All Questions
Tagged with prime-factorization abstract-algebra
97
questions
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32
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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,...
- 37
2
votes
2
answers
45
views
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 ...
- 1,649
0
votes
1
answer
69
views
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 $\mathbb{Z}/p \mathbb{Z}$ ...
- 11
0
votes
2
answers
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 ...
- 103
2
votes
1
answer
44
views
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 ...
- 185
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 ...
- 43
3
votes
1
answer
93
views
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 ...
- 3,073
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
views
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 $z=a+bi$. ...
- 527
0
votes
0
answers
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 $K$....
- 527
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 ...
- 2,230
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 $\...
- 1,349
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 ...
- 603
0
votes
1
answer
82
views
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?
- 55
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0
answers
33
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A question about Gaussian integers [duplicate]
I am working on the following exercise:
Show that if $p \equiv 1 \mod 4$ then $p$ is not prime in $\mathbb{Z}[i]$, but instead splits as the product of two distinct prime. [Hint: Show that $p|(a^2+1)$ ...
- 1,065
1
vote
1
answer
314
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Clarify about prime decomposition of a radical ideal
My teacher stated the following result:
Let $R$ be a ring, and consider $I=\sqrt I\subseteq A$ a radical ideal. $I$ is equal to the intersection of the primes minimal over $I$, i.e. the minimal (with ...
- 2,523
3
votes
0
answers
33
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least upper bounds that are coprime
Given $n$ natural numbers $p_1$, $p_2$, ... $p_n$ find numbers $q_1$, $q_2$, ... $q_n$ that are pairwise coprimes such that $p_i$ ≤ $q_i$ and such that $\prod_{i=1..n} q_i$ is smallest possible.
I ...
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1
vote
2
answers
365
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If $R$ is a UFD, then $R[x]$ is Noetherian?
If $R$ is Noetherian, then $R[x]$ is Noetherian. However, if $R$ is a UFD, then $R$ might not be Noetherian. $\DeclareMathOperator{\Frac}{Frac}$
I want to show that if $R$ is a UFD, then $R[x]$ is a ...
- 1,751
2
votes
0
answers
92
views
Non-constant polynomial over an integral domain without any irreducible factors.
Let $R$ be an integral domain. I am trying to find a $f \in R[x]$, such that $\deg(f) \geq 1$, and $f$ does not have any irreducible factors in $R[x]$.
Does such $f$ exist?
Though I haven't been able ...
- 69
3
votes
3
answers
764
views
Is $\mathbb{Z}[√13] $ a Unique factorization domain?
I think it is not so as $12 $ can be written in two ways $12=2.6=(1+\sqrt{13})(-1+\sqrt{13})$. Are these two factorization unique upto irreducibles? Please help.
- 241
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0
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66
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Why $r \notin R^*$ if it is a prime? [duplicate]
My professor gave us the following definition:
Let $R$ be a commutative ring. $r \in R$ is prime if $r.R$ is a prime ideal. Then he concluded that this definition tells us $r \notin R^*$ but I do not ...
user778657
0
votes
1
answer
501
views
Polynomial in $\mathbb{Z}[x]$ which is irreducible but not prime. [duplicate]
I know that in an integral domain, prime implies irreducible. Moreover, in a principal integral domain, these notions are equivalent. The ring of polynomials $\mathbb{Z}[x]$ is not a principal ...
- 1,750
0
votes
1
answer
186
views
Prime elements of $\mathbb{Z}[i\sqrt5]$.
I was studying the Gaussian integers and I proved that every composite number in $\mathbb{N}$ is not a prime in $\mathbb{Z}[i]$. This is true because this ring is an Euclidean domain, and if $n=ab$ is ...
user723846
0
votes
1
answer
94
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Disproving unique factorization of $\mathbb{Q}(\sqrt{-6})$
For homework, we were given a problem that asked to explain why $(\sqrt{-6})(-\sqrt{-6})$ implies $\mathbb{Q}(\sqrt{-6})$ does not have a unique factorization.
I understand the unique part in this ...
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0
votes
1
answer
81
views
Why does $(5+\sqrt{3})(5-\sqrt{3})$ not conflict with $\mathbb{Q}(\sqrt{3})$ having unique factorization?
