Timeline for Pushout and pullback in the category of rings

Current License: CC BY-SA 4.0

12 events

when toggle format	what		by	license	comment
Dec 19, 2020 at 14:53	comment	added	მამუკა ჯიბლაძე		Do you mean all associative rings or only commutative ones?
Dec 18, 2020 at 18:56	answer	added	Bertram Arnold		timeline score: 8
Dec 18, 2020 at 18:07	comment	added	Bertram Arnold		I claim that the multiplication map $\mathbb Q\otimes_{\mathbb Z}\mathbb Q\to\mathbb Q$ and the "left inclusion" $\mathbb Q\to\mathbb Q\otimes_{\mathbb Z}\mathbb Q$ are inverse bijections. The composition $\mathbb Q\to\mathbb Q$ is obviously the identity, so that the other composition is surjective since $\frac{m_1}{n_1}\otimes \frac{m_2}{n_2} = n_2\frac{m_1m_2}{n_1n_2}\otimes \frac{1}{n_2} = \frac{m_1m_2}{n_1n_2}\otimes 1$. But a retraction where the inclusion map is surjective is already an isomorphism.
Dec 18, 2020 at 15:37	comment	added	Bertram Arnold		The statement is wrong: Take $A = \mathbb Z, B = C = \mathbb Q$. Then $D = \mathbb Q$, and the maps from $B$ and $C$ are isomorphisms, so that the pullback is $\mathbb Q$.
Dec 18, 2020 at 15:09	history	edited	Nil123	CC BY-SA 4.0	added 25 characters in body
Dec 18, 2020 at 15:01	comment	added	Nil123		Yes, I'm trying to prove that it's a pullback diagram @HarryWilson
Dec 18, 2020 at 14:59	comment	added	Harry Wilson		What are you trying to prove? That it's also a pullback diagram?
Dec 18, 2020 at 14:52	comment	added	Nil123		I believe that the statement is true. I did not find any argument to prove it. If it is not true, please give a counter-example.
Dec 18, 2020 at 14:49	history	edited	Nil123	CC BY-SA 4.0	added 23 characters in body
Dec 18, 2020 at 14:36	review	Close votes
Dec 25, 2020 at 3:07
Dec 18, 2020 at 13:41	review	First posts
Dec 18, 2020 at 14:09
Dec 18, 2020 at 13:38	history	asked	Nil123	CC BY-SA 4.0