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Let $C[0,1]$ be the space of continuous real-valued functions on the interval $[0,1]$. This is a ring under point-wise addition and multiplication. Which of the following are true:

(a) For any $x ∈ [0,1]$, the ideal $M (x) = \{f ∈ C[0, 1] : f (x) = 0\}$ is maximal.

(b) $C[0, 1]$ is an integral domain.

(c) The group of units of $C[0, 1]$ is cyclic.

(d) The linear functions form a vector-space basis of $C[0, 1]$ over $\mathbb R$.

i know a statement is true because it is for any x not $every x$ therefore (a) is maximal ideal.

Because space is of real continuous real valued functions therefore (b) is true. I don't know how to prove or disprove (c) (d). thanx in advance

user26857
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Gautam Singla
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    I believe (a) is true, (b) is false. I'm pretty sure (c) is also false, how could it be cyclic? It's way too big. – Gregory Grant Feb 02 '16 at 16:39
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    For b), consider the function $f$ that is $0$ up to $x=1/2$ and then climbs linearly, say $f(x)=x-1/2$. Let $g(x)$ be the function which is $1/2-x$ up to $1/2$ and then $0$. – André Nicolas Feb 02 '16 at 16:39
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    @gautam you ought to try searching for these questions individually. That would definitely get answers for a) and b). It is also just a bad idea in general to post a burst of more than three questions in a single post. Simple is better than complex. – rschwieb Feb 02 '16 at 16:44
  • d) is false for all sorts of reasons. You should be able to come up with some thoughts on that one. – rschwieb Feb 02 '16 at 16:46
  • @gregory grant sir if i am not wrong here we are talking about continuous function so $C[0,1]$ is integral domain – Gautam Singla Feb 02 '16 at 16:48
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    @GautamSingla No, consider the function $f(x)=\left{\begin{array}[ll]\ 0 & 0\leq x\leq 1/2\ x-1/2 & 1/2\leq x\leq 1\end{array}\right.$. And let $g(x)=\left{\begin{array}[ll]\ x-1/2 & 0\leq x\leq 1/2\ 0 & 1/2\leq x\leq 1\end{array}\right.$. Then $f$ and $g$ are both continuous and not identically zero, but $f\cdot g$ is identically $0$. – Gregory Grant Feb 02 '16 at 17:08
  • Maximal ideals in the ring of real functions on $[0,1]$ – Jesse P Francis May 11 '16 at 14:52

1 Answers1

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a),b) and d) have been treated in the comments. For c) you should notice that the constant functions give you a natural inclusion $\mathbb R^* \subset C[0,1]^*$ and $\mathbb R^* = \mathbb R \setminus \{0\}$ is not even countable, in particular not cyclic.

MooS
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