Tagged Questions

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Help understanding statement relating to structure of modules over PIDs

Lemma IV.6.11 of Hungerford's Algebra is Lemma 6.11. Let $R$ be a principal ideal domain. If $r \in R$ factors as $r = p_1^{n_1} \cdots p_k^{n_k}$ with $p_1,\ldots,p_k \in R$ distinct primes and ...
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Quotient of a PID by a prime ideal [duplicate]

Prove that quotient of a PID by a prime ideal is PID.
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Specific way of showing $\Bbb Z[\sqrt{-d}]$ is not a Euclidean Domain when $d>2$

Is it true that if a ring is not a UFD then it's not a Euclidean Domain? I have a ring $R=\mathbb{Z}[\sqrt{-d}]=\{ a+b\sqrt{-d} \mid a,b \in \mathbb{Z} \}$ where $d$ is a square free integer. I want ...
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Describe units and maximal ideals in these two PIDs

If $p$ is a fixed prime integer, let $R$ be the set of all rational numbers that can be written in the form $(a)$ $\frac{a}{b}$ with $b$ not divisible by $p$. $(b)$ $\frac{a}{b}$ with $b=p^k$ for a ...
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Showing that an integral domain is a PID if it satisfies two conditions

This is just a textbook problem from Dummit and Foote, but the issue is that our class barely touched on PIDs and the preceding material, so I don't really know or understand much. Anyway, Let ...
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Why $\langle I, J\rangle =R$ for distinct prime ideals $I$, $J$ of a principal ideal domain $R$?

Let $R$ be a principal ideal domain with identity and $I$, $J$ be distinct prime ideals of $R$. Prove that $1 \in \langle I, J\rangle$ hence $\langle I, J\rangle = R$. How to prove?
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Subring of the field of rational numbers

Let $R=\{a\cdot2^n\mid a,n \in \mathbb{Z}\}$ be a subring or the field of rational numbers $\mathbb Q$. i) What kind of elements are invertible in $R$? ii) Prove that $R$ is a principal ...
Quotient ring of $\Bbb Z[x]$ by an irreducible polynomial is a PID
I don't know what can I do with this problem. How can I prove that $\mathbb{Z}[x]/(x^{3}-4x+2)$ is PID?
$\mathbb Z\times\mathbb Z$ is principal but is not a PID
I need to find an example of a ring that is not a PID but every ideal is principal. I know that $\mathbb Z\times\mathbb Z$ is not an integral domain, so certainly is not a PID, but here every ideal is ...