What is an Integral Domain?

Okay, so virtually 3 months right into my abstract algebra, we simply began rings. I have a couple of inquiries.

A "trivial ring" is a ring with just one component. So $R={0}$ is an unimportant ring. Easy to understand.

After that a definition mentions : Let $R$ be a ring. If there is a component $x \in R$ s.t. for all $a \in R$ you have $a * x = a = x * a$ after that R is called a "ring with identity". The symbols is $1_{R}$. I additionally recognize it is feasible to have an appropriate identification and also not a left identification (when it comes to a 2 x 2 matrix).

After that we have a definition for an indispensable domain name. An indispensable domain name is a commutative ring $R$ with identification $1_{R} \neq 0_{R}$. Okay so this is where I am perplexed currently. What does this suggest?

Is it claiming that the identification component can not be the absolutely no component? And also the identification component could be anything right? For integers its 1. So primarily it mentions that if you have a TRIVIAL RING and also if the component (given that there can just be one) is not the absolutely no component after that its an ID?

0
2019-12-02 02:49:50
Source Share
Answers: 4

There is just one unimportant phone to isomorphism. It has one component $x$ which pleases $x + x = x \cdot x = x$. That suggests that $x$ is both the additive identification and also the multiplicative identification ; symbolically, $x = 1_R = 0_R$. This is an if - and also - just - if, also : if a ring $R$ has the building that $1_R = 0_R$, after that every component amounts to $0_R$, given that $x \cdot 1_R = x = x \cdot 0_R = 0_R$. So a ring is the unimportant ring if and also just if $1_R = 0_R$.

In the definition of an indispensable domain name, we call for that the ring is nontrivial. There are numerous excellent factors for this, yet they are type of tough to encourage at the degree of a first training course in abstract algebra. (But as Joe Johnson mentions, this is not the major component of the definition of an indispensable domain name.)

0
2019-12-03 04:21:49
Source

An indispensable domain name is a commutative ring with device $1\neq 0$ such that if $ab=0$ after that either $a=0$ or $b=0$. The suggestion that $1\neq 0$ suggests that the multiplicative device, the component $x$ such that $xa=a$ for all $a\in R$ is not the very same component as the additive device, the component $y$ such that $a+y=a$ for all $a\in R$. We represent the multiplicative device by 1 and also the additive device by 0, given that they resemble the very same point in $\mathbb{Z}$. The 2nd component claims that if $a,b\neq 0$, after that $ab\neq 0$. As an example $\mathbb{Z}/6$ has 2 components $[2]$ and also $[3]$ such that $[2]\cdot [3]=[0]$. So, it is not an indispensable domain name.

0
2019-12-03 04:20:43
Source

Some instances could be in position :

INTEGRAL DOMAINS : $\mathbb{Z},\mathbb{Z}/(p) \mathbb{Q}, \mathbb{R}, k[x], \mathbb{Z}[x], \mathbb{Z}[x_1,x_2,\ldots]$ (where the last represents a polynomial ring with countably several indeterminates.

NON - INTEGRAL DOMAINS : (i.e. every ring in which you can increase a non - absolutely no number to get absolutely no)

$\mathbb{Z}/(6), k[x]/(x^2)$ ($x \times x = x^2=0$ and also $x \neq 0$)

0
2019-12-03 04:16:27
Source

An indispensable domain name is a ring without absolutely no divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0\:.\:$ Additionally it is a prevalent convention to forbid as a domain name the unimportant one - component ring (or, equivalently, the ring with $\: 1 = 0\:$). It is the nonexistence of absolutely no - divisors that is the vital theory in the definition. The exemption of the unimportant ring is just a convention that confirms hassle-free in several contexts (see my post here for one reason or another why the convention confirms hassle-free).

One might consider a domain name as a ring - logical analog of an area, given that a ring is a domain name iff it is a subring of some area. Without a doubt, an area plainly has no absolutely no - divisors so it's the same for every one of its subrings. Alternatively, any kind of domain name might be bigger to its area of portions by adjacent an inverted for every single nonzero component. Although domain names need not have inverses for every single $\rm\ c\ne 0\:,\:$ they do preserve a trace of this building. Particularly, nonzero components $\rm\:c\:$ are cancellable : $\rm\ c\ x = c\ y\ \Rightarrow\ x = y\:.\ $ This makes it possible for transfer of several (yet not all) buildings of areas to domain names. Below is one valuable instance : $\ $ a polynomial $\rm\ f(x)\in D[x]\ $ contends the majority of $\rm\ deg\ f\ $ origins in the ring $\rm\:D\ $ iff $\rm\ D\:$ is a domain name. For the straightforward evidence see my post here, where I highlight it constructively in $\rm\ \mathbb Z/m\ $ by revealing that, offered any kind of $\rm\:f(x)\:$ with even more origins than its level, we can promptly calculate a nontrivial variable of $\rm\:m\:$ using a $\rm\:gcd\:$. The square instance of this outcome goes to the heart of several integer factorization formulas, which try to variable $\rm\:m\:$ by looking for a nontrivial square origin in $\rm\: \mathbb Z/m\:,\:$ as an example a square origin of $1$ that is not $\:\pm 1$.

0
2019-12-03 04:10:20
Source