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Simple problem concerning R[x], R and a prime ideal I

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Dummit and Foote Section 15.4: Localization.

On page 710, D&F make the following statement:

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"In general, suppose R is a commutative ring. If P is a prime ideal in R[x] then \(\displaystyle P \cap R \) is a prime ideal in R ... ..."

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In checking out the validity of this assertion I wish to reason the following way:

Elements of \(\displaystyle P \cap R \) must be constant terms in P of the form \(\displaystyle a,b \in R\).

Given this, we reason as follows:

\(\displaystyle ab \in P \cap R \Longrightarrow ab \in P \) and \(\displaystyle ab \in R \)

Now \(\displaystyle ab \in P \Longrightarrow a \in P \) or \(\displaystyle b \in P \) since P is a prime ideal ... ... ... (1)


But we also have that:

\(\displaystyle ab \in R \Longrightarrow a \in R \) and \(\displaystyle b \in R \) ... ... ... (2) [... ? but why exactly ... does this actually follow from the closure of multiplication ... or not ?]

Then I wish to claim that (1) and (2) give me the following:

\(\displaystyle ab \in P \cap R \Longrightarrow a \in P \cap R \) or \(\displaystyle b \in P \cap R \)

Can someone please confirm that this reasoning is correct?

My problem or uncertainty is with my claim in the above that:

\(\displaystyle ab \in R \Longrightarrow a \in R \) or \(\displaystyle b \in R \)

I am not sure how to justify this reasoning.

I would appreciate a clarification.

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
You don't have to show (2).

To show $P$ (in an arbitrary ring $R$) is a prime ideal, you show that for any $a,b \in R$ such that $ab \in P$, that one of $a,b \in P$.

In other words, we already have, by assumption that $a,b \in R$. We can assume (why?) that for $ab \in P \cap R$, that $a \not\in P \cap R$.

Since $a$ is already GIVEN to be an element of $R$, the only way it can fail to be in $P \cap R$ is if $a \not \in P$ (here we are considering $a$ as belonging to two rings:

the first is as an element of $R$, and the second is as an element of the embedding of $R$ (as constant polynomials) in $R[x]$).

Since $a$ (as an element of $R[x]$) is not in $P$, and since $ab$ (as likewise an element of $R[x]$) IS in $P$, and $P$ is prime, we must have $b \in P$.

But $b$ is also given as belonging to $R$, so $b \in P \cap R$, showing $P \cap R$ is a prime ideal of $R$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
You don't have to show (2).

To show $P$ (in an arbitrary ring $R$) is a prime ideal, you show that for any $a,b \in R$ such that $ab \in P$, that one of $a,b \in P$.

In other words, we already have, by assumption that $a,b \in R$. We can assume (why?) that for $ab \in P \cap R$, that $a \not\in P \cap R$.

Since $a$ is already GIVEN to be an element of $R$, the only way it can fail to be in $P \cap R$ is if $a \not \in P$ (here we are considering $a$ as belonging to two rings:

the first is as an element of $R$, and the second is as an element of the embedding of $R$ (as constant polynomials) in $R[x]$).

Since $a$ (as an element of $R[x]$) is not in $P$, and since $ab$ (as likewise an element of $R[x]$) IS in $P$, and $P$ is prime, we must have $b \in P$.

But $b$ is also given as belonging to $R$, so $b \in P \cap R$, showing $P \cap R$ is a prime ideal of $R$.
Thanks for the help Deveno.

But, just one further clarification ... you write:

"To show $P$ (in an arbitrary ring $R$) is a prime ideal, you show that for any $a,b \in R$ such that $ab \in P$, that one of $a,b \in P$."

The definition of a prime ideal (D&F page 255) is as follows:

Definition. Assume R is commutative. An ideal P is called a prime ideal if \(\displaystyle P \ne R \) and whenever the product of ab of two elements \(\displaystyle a, b \in R\) is an element of P, then at least one of a and b is an element of P.

Now I am assuming that since the definition of a prime ideal P reads "at least one of a and b is an element of P" that we may have the possibility of both elements being contained in P.

Thus I am having trouble following you when you write:

"In other words, we already have, by assumption that $a,b \in R$. We can assume (why?) that for $ab \in P \cap R$, that $a \not\in P \cap R$."

if the situation is that both a and b may be in P, how can we assume that $a \not\in P \cap R$?

Can you help?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
A more complete elaboration goes as follows:

Suppose $a,b \in R$ are such that: $ab \in P$.

If we wish to prove $P$ is prime, we must show that one (or both) of $a,b \in P$.

If $a \in P$, there is nothing to prove.

Thus, we may assume without loss of generality that $a \not \in P$.

(For if $a \in P$ for any such $a$, then $P$ is prime).

In this case, to prove $P$ prime, we must show $b \in P$.

In a more familiar setting:

suppose we have INTEGERS $a,b$ such that $ab = pd$ for some prime integer $p$.

Clearly, $p$ divides $pd$, which means $p$ divides $ab$ (since these are equal).

If $p$ divides $a$, then clearly $a = kp$, that is: $a \in P = (p)$.

If $p$ does NOT divide $a$, then we must have (by the primality of $p$, by definition) $p$ divides $b$.

We have 4 possible cases:

$p|a,p|b$
$p|a,p\not\mid b$
$p \not\mid a,p|b$
$p \not\mid a,p\not\mid b$

The case we want to rule out is the last one. To rule it out, assume $p$ doesn't divide one of them (it doesn't really matter which one, $a$ comes first, and by commutativity of multiplication in integers we can switch their roles). Then we want to show the NEGATION of:

$p\not\mid b$

which is, of course, the same as showing $p|b$.

I hope this helps.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
A more complete elaboration goes as follows:

Suppose $a,b \in R$ are such that: $ab \in P$.

If we wish to prove $P$ is prime, we must show that one (or both) of $a,b \in P$.

If $a \in P$, there is nothing to prove.

Thus, we may assume without loss of generality that $a \not \in P$.

(For if $a \in P$ for any such $a$, then $P$ is prime).

In this case, to prove $P$ prime, we must show $b \in P$.

In a more familiar setting:

suppose we have INTEGERS $a,b$ such that $ab = pd$ for some prime integer $p$.

Clearly, $p$ divides $pd$, which means $p$ divides $ab$ (since these are equal).

If $p$ divides $a$, then clearly $a = kp$, that is: $a \in P = (p)$.

If $p$ does NOT divide $a$, then we must have (by the primality of $p$, by definition) $p$ divides $b$.

We have 4 possible cases:

$p|a,p|b$
$p|a,p\not\mid b$
$p \not\mid a,p|b$
$p \not\mid a,p\not\mid b$

The case we want to rule out is the last one. To rule it out, assume $p$ doesn't divide one of them (it doesn't really matter which one, $a$ comes first, and by commutativity of multiplication in integers we can switch their roles). Then we want to show the NEGATION of:

$p\not\mid b$

which is, of course, the same as showing $p|b$.

I hope this helps.
You write: "I hope this helps"

Indeed, most definitely helpful, Deveno.

Thank you.

Peter