable to assume Zorn's lemma in this context. Talking about the prime ideals of a ring of continuous functions (other than the easy maximal ones of the form $M_c={f\mid f(c)=0}$) is nearly impossible without doing so. arbitrarily excluding it makes the post fishy. – rschwieb Nov 01 '23 at 17:04
  • why this maximal ideal is prime? Is this a solution for letter a or b? – Emptymind Nov 01 '23 at 18:00
  • @rschwieb the question was an exam question out of 11 questions that have to be solved in one hour, so I am guessing that the solution should be easy and only take not more than 5 minutes – Emptymind Nov 01 '23 at 18:02
  • Is this a solution for letter a or b? The first five characters of the post are For b. – rschwieb Nov 01 '23 at 18:16
  • why this maximal ideal is prime? Every maximal ideal is prime. – rschwieb Nov 01 '23 at 18:17
  • if you are expected to solve this question in 5 minutes then I don't think there's any doubt that Zorn's lemma is allowed. – rschwieb Nov 01 '23 at 18:18
  • Here’s an example of what happens if you try to do this without Zorn’s lemma (or without appealing to the general fact, proven using Zorn’s lemma, that every ideal is contained in a maximal ideal): https://math.stackexchange.com/a/2353008/881155. If you want to do this question quickly, I think it’s most likely the intended solution is to say that every ideal is contained in a maximal ideal, and all maximal ideals are prime. – Kenanski Bowspleefi Nov 01 '23 at 20:17
  • @rschwieb yes you are correct ... we should use zorn's Lemma .... my professor said that ..... but what is the solution of the question? why is this maximal ideal prime? – Emptymind Nov 01 '23 at 20:41
  • @Emptymind Again, Every maximal ideal is prime. – rschwieb Nov 01 '23 at 20:43
  • @rschwieb Proposition 11, on pg.254 in Dummit & Foote said that in a ring with identity every proper ideal is contained in a maximal ideal ..... but our ring is without an identity, so we can not find $J$ .... what do you think? – Emptymind Nov 01 '23 at 23:15
  • @Emptymind I think you have not read the comments. As pointed out to you in the first comment, this ring does have an identity. The function that is constantly $1$ is a continuous function from $\mathbb R$ into $\mathbb R$. – rschwieb Nov 02 '23 at 03:19
  • @rschwieb but the identity function does not have a compact support, that is what is said on pg.225 of Dummit & Foote third edition at the end of Example (7). – Emptymind Nov 02 '23 at 09:45
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    @Emptymind The ring has an identity element. The ideal $I$ does not contain this element. – Arthur Nov 02 '23 at 12:28