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3 votes
2 answers
682 views

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 ...
abeliangrape's user avatar
8 votes
5 answers
14k views

Proving gcd($a,b$)lcm($a,b$) = $|ab|$

Let $a$ and $b$ be two integers. Prove that $$ dm = \left|ab\right| ,$$ where $d = \gcd\left(a,b\right)$ and $m = \operatorname{lcm}\left(a,b\right)$. So I went about by saying that $a = p_1p_2......
SalmonKiller's user avatar
  • 2,148
7 votes
2 answers
4k views

Prove that every nonzero prime ideal is maximal in $\mathbb{Z}[\sqrt{d}]$

$d \in \mathbb{Z}$ is a square-free integer ($d \ne 1$, and $d$ has no factors of the form $c^2$ except $c = \pm 1$), and let $R=\mathbb{Z}[\sqrt{d}]= \{ a+b\sqrt{d} \mid a,b \in \mathbb{Z} \}$. Prove ...
PandaMan's user avatar
  • 3,269
35 votes
3 answers
17k views

What are examples of irreducible but not prime elements?

I am looking for a ring element which is irreducible but not prime. So necessarily the ring can't be a PID. My idea was to consider $R=K[x,y]$ and $x+y\in R$. This is irreducible because in any ...
MyNameIs's user avatar
  • 1,057
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 ...
Ishan Levy's user avatar
3 votes
1 answer
2k views

About the ways prove that a ring is a UFD.

I'm doing an exercise where I have a commutative ring with unity $R$. We had to find that the nonunits formed an ideal (maximal). After that, we found the irreducible elements, and then we saw that ...
Relure's user avatar
  • 4,225
2 votes
1 answer
306 views

A simple algebraic ring extension of a UFD having no prime elements

Let $D$ be a UFD over a field $k$ of characteristic zero. Assume that $w$ is algebraic over $D$. Denote $R=D[w]$. Observe that $R$ is not necessarily a UFD. Can one find an example in which $R$ ...
user237522's user avatar
  • 6,705
2 votes
3 answers
236 views

Number of solutions of polynomials in a field

Consider the polynomial $x^2+x=0$ over $\mathbb Z/n\mathbb Z$ a)Find an n such that the equation has at least 4 solutions b)Find an n such that the equation has at least 8 solutions My idea is to ...
Linh Phan's user avatar
  • 171
13 votes
2 answers
3k views

$K[x_1, x_2,\dots ]$ is a UFD

I wonder about how to conclude that $R=K[x_1, x_2,\dots ]$ is a UFD for $K$ a field. If $f\in R$ then $f$ is a polynomial in only finitely many variables, how do I prove that any factorization of $f$...
harajm's user avatar
  • 2,157
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 ...
Smiley1000's user avatar
  • 1,649
2 votes
0 answers
202 views

Non-unique factorization in $\mathbb{Z}[\sqrt{-5}]$

I want to show that the decomposition into irreducible factors in the ring $$\mathbb{Z}[\sqrt{-5}] = \{a + b\sqrt{-5}|\space a, b \in \mathbb{Z}\}$$ is not unique, except for the order of factors ...
moran's user avatar
  • 3,057
0 votes
1 answer
175 views

Prove that the elements $2x$ and $x^2$ have no LCM in the ring of integral polynomials with even coefficient of $x$

Let $A$ be the subring of $\Bbb Z[x]$ consisting of all polynomials with even coefficient of $x$. Prove that the elements $2x$ and $x^2$ have no lowest common multiple. Hints please!
Open Season's user avatar
  • 1,332
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 ...
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