# Tagged Questions

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

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### Principal Ideal Domain and Factorization

If $A$ is a local domain such that each non-trivial ideal factors uniquely into primes then does it follow that $A$ must be a principal ideal domain?
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### Application of Structure Theorem to Prove Simultaneous Diagonalizability and Group of Units of Cyclic Groups

I am reading these notes on Modules over PID. Exercise 67 (pg 24) asks to prove that: Problem. Let $A$ and $B$ be $n\times n$ matrices with complex entries. Then $A$ and $B$ are simultaneously ...
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### $\Bbb Z[\sqrt{-5}]$ is not a PID [duplicate]

I want to show: In a PID $R$ two elements $a,b\in R$ always have a greatest common divisor. Therefore $\Bbb Z[\sqrt{-5}]$ is not a PID. For the first part: $I=\{ax+by:x,y\in R\}$ is an ideal, ...
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### Is $\Bbb Z[i]$ a Euclidean ring? [duplicate]

Is $\Bbb Z[i]$ a Euclidean ring? If not, what would be the simplest way of seeing that $\Bbb Z[i]$ is a PID?
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### $R/Rg$ is a field iff $g\in R$ is irreducible.

Let $R$ be a PID and $g\in R$. I want to show: $R/Rg$ is a field iff $g\in R$ is irreducible. I.e. I want to show that all $a\notin Rg$ are invertible modulo $g$ iff $g$ is irreducible. So if I ...
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### In a local ring, the maximal ideal $\mathfrak{m}$ is principal $\implies \dim_k(\mathfrak{m}/\mathfrak{m}^2)\leq 1$

This is Proposition 8.8, $ii)\implies iii)$ in Atiyah and Macdonald and it says there that this is clear. It isn't for me though. How would I prove this?
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### Why is a discrete valuation ring's unique prime ideal generated by a non-nilpotent element?

Serre gives first the definition of a discrete valuation ring as a "principal ideal domain that has a unique non-zero prime ideal $m(A)$." Next, he says: Let $A$ be a commutative ring. In order that ...
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### Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$

Is it true that all modules over $\mathbb{Q}$ are of form $\mathbb{Q}^n$ where $n\geq 0$ I feel like this is true because $\mathbb{Q}$ is a principal ideal domain
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### Can we characterize all infinite Euclidean-domains having exactly one invertible element?

$\mathbb Z_2$ and $\mathbb Z_2[x]$ are two euclidean-domains having exactly one invertible element ; my question is ; Can we characterize all euclidean domains $D$ having exactly one invertible ...
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### Luröth's Theorem

I've been struggling trying to understand the Jacobson's Basic Algebra vol. II proof of the Luröth's theorem. Let $K$ be a field, $K(X)$ the field of rational fonctions and take $L$ to be a sub-...
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### Finding the structure of an $F_p[X]$module

$p$ is a prime and $M$ is an $F_p[X]-$ module. Given $(X-1)^3M=0, \vert (X-1)^2M\vert=p, \vert (X-1)M\vert=p^3$ and $\vert M\vert =p^7$, determine $M$ as an $F_p[X]-$module, up to isomorphism. Using ...
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### Cardinal of quotient rings of gaussian integers. [duplicate]

It is known that $\mathbb{Z}[i]$ is a PID and that $\mathbb{Z}[i]/(a+bi)\mathbb{Z}[i]$ is finite for all $(a,b) \in \mathbb{Z}^2\backslash \{(0,0)\}$. My question : Is there any result on the ...
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### In a P.I.D., if $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$ with $\gcd(m,n) = 1$, then $a=b$

Let $R$ be a principal ideal domain, and $a, b \in R$ with $a^m = b^m$ and $a^n = b^n$ for $m, n \in \mathbb{N}$, and $\gcd(m, n) = 1$. I now want to show that we then already have $a = b$. I think ...
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### The principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal.

I'm trying to prove that the principal ideal $(1+i)$ in $\mathbb{Z[i]}$ is maximal. My approach: $m+in \equiv (m-n)$ mod $(1+i), \forall m,n \in \mathbb{Z}$ and $2 \equiv 0$ mod $(1+i)$ then what?
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### Decomposition of Free Module over PID

Let $M$ be a free module with rank $2$ over the PID $R$ having basis $B=\{b_1,b_2\}$. I have a few questions regarding the submodule $Rm$, where $m$ is some element of our module $M$. Suppose $m$ is ...
Let $R$ be a field. Show that the polynomial ring $R[x_{1},...x_{n}]$ is not a PID if $n>1$. How do I show this? So far, I've shown that $(2)+(x)$ wouldn't be principal in $\mathbb{Z}[x]$ so I ...
### Is $\mathbb{Z}_p[\mathbb{Z}_p]$ a PID?
Is $\mathbb{Z}_{p}[G]$ a PID, where $G=(\mathbb{Z}_{p},+)$ is the additive group of the $p$-adics $\mathbb{Z}_{p}$? I am studying a paper where the authors implicitly use that claim, but it is ...