Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am working on the following problem:

Suppose that $R$ is the direct product of a family of fields indexed by a set $X$ (and hence is easily seen to be a ring). Show that $I \mapsto F_I := \{U \subset X : \exists f \in I$ s.t. $f(x)=0 \iff x \in U \} $ defines a bijection between the set of proper ideals in R and the proper filters $F$ on $X$. Also show that the following are equivalent:

(i) $R/I$ is an integral domain in which $0 \neq 1$

(ii) $R/I$ is a field

(iii) $F_I$ is a proper ultrafilter

All I have shown so far is that $F_I$ is indeed a filter. However, I am not sure about the rest of this problem. I would greatly appreciate help with this. Thank you.

share|improve this question
1  
Let $J$ be an ideal of $R$. Then $J$ restricted to one of the fields $F_x$ is an ideal of $F_x$, and therefore trivial, either $0_x$ or all of $F_x$. That observation will get you the bijection between proper ideals and proper filters. –  André Nicolas Nov 21 '12 at 20:18
    
@AndréNicolas: I don't see how this gives the bijection? Could you please explain this? Thank you very much. –  user49097 Nov 21 '12 at 20:46
add comment

1 Answer

up vote 1 down vote accepted

$\def\Pot{\mathscr P}\def\ideal#1{\left\langle#1\right\rangle}$For a Filter $F \subseteq \Pot(X)$, define an ideal $I_F$ by $$ I_F := \ideal{\chi_U \bigm| U \in F} $$ where $\ideal{}$ denotes the generated ideal and $\chi_U\in \prod_x F_x$ (one minus) the characteristic function of $U$, that is $\chi_U(x) = 0_x \iff x \in U$ and $\chi_U(x) = 1_x\iff x \not\in U$. $I_F$ is an ideal by definition and it is not proper if $1 \in J_F$, that is there exist $f_i \in R$, $U_i \in F$ with $1 = \sum_{i=1}^n f_i \chi_{U_i}$, that is for every $x$ $$ 1 = \sum_{i=1}^n f_i(x)\chi_{U_i}(x) $$ this gives $x \not\in \bigcap_{i=1}^n U_i$. As $x$ was arbitrary, $\emptyset = \bigcap_{i=1}^n U_i \in F$, so $F$ wasn't proper.

Now let's show that the operations $I \mapsto F_I$ and $F \mapsto I_F$ are inverse to each other, we have:

  • Let $U \in F_{I_F}$, the there is an $f \in I_F$ such that $x \in U$ iff $f(x) = 0$. As $f \in I_F$ we can write $f = \sum_i f_i \chi_{U_i}$ for finitely many $U_i \in F$. If $x \in \bigcap_i U_i$ we have $f(x) = 0$, so $\bigcap_i U_i \subseteq U$. As $F$ is a filter $U \in F$.
  • Let $U\in F$, then $\chi_U \in I_F$ and hence $U \in F_{I_F}$ for we can take $\chi_U$ as $f$ in the definition of $F_{I_F}$.
  • Let $f \in I_{F_I}$, then there are $f_i \in R$ and $U_i \in F_I$ with $f = \sum_i f_i U_i$. Now for each $U_i$, there is an $g_i \in I$ with $g_i(x) = 0$ iff $x \in U_i$. Define $h_i\in R$ by $$ h_i(x) =\begin{cases} g_i(x)^{-1} & g_i(x) \ne 0\\ 0 & g_i(x) = 0\end{cases} $$ then $\chi_{U_i} = h_i\cdot g_i \in I$ and hence $f = \sum_i f_i\chi_{U_i} \in I$.
  • Let $f \in I$, let $U = \{x : f(x) = 0\}$, we have $\chi_U \in I_{F_I}$, but $f = f\chi_U \in I_{F_I}$.

So $I \mapsto F_I$ is bijective.

(i) $\to$ (ii): Let $f \in R$ with $f \not\in I$. Define $g \in R$ by $$ g(x) = \begin{cases} f(x)^{-1} & f(x) \ne 0\\ 0& f(x) = 0\end{cases} $$ Then $f\cdot g \cdot f = 1 \cdot f$, and as $R/I$ is integral $g\cdot f + I =1 + I$, that is every $f + I\in R/I$, $f \not\in I$ is invertible and hence $R/I$ is a field.

(ii) $\leftrightarrow$ (iii): Note that $F \mapsto I_F$ is monotone, hence bigger filters give bigger ideal and $I\mapsto F_I$ is also. As $R/I$ is a field iff $I$ is maximal, this gives $R/I$ is a field iff $F_I$ is a maximal proper filter, that is an ultrafilter.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.