4
$\begingroup$

All rings below are assumed to be commutative with unity.

Let $A\subseteq B$ be an integral extension of integral domains ($A,B$ are both integral domains and $B$ integral over $A$). If $p\in A$ is a prime element in $A$ then is $p$ a prime element in $B$ also ? If this is not true in general, what if we take $B$ to be the normalization of $A$ (integral closure of $A$ in its fraction field), is the result true then ?

$\endgroup$
1

1 Answer 1

Reset to default
3
$\begingroup$

In general prime elements do not remain prime in integral extensions.
For example in the situation $A=\mathbb Z\subset B=\mathbb Z[i]$, the prime $2\in A$ does not remain prime in $B$ since $2$ divides $2=(1+i)(1-i)$ without dividing any of its two factors $1\pm i$.

$\endgroup$
1
  • $\begingroup$ Thanks, really simple example. Please give the normalization question some thought if possible $\endgroup$
    – user495643
    Jan 26, 2018 at 13:30

You must log in to answer this question.