2
$\begingroup$

For a field $k$, let $R:=k[x_1, y_1, x_2, y_2...,x_n, y_n...]$, the polynomial ring over $k$ with infinitely many variables. Now let $I_{(a_1, a_2, a_3,...)}$,  $a_i\in \{0, 1\}$ for all $i$, be the family of ideals of $R$ generated by a subset of $\{x_1, y_1, x_2, y_2...,x_n, y_n...\}$ as follow:

If $a_1=0$, then $x_1$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$, and if  $a_1=1$, then $y_1$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$.

If $a_2=0$, then $x_2$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$, and if  $a_2=1$, then $y_2$ is in the generating set of $I_{(a_1, a_2, a_3,...)}$.

\begin{array}{c}   . \\   . \\   . \end{array}

For example $I_{(0, 0, 1, 0, 1 ,1, 0 ...)}=\langle x_1, x_2, y_3, x_4, y_5, y_6, x_7...\rangle$.

Now let $\Gamma$ be an infinite subset of $\{0, 1\}^\mathbb{N}$, how can we show that there exists $\alpha\in \Gamma$ such that $I_\alpha\subseteq\cup_{\alpha\neq\beta\in\Gamma}I_\beta$.

$\endgroup$
  • $\begingroup$ WhatsUp@ When $\Gamma$ is the whole set $\{0, 1\}^{\mathbb{N}}$ the result is true. Actually, for every $\alpha\in\{0, 1\}^{\mathbb{N}}$, we have $I_\alpha\subseteq\cup_{\alpha\neq\beta\in $\{0, 1\}^{\mathbb{N}}$}I_\beta$. $\endgroup$ – Rostami Aug 14 '16 at 12:46
  • 2
    $\begingroup$ Let $\gamma_i \in \{0, 1\}^{\mathbb{N}}$ be the sequence whose $i$-th term is $1$ and all other terms are $0$. Take $\Gamma$ to be the set $\{\gamma_i: i \in \mathbb{N}\}$. Then there is no $\alpha$ in $\Gamma$ satisfying the required property, since every $\gamma_i$ contains a unique member, namely $y_i$. $\endgroup$ – WhatsUp Aug 14 '16 at 12:53
1
$\begingroup$

If $\Gamma$ is uncountable then there is such $\alpha$. Otherwise, (as WhatsUp commented) there is a counterexample.

Assume that $\Gamma$ is a counterexample, i.e. for every $\alpha \in \Gamma$, $I_\alpha \not\subseteq \bigcup_{\alpha \neq \beta \in \Gamma} I_{\beta}$.

For a given $\alpha \in \{0, 1\}^\mathbb{N}$, $I_\alpha$ is the vector field $$Sp_k \{ \prod_{i \in s} x_i \cdot \prod_{j \in t} y_j \mid s \subseteq \alpha^{-1}(0) \text{ finite }, t \subseteq \alpha^{-1}(1) \text{ finite}\}.$$ Since $I_\alpha \not\subseteq \bigcup_{\alpha \neq \beta \in \Gamma} I_{\beta}$, then there is finite $s, t$, such that $s\subseteq \alpha^{-1}(0), t\subseteq \alpha^{-1}(1)$, and $\prod_{i \in s} x_i \cdot \prod_{j \in t} y_j \notin I_\beta$ for every $\beta \in \Gamma$, $\beta \neq \alpha$.

Thus, we can define a one to one function sending $\alpha$ to any such pair $(s, t)$. Since there are only countably many pairs of finite subsets of $\mathbb{N}$, $\Gamma$ must be countable.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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