Remark 61.8.2. Let $A$ be a ring. Let $\kappa $ be an infinite cardinal bigger or equal than the cardinality of $A$. Then the cardinality of $T(A)$ is at most $\kappa $. Namely, each $B_ E$ has cardinality at most $\kappa $ and the index set $I(A)$ has cardinality at most $\kappa $ as well. Thus the result follows as $\kappa \otimes \kappa = \kappa $, see Sets, Section 3.6. It follows that the ring constructed in the proof of Lemma 61.8.1 has cardinality at most $\kappa $ as well.

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (0)