On cardinality of generators

Question

I've read this paper and found a problem to prove Proposition 3.7. The proposition said that "from $\mu(I + aR) < \alpha$ it is straightforward to verify that there is a right ideal $I_0 \subseteq I$ with $\mu(I_0 ) < \alpha$ such that $$I + aR = I_0 + aR."$$ Where $\mu(M)$ denote the smallest cardinal $\mu$ such that the module $M_R$ can be generated by a set of cardinality $\mu$.

I've tried to construct $I_0 = I \cap S$, where $\langle S \rangle = I + aR$. But it failed because I cannot prove $I + aR = I_0 + aR$.

How to find $I_0$?

Answer 1

What you need is an observation that can be formulated in a more general setting (for algebraic closure systems), however we shall limit ourselves to the closure system of left ideals in a given ring $A$. For any left ideal $I \subseteq A$ we shall denote the smallest cardinal of a generating system for $I$ by $\mathrm{rk}\ I$.

Lemma. Let $\mathbf{a}$ be an infinite cardinal, $I$ a left ideal such that $\mathrm{rk}\ I < \mathbf{a}$ and $S \subseteq I$ a generating system for $I$. Then there exists a generating system (for $I$) $T \subseteq S$ such that $|T| < \mathbf{a}$.

Proof: Fix a generating system $R \subseteq I$ such that $|R|=\mathrm{rk}\ I$; for any $x \in I$ there will exist a finite subset $P_x \subseteq S$ such that $x \in (P_x)_{\mathrm{s}}$ (I use the notation $(X)_{\mathrm{s}}$ to denote the left ideal generated by subset $X$). Consider now the set

$$T=\bigcup_{x \in R} P_x$$

and let us argue that it satisfies indeed the desired properties:

• since $P_x \subseteq S$ for any $x \in I$, clearly $T \subseteq S \subseteq I$
• since for each $x \in R$ we have $x \in (P_x)_{\mathrm{s}} \subseteq (T)_{\mathrm{s}}$ we conclude that $R \subseteq (T)_{\mathrm{s}}$ and hence $I=(R)_{\mathrm{s}} \subseteq (T)_{\mathrm{s}} \subseteq I$, so $R$ indeed generates $I$
• since all subsets $P_x$ are finite and thus $|P_x| < \aleph_0$ we have the bound $$|T| \leqslant \sum_{x \in R}|P_x| \leqslant \sum_{x \in R} \aleph_0 =|R| \aleph_0 \tag{*}$$

on the basis of which we make the following discussion by cases: if $R$ were infinite to begin with, then $|R| \aleph_0=|R| < \mathbf{a}$, so by virtue of relation (*) we have $|T| < \mathbf{a}$; if on the other hand $R$ were finite, then $T$ is also finite as a finite union of finite sets and thus once again $|T| <\aleph_0 \leqslant \mathbf{a}$. $\Box$

In your case you can apply the proposition to the ideal $J=I+Aa$ (I prefer to denote rings with $A, B, C$ etc, so I changed the notation a bit) and to its generating system $I \cup \{a\}$ in order to infer the existence of a generating system (for $J$) $S \subseteq I \cup \{a\}$ such that $|S| < \mathbf{a}$. If you proceed to consider $I'=(S \setminus \{a\})_{\mathrm{s}}$ then it follows effortlessly that:

• $I' \subseteq I$, since $S \setminus \{a\} \subseteq I$
• $\mathrm{rk}\ I' \leqslant |S \setminus \{a\}| \leqslant |S| < \mathbf{a}$
• $I'+Aa=(S \setminus \{a\})_{\mathrm{s}}+(a)_{\mathrm{s}}=(S \cup \{a\})_{\mathrm{s}} \supseteq (S)_{\mathrm{s}}=J$, the reverse inclusion being obvious.