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Not every ideal is prime

Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.4 Exercise 7.4.34


(1) First we show that $I$ is an ideal. Let $f,g \in I$; then for some real numbers $m$ and $n$, we have $f(x) = 0$ when $|x| \geq m$ and $g(x) = 0$ when $|x| \geq n$. Then whenever $|x| \geq \max(m,n)$, $(f-g)(x) = 0$, so that $f-g \in I$. Certainly the zero function has compact support, so that $I$ is nonempty. Now let $h \in R$; when $|x| \geq m$, $(hf)(x) = 0$. Thus $I$ absorbs $R$, and so $I$ is an ideal. We now show that $I$ is not prime. Define $f(x) = x+1$ if $x \geq -1$ and 0 otherwise, and define $g(x) = -x+1$ if $x \leq 1$ and 0 otherwise. Certainly $f$ and $g$ are continuous and $fg$ has compact support, but neither $f$ nor $g$ is in $I$; thus $I$ is not prime.

(2) Let $M \subseteq R$ be a maximal ideal containing $I$. Let $c \in \mathbb{R}$. Consider $f_c(x) = x+1-c$ if $x \in (c-1,c)$, $-x+1+c$ if $x \in [c,c+1)$, and 0 otherwise. Certainly $f_c$ has compact support, so $f_c \in I \subseteq M$. But $f_c(c) \neq 0$, so $f_c \notin M_c$; thus $M \neq M_c$.


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