Question

(7) Let R= {f [0,1] - R | f continuous} be the ring of all continuous functions from the interval [0,1] to the real numbers. (a) For cE [0, 1, prove that Me := {feR | f(c) = 0} is a maximal ideal of R. Hint: consider the evaluation map ec- (b) Show that if M is any maximal ideal of R then there exists a cE [0,1 such that M = Me. Hint: show that any maximal ideal M contains Mc for some c E [0,1]

Answer 1

7. a) Recall for a commutative ring with unity, an ideal $m$ is maximal iff $R/m$ is an field. We will show that $R/M_a$ is a field.

Let $\bar{f}\in R/M_a$ be a non-zero element. To show $\bar{f}$ has an inverse. Note that since $\bar{f}\in R/M_a$ be a non-zero element, $f(c)\not=0$. Define $h(x)=\frac{f(x)}{f(c)}$ , Then note that $h(x)-1\in M_c$, as $(h-1)(c)=\frac{f(c)}{f(c)}-1=0$ . Hence we get

$f(x)\frac{1}{f(c)}=1+(h(x)-1)$, hence going modulo $M_c$ we get $\bar{f(x)}{\frac{1}{\bar{f(c)}}}=\bar{1}$ , Hence $\bar f$ is an unit. hence done the maximal ideal part.

b) Using the compactness of $[0,1]$ we will show that all maximal ideals of R is of the form $M_c$, for some $c\in [0,1]$.

Proof by contradiction. Suppose there exists some maximal ideal which is not of the form $M_c$ for any $c\in[0,1]$. Then for all $c\in[0,1]$ there exists $g_c\in M$ be such that $g_c(c)=0$, note that since $g_c$ is continuous there exists an open neighbourhood $U_c$ of c, such that $x\in U_c=> g_c(x)\not=0$ . Thus we get an open cover $\{U_c\}_{c\in [0,1]}$ of $[0,1]$, hence by compactness will get a finite sub cover say $\{U_1,...,U_n\}$ , and functions $g_1,...,g_n\in M$, such that $x\in U_i=>g_i(x)\not=0$ . Consider $g(x)=g_1(x)^2+...+g_n^2(x)\in M$ . Note that $g\in M$as each $g_i\in M$, and $g(x)\not=0$ for all $x\in [0,1]$, as if for some $c\in [0,1]$, $g(c)=0$, means we can find $U_j\subset \{U_1,...,U_n\}$ be such that $c\in U_j$, and $g(c)=0=>g_j(c)=0$ , which is a contradiction. Hence there will not exists any maximal ideal $M$ other than $M_c$.

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