My intuition is to try to show that there is some irreducible element $p \in \mathbb{Q}(5)$ that divides $(5+\sqrt{3})(5-\sqrt{3})$, but I'm having trouble finding it. Is there an easier way to do ...
- 1
1
vote
0
answers
117
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Factorization in a principal ideal ring/rng
It is known that every PID is a UFD.
Is it true that every element of a commutative principal ideal ring (PIR) or rng that is not zero and not a unit is a product of a finite number of irreducible ...
- 1,120
0
votes
2
answers
138
views
Number of irreducible divisors in a UFD
The following fact is totally obvious, but I cannot find a way to prove it.
Let $R$ be a UFD and $a \in R$ be non zero and non invertible. Factor it as a product of irreducible elements: $a=p_1 \cdot ...
- 579
2
votes
2
answers
267
views
Prime element in $\mathbb{Z}[\sqrt{p}]$ [duplicate]
For which prime integers $p$ is $5$ a prime in $\mathbb{Z}[\sqrt{p}]$, the subring of the reals generated by the integers and $\sqrt{p}$ ?
I know that $\mathbb{Z}[\sqrt{p}] = \{a + b\sqrt{p} | a,b \...
- 1,503
2
votes
0
answers
32
views
reducuble polynomial in two variables
Let $f(x,y)=x^4-y^3 \in \mathbb{R}[x,y]$. I claim that $f$ is reducible but i don`t know how to find its irrdeucible factors. Can anyone help me?
Thanks in advance.
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1
vote
3
answers
500
views
$(2, 1 + \sqrt[]{−17})$ is a prime ideal in $\Bbb Z[\sqrt[]{-17}]$
How can we see that $(2, 1 + \sqrt[]{−17})$ is prime ideal in $\Bbb Z[\sqrt[]{−17}]$?
We have \begin{align*}
\frac{\mathbb{Z}\left[\sqrt{-17}\right]}{\left(2, 1 + \sqrt{-17}\right)} &\cong \frac{\...
- 3,270
1
vote
1
answer
106
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Infinite primes proof based on natural logarithm
I'm trying to understand the proof outlined in this question. There are related questions, but those concern different parts of the proof.
For completeness, here it is:
I have trouble understanding ...
- 404
2
votes
1
answer
94
views
Ideal factorization in cubic extension
Let $a:=\sqrt[3]3$
Factoring the ideal $(5)$ in $\Bbb Z[a]$, I used reduction of $X^3-3$ mod $5$ to find gives $(5)=(5,a-2)(5,a^2+2a+4)$
Checking my result under sage math gives $(5)=(a-2)(a^2+2a+4)$...
- 3,270
2
votes
0
answers
59
views
Characterizing commutative semigroups with a factorization property.
Let $(N, \times)$ be a commutative semigroup and assume that a countably infinite subset $P$ of $N$ algebraically generates $N$, and let ${\mathcal F}(P)$ denote the set of all non-empty finite ...
- 11.5k
0
votes
1
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78
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The prime factorization of $15$ when finding the number of solutions to $15=a^{2}+b^{2}$.
Find the number of solutions to $15=a^{2}+b^{2}$.
My professor told us to write $15$ in the form $2^{a}p_{1}^{t_{1}}\cdots p_{n}^{t_{n}}q_{1}^{c_{1}}\cdots q_{m}^{c_{m}}$, and if any $t_{i}$ is odd, ...
user482939
1
vote
1
answer
177
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Ring with infinitely reducible elements
Can you give or construct an elementary example of a factorial ring with elements which are product of infinitely many irreducible elements? i.e. there are reducible elements that can't be written as ...
- 192
3
votes
1
answer
480
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Why a certain integral domain is not a UFD.
Let
$$\mathbb{Z}[q]^{\mathbb{N}} = \varprojlim_j \mathbb{Z}[q]/((1-q)\cdots (1- q^j))$$
Why isn't $\mathbb{Z}[q]^{\mathbb{N}}$ a unique factorization domain?
The author proposes a proof whose ...
- 2,236
3
votes
1
answer
386
views
Number of prime elements in a ring
Is there any way to count the prime elements in a ring?
More precisely a way to count prime elements in a UFD? Which would be the same as counting irreducibles. Are there even UFD with finitely many ...
