$B\otimes_A A/\mathfrak{m}$ a domain with $B$ not a domain

Question

I want to find an example of where $A$ is a dedekind domain, $B$ is a flat algebra over $A$, $m$ is some maximal ideal of $A$ with $B\otimes_A A/\mathfrak{m}$ a domain but $B$ is not a domain.

Specifically could you hint me towards how you would come up with one of these. This is my thought process so far: We want $B$ to have some zero divisors, $ab=0$. But since we want $B\otimes_A A/\mathfrak{m}$ a domain we do not want $(a\otimes 1)$ to be non zero since then $(a\otimes 1)(b\otimes 1)=0$ and then $B\otimes_A A/\mathfrak{m}$ is not a domain. But how to try an construct a ring so that $(a\otimes 1)$ is zero? My best guess would be to have all the zero divisors in $B$ to be elements in $\mathfrak{m}$. But I don't see how we construct this?

Lukas Heger · Accepted Answer · 2020-12-18T00:31:57.787

2

Consider $A=\Bbb Z$, $\mathfrak{m}=(p)$ for some prime number and $B=\Bbb Z \times\Bbb Q$. This is a flat algebra. We have $B \otimes_A A/\mathfrak{m} \cong \Bbb Z/(p)$ which is a field, but $B$ is not a domain.

edited Dec 18 '20 at 00:31

answered Dec 18 '20 at 00:21

Lukas Heger

20,801

Its not clear at all to me why $B$ is not a domain? Oh $(1,0)\cdot(0,1)$ – samlanader Dec 18 '20 at 00:43
Could you elaborate on your thinking process behind this? – samlanader Dec 18 '20 at 00:47

$B\otimes_A A/\mathfrak{m}$ a domain with $B$ not a domain

1 Answers1