All Questions
Tagged with abstract-algebra prime-factorization
97
questions
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Primes (i.e irreducibles) have no nontrivial factorizations. [duplicate]
I am reading Herstein and it makes the following claim.
The sentence followed by the definition is what I don't get.
A prime element $\pi \in R$ has no non-trivial factorisation in $R$.
By definition,...
2
votes
2
answers
45
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Does the monoid of non-zero representations with the tensor product admit unique factorization?
Let $(M, \cdot, 1)$ be a monoid. We will now define the notion of unique factorization monoid. A non-invertible element in $M$ is called irreducible if it cannot be written as the product of two other ...
0
votes
1
answer
69
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Reducible/Irreducible Polynomials in Ring Theory
I have this following exercise I've been trying to solve for a while now.
We are supposed to study the irreducibility of the polynomial $A=X^4 +1$ in $\mathbb{Z}[X]$ and in $\mathbb{Z}/p \mathbb{Z}$ ...
0
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2
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65
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prime factorization in $\mathbb{Z}[i]$ [duplicate]
We were asked to show where the following reasoning goes wrong. Since $1+i$ and $1-i$ are prime elements in $\mathbb{Z}[i]$, the equation $$(-i)(1+i)^2=(1+i)(1-i)=2$$ show that unique prime ...
2
votes
1
answer
44
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Factorizaton in an Euclidean ring
I have a doubt concerning Lemma 3.7.4 from Topics in Algebra by I. N. Herstein.
The statement of the Lemma is:
Let $R$ be a Euclidean ring. Then every element in $R$ is either a unit in $R$ or can be ...
0
votes
0
answers
44
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Existence of prime elements in an atomic integral domain
Let $R$ be an integral domain, is it true that if $R$ is atomic, then it must contain a prime element?
If not, what is a counterexample?
I know that if an element is prime, then if $I$ is the ideal ...
3
votes
1
answer
93
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Problem in understanding the unique factorization theorem for Euclidean Rings.
Unique Factorisation Theorem: Let $R$ be a Euclidean ring and $a\neq 0$ non-unit in $R.$ Suppose that $a =\pi_1\pi_2\cdots\pi_n=\pi_1'\pi_2'\cdots\pi_m'.$ where the $\pi_i$ and $\pi_j'$ are prime ...
1
vote
1
answer
118
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Prove that $\sqrt{-5}$ is a prime in the ring $R=ℤ[\sqrt{-5}]$.
If $R=ℤ[\sqrt{-5}]$ is a ring but not a UFD, prove that the irreducible element $\sqrt{-5}$ is a prime.
This is what I have so far.
Proof: Let $R=ℤ[\sqrt{-5}]$ be a ring but not a UFD. Since $\sqrt{-5}...
2
votes
1
answer
201
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Is there an effective way to decompose gaussian integers into prime factors?
We define $\mathbb{Z}[i] := \{a + bi \mid a, b \in \mathbb{Z}\}, i = \sqrt{-1},$ which is an euclidean ring together with $N: \mathbb{Z}[i] \to \mathbb{N}_0, z \mapsto z\bar{z}=a^2+b^2$ for $z=a+bi$. ...
0
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0
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37
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Confusing example of prime and irreducible elements from my lecture script in abstract algebra [duplicate]
Could you please help me to understand the following "example" from my lecture script in the abstract algebra?
Example 12.34. Let $R = K[[x]]$ be a formal power series ring over a field $K$....
1
vote
1
answer
62
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Generating an element of a specific order if I know the prime factors of $N$
Let $N$ be an integer and suppose we know the prime factorization of $N$.
Will there then be a way of finding an element of a desired order in the multiplicative group of integers modulo $N$?
Let's ...
-2
votes
1
answer
54
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What do you call rings that have unique factorizations?
For example, integers, gaussian integers, and polynomials all have unique factorizations. What are these rings (or this property) referred to as? Or is unique factorization a ubiquitous property that ...
1
vote
0
answers
67
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Let $\mathbb{Z}[i]$ denote the *Gaussian integers*. Factor both $3+i$ and its norm into primes in $\mathbb{Z}[i]$
Question: Let $\mathbb{Z}[i]$ denote the Gaussian integers.
(a) Compute the norm $N(3+i)$ of $3+i$ in $\mathbb{Z}[i]$
(b) Factor both $3+i$ and its norm into primes in $\mathbb{Z}[i]$
(c) Compute $\...
3
votes
2
answers
682
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Definition of UFD and the fact that UFDs are integrally closed
I am trying to understand the proof of the fact that UFDs are integrally closed. In addition to the lecture notes I have, there are at least two solutions on MSE:
One is here: How to prove that UFD ...
0
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1
answer
82
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An example showing $\mathbb{Z}[\sqrt[3]{7}]$ is not a UFD [closed]
It cannot be a UFD because it's the ring of integers of $\mathbb{Q}(\sqrt[3]{7})$ and has class number 3. How can we give an example showing this?