Show Finite Principal Ideals Contain $(d)$ in U.F.D.

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In summary, the conversation discusses the idea of showing that there are finitely many different principal ideals that contain a given ideal in a U.F.D. It is stated that for an element to be in an ideal, it must divide the generator of the ideal. It is also mentioned that in a U.F.D., the irreducible elements are prime. The conclusion is that the divisors of a given element can be represented in a finite set.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

Let $R$ be a U.F.D. and $0\neq d\in R$.
I want to show that there are finitely many different principal ideals that contain the ideal $(d)$.

A principal ideal is generated by a single element, say $i$, and so that it contains the ideal $(d)$, $i$ must divide $d$, right? (Wondering)
We have that $d=a_1^{k_1}\cdots a_r^{k_r}$ with $a_i$ irreducible.
Since $R$ is a U.F.D. the irreducible elements are prime. Does it follow from that that the divisors of $d$ are of the form $a_1^{j_1}\cdots a_r^{j_r}$ with $0\leq j_i\leq k_i$ ? (Wondering)
From that we get that the set of the divisors is finite, right? (Wondering)
 
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  • #2
Yes.
 
  • #3
Deveno said:
Yes.

Nice... Thank you! (Yes)
 

FAQ: Show Finite Principal Ideals Contain $(d)$ in U.F.D.

What is a U.F.D.?

A U.F.D. is a unique factorization domain, which is a type of commutative ring in abstract algebra where every non-zero and non-unit element can be expressed as a unique product of irreducible elements.

What is a principal ideal?

A principal ideal is an ideal in a ring that is generated by a single element. In other words, it is the set of all elements that can be obtained by multiplying the generator by any element in the ring.

What does it mean for a principal ideal to contain $(d)$?

When a principal ideal contains $(d)$, it means that the generator of the ideal is a multiple of $d$. In other words, every element in the ideal can be obtained by multiplying $d$ by an element in the ring.

Why is it important for finite principal ideals to contain $(d)$ in a U.F.D.?

In a U.F.D., every non-zero element can be expressed as a unique product of irreducible elements. Therefore, if a finite principal ideal contains $(d)$, it means that every element in the ideal can be expressed as a multiple of $d$, which helps ensure that the ideal is closed under multiplication.

How can one prove that finite principal ideals contain $(d)$ in a U.F.D.?

This can be proven using the unique factorization property of U.F.D.s. By expressing the generator of the principal ideal as a product of irreducible elements, and then expressing $d$ as a product of those same irreducible elements, it can be shown that every element in the ideal is a multiple of $d$.

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