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-1
votes
2
answers
180
views
fundamental theorem of arithmetic word problem [duplicate]
Hi here is the question I have in hand:
There are $1000$ empty baskets lined up in a row. A monkey walks by, and puts a banana in each basket, because this is a word problem,
and that is what a ...
0
votes
1
answer
99
views
Working with divisor function
So by Fundamental Arithmetic Theorem, any integer has a unique prime factorization into primes, written as:
$$n=p_1^{k_1}p_2^{k_2}p_3^{k_3}...p_r^{k_r}$$
From exponents $k_1,...k_r$ it is possible to ...
0
votes
1
answer
43
views
Proof check and Are these conditions sufficient for p prime to imply p irreducible?
Let $D$ be an integral domain with a multiplicative norm $N:D \rightarrow \mathbb{N}$ such that
$$
p|b \implies N(p) \leq N(b) \ \ \ \forall p,b \in D.
$$
with equality only if $b,p$ are associates....
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3
votes
1
answer
293
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Confusion about ideal class group computation
I am attempting to compute the ideal class group of the real quadratic field $K = \Bbb Q(\sqrt{65})$, which has ring of integers $\mathcal{O}_K = \Bbb Z\left[\frac{1 + \sqrt{65}}{2}\right]$.
The ...
- 4,608
3
votes
0
answers
60
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What are the primes of $\mathbb{Z}[\sqrt{-p}\colon \text{$p$ prime}]$?
Let $R$ denote subdomain of $\mathbb{C}$ which is isomorphic to the direct limit of the diagram of commutative rings $\mathbb{Z}\hookrightarrow\mathbb{Z}[\sqrt{-2}]\hookrightarrow\mathbb{Z}[\sqrt{-2},\...
- 1,278
1
vote
2
answers
233
views
Is there always a prime having only one prime above it?
Given a number field $K / \Bbb Q$, can we find a prime $p \in \Bbb Z$ which has only one prime of $K$ above it (e.g. $p$ is inert or $p$ is totally ramified)?
For instance, if $K$ is Galois over $\...
- 6,372
17
votes
1
answer
963
views
What is the correct notion of unique factorization in a ring?
I was recently writing some notes on basic commutative ring theory, and was trying to convince myself why it was a good idea to study integral domains when it comes to unique factorization.
If $R$ is ...
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0
votes
1
answer
184
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factor $29$ into irreducible elements in $\mathbb{Z}[i\sqrt{ 2}]$ and $\mathbb{Z}[ \sqrt {7}]$
The statement goes as the title.
So I figured that $29 = (5+2i)(5-2i)$ in $\mathbb{Z}[i]$. However I'm not sure how to continue since the respective rings look like this:
$\mathbb{Z}[i\sqrt{ 2}]=\{a+...
- 77
0
votes
0
answers
233
views
Prime elements in the ring of integers adjoin a complex number.
Let $R=\mathbb Z[e^{i\pi/3}]=\{a+be^{i\pi/3}\mid a,b\in \mathbb Z\}\subseteq \mathbb C$
(a) Show that $R$ is a Euclidean domain using the Euclidean norm $N(u)=|u|^2$.
(b) Show that if $p$ is a prime ...
- 149
1
vote
1
answer
33
views
Why is it that the degree of a subextension of $K(a^{1/p})/K$ must have a degree dividing $p$?
$p$ is a prime
$K$ is of characteristic not $p$
$a∈K$
$a^{1/p}∉K$
$a$ is not a root of unity
"Then if we pick an element $b∈K(a^{1/p})$, $b∉K$, then $K(b)/K$ is a non-trivial subextension, thus of ...
- 5,166
1
vote
2
answers
136
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Justifying the representation of a monic polynomial over a UFD
Let $D$ be a UFD and $f(x) \in D[x]$ be monic. The book I'm reading from claims that $$f(x) = p_1(x)^{e_1} \cdots p_n(x)^{e_n}$$
where $p_i(x)$ are distinct, irreducible, and monic, and $e_i >0$.
...
- 4,039
3
votes
1
answer
688
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Primary Decomposition
The standard primary decomposition theorem in algebra is about being able to write an ideal uniquely as an intersection of primary ideals. In linear algebra the theorem is about how a vector space can ...
- 1,457