Mathematics

Questions tagged [principal-ideal-domains]

For questions about principal ideal domains: rings without zero divisors where every ideal is principal.

A PID is a type of integral domain where every proper ideal can be generated by a single element. Every Euclidean ring is a PID but the converse is not true. Every PID is a unique factorisation domain but not conversely.

Examples of PIDs are any field, $\mathbb{Z}$, rings of polynomials, Gaussian integers, and Eisenstein integers.

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Submodule of free module over a p.i.d. is free even when the module is not finitely generated?

I have heard that any submodule of a free module over a p.i.d. is free. I can prove this for finitely generated modules over a p.i.d. But the proof involves induction on the number of generators, so it does not apply to modules that are not finitely…
Ben Blum-Smith
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Proving some number rings are PIDs but not Euclidean, e.g. $\Bbb{Z}[\frac{1+\sqrt{-19}}{2}]$

I have noticed that to prove fields like $\mathbb{Q}(i)$ and $\mathbb{Q}(e^{\frac{2\pi i}{3}})$ have class number one, we show they are Euclidean domains by tessellating the complex plane with the points $a+bv : a, b \in \mathbb{Z}$ , where $1, v$…
D_S
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Is $\mathbb{Z}[x]$ a principal ideal domain?

Is $ \mathbb{Z}[x] $ a principal ideal domain? Since the standard definition of principal ideal domain is quite difficult to use. Could you give me some equivalent conditions on whether a ring is a principal ideal domain?
KunXian Xia
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4 answers

An integral domain whose every prime ideal is principal is a PID

Does anyone has a simple proof of the following fact: An integral domain whose every prime ideal is principal is a principal ideal domain (PID).
fmoura2005
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A subring of the field of fractions of a PID is a PID as well.

Let $A$ be a PID and $R$ a ring such that $A\subset R \subset \operatorname{Frac}(A)$, where $\operatorname{Frac}(A)$ denotes the field of fractions of $A$. How to show $R$ is also a PID? Any hints?
CC_Azusa
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2 answers

Quotient of polynomials, PID but not Euclidean domain?

While trying to look up examples of PIDs that are not Euclidean domains, I found a statement (without reference) on the Euclidean domain page of Wikipedia that $$\mathbb{R}[X,Y]/(X^2+Y^2+1)$$ is such a ring. After a good deal of searching, I have…
Jessica B
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Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal

Prove that a UFD is a PID if and only if every nonzero prime ideal is maximal. The forward direction is standard, and the reverse direction is giving me trouble. In particular, I can prove that if every nonzero prime ideal is maximal then every…
JeremyKun
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$A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ is a PID.

I was given an exercise to show $A=\frac{\mathbb{C}[X,Y]}{(X^2+Y^2-1)}$ is a PID. But I wonder if it is true. Note that PID $\implies$ UFD. But we have $$X\cdot X = 1-Y^2 =(1-Y)(1+Y)$$ in $A$ which contradicts UFD. Is there something wrong in the…
Sayan
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In a PID every nonzero prime ideal is maximal

In a principal ideal domain, prove that every non trivial prime ideal is a maximal ideal Attempt: Let $R$ be the principal ideal domain. A principal ideal domain $R$ is an integral domain in which every ideal $A$ is of the form $\langle a \rangle =…
MathMan
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Prove that $n^2+n+41$ is prime for $n<40$

Here's a problem that showed up on an exam I took, I'm interested in seeing if there are other ways to approach it. Let $n\in\{0,1,...,39\}$. Prove that $n^2+n+41$ is prime. I shall provide my own solution, though I am curious does anyone know how…
Stella Biderman
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6 answers

The ring $\Bbb Z\left [\frac{-1+\sqrt{-19}}{2}\right ]$ is not a Euclidean domain

Let $\alpha = \frac{1+\sqrt{-19}}{2}$. Let $A = \mathbb Z[\alpha]$. Let's assume that we know that its invertibles are $\{1,-1\}$. During an exercise we proved that: Lemma: If $(D,g)$ is a Euclidean domain such that its invertibles are $\{1,-1\}$,…
Jacopo Notarstefano
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3 answers

Is the quotient ring of a PID a PID?

Let $A$ be a commutative ring and $S$ a multiplicative closed subset of $A$. If $A$ is a PID, show that $S^{-1}A$ is a PID. I've taken an ideal $I$ of $S^{-1}A$ and I've tried to see that is generated by one element; the ideal $I$ has the form…
Sadi
21
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7 answers

Show that every ideal of the ring $\mathbb Z$ is principal

Let $\mathbb Z$ be the ring of integers. The question asks to show that every ideal of $\mathbb Z$ is principal. I beg someone to help me because it is a new concept to me.
Mwanginde Geofrey
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Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field.

I've just read a proof of the statement : Let $R$ be a commutative ring. If $R[X]$ is a principal ideal domain, then $R$ is a field. In one part of the proof there is a step which I don't understand. I'll copy the proof : Let $u\in R$ be non-zero.…
user100106
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Projective module over a PID is free?

A common result is that finitely generated modules over a PID $R$ are projective iff they are free. Is the same true that an arbitrary projective module over a PID is free? I can't find this fact anywhere, so I suspect it is false, but I can't…
Hana Bailey
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