Question

11. Show that if P is a prime ideal of D, then Dp = {a/b € Q(D) | b & P} is an integral domain with DCDP SQ(D).

All I know is that this problem entails localization of a prime...

Answer 1

Solution:-

Given that

So, you need to prove that, is prime

Pc D

iff $\frac{D}{P}$ is an integral domain

a De b

Now, firstly let suppose
Pc D
, & try to prove that $\frac{D}{P}$ is an integral domain.

we have to prove commutative ving and ith multiplicative identity & non zero divisor

but $D_p$ is already an commutativering with identity D.

$D_p=\left \{ \frac{a}{b}\in Q(D) \right \},a,b\in Q(D)$

i.e., $a,b\in P$ for non zero element,

a + P & b + P be non zero elements, $a,b\in P$

Since P is an prime, $a\cdot b\notin P$

$a\cdot b+ P$ is also an non zero as we started.

any two non zero elements of $D_p$ have non zero product, i.e., non zero divisor.

Therefore $D_p$ is an integral domain.

conversly; we assume $D\subseteq D_p$

Such that, $D_p=\frac{D}{p}$ is an integral domain.

So, we need to prove that P is an prime ideal.

for this we need to show;

if $ab\in p$ then neither $a\in p$ or $b\in p$

but if $ab\in p$ then $ab+p=p$

but we know, $ab+p=(a+p)(b+p)$ is zero in $D_p$

but $D_p$ is an integral domain by assumption

It can't have divisors

This proves that P is prime ideal for D

Hence proved.

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