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Okay, so almost 3 months into my abstract algebra, we just started rings. I have a few questions.

A "trivial ring" is a ring with only one element. So $R={0}$ is a trivial ring. Understandable.

Then a definition states: Let $R$ be a ring. If there is an element $x \in R$ s.t. for all $a \in R$ you have $a * x = a = x * a$ then R is called a "ring with identity". The notation is $1_{R}$. I also know it is possible to have a right identity and not a left identity (in the case of a 2 x 2 matrix).

Then we have a definition for an integral domain. An integral domain is a commutative ring $R$ with identity $1_{R} \neq 0_{R}$. Okay so this is where I am confused now. What does this mean?

Is it saying that the identity element can not be the zero element? And the identity element could be anything right? For integers its 1. So basically it states that if you have a TRIVIAL RING and if the element (since there can only be one) is not the zero element then its an ID?

Mike Spivey
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Tyler Hilton
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    How do 2 by 2 matrices have a right identity and not a left identity? – Sean Tilson Dec 07 '10 at 03:45
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    @Sean Tilson: I note that Tyler said a $2\times 2$ matrix, not the ring of all $2\times 2$ matrices. You can construct a subring of the $2\times 2$ matrices that have a right identity but not a left identity. For example, take the matrices with first column all $0$s. – Arturo Magidin Dec 07 '10 at 16:06
  • ok, I didn't parse it that way. I thought he meant that there is a 2 by 2 matrix that is the right identity and not a left identity. Quite right though if we look at a subring like the one you suggested. – Sean Tilson Dec 07 '10 at 16:49

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There is only one trivial ring up to isomorphism. It has one element $x$ which satisfies $x + x = x \cdot x = x$. That means that $x$ is both the additive identity and the multiplicative identity; symbolically, $x = 1_R = 0_R$. This is an if-and-only-if, too: if a ring $R$ has the property that $1_R = 0_R$, then every element is equal to $0_R$, since $x \cdot 1_R = x = x \cdot 0_R = 0_R$. So a ring is the trivial ring if and only if $1_R = 0_R$.

In the definition of an integral domain, we require that the ring is nontrivial. There are several good reasons for this, but they are sort of hard to motivate at the level of a first course in abstract algebra. (But as Joe Johnson points out, this is not the main part of the definition of an integral domain.)

Qiaochu Yuan
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  • Moreover, the trivial ring is the terminal object in the category of rings, so it is unique up to (unique) isomorphism. – Akhil Mathew Dec 07 '10 at 18:57
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An integral domain is a commutative ring with unit $1\neq 0$ such that if $ab=0$ then either $a=0$ or $b=0$. The idea that $1\neq 0$ means that the multiplicative unit, the element $x$ such that $xa=a$ for all $a\in R$ is not the same element as the additive unit, the element $y$ such that $a+y=a$ for all $a\in R$. We denote the multiplicative unit by 1 and the additive unit by 0, since they are similar to the same thing in $\mathbb{Z}$. The second part says that if $a,b\neq 0$, then $ab\neq 0$. For instance $\mathbb{Z}/6$ has two elements $[2]$ and $[3]$ such that $[2]\cdot [3]=[0]$. So, it is not a integral domain.

Arturo Magidin
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J126
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  • Oh, so it doesn't have to be the trivial ring. The integers are an example of a non-trivial integral domain. – J126 Dec 07 '10 at 02:02
  • But dosn't $[0]$ imply all infinite integers (assuming we are in $Z_1$ – Tyler Hilton Dec 07 '10 at 02:06
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    Er... it can't be the trivial ring. @Tyler: I am not sure what you mean. – Qiaochu Yuan Dec 07 '10 at 02:06
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    An integral domain can't be a trivial ring - because that only has one element and hence can't satisfy $1_R \ne 0_R$ – Juan S Dec 07 '10 at 02:07
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    The integers $\mathbb{Z}$ is not $\mathbb{Z}/1$. It is $\mathbb{Z}/0$. – J126 Dec 07 '10 at 02:08
  • Qwirk, that is my question kinda. If the element is $2_R$ then what is it? a trivial ring, an ID or both? – Tyler Hilton Dec 07 '10 at 02:09
  • @Tyler: "doesn't [0] imply all infinite integers" makes no sense. "[x]" is standard notation for "the equivalence class containing x" – kahen Dec 07 '10 at 02:09
  • @Joe, but doesnt $Z_1$ imply just one element of the set, namely $[0]$ which can be written as [0 + 1k] – Tyler Hilton Dec 07 '10 at 02:10
  • @Tyler: [0] is the set of integers divisible by 6. Again, I'm not sure what you're trying to say here. – Qiaochu Yuan Dec 07 '10 at 02:11
  • @Qiaochu: You're right, it can't be the trivial ring. I had a lapse in judgment, since to me the two element ring is trivial. I don't think too much about one element rings. – J126 Dec 07 '10 at 02:12
  • $Tyler: The set {2} is a trivial ring, with the zero element as 2. Don't confuse the zero over the integers with the zero of an arbitrary ring – Juan S Dec 07 '10 at 02:13
  • @Tyler: Yes, $\mathbb{Z}/1$ does imply just one element. Though I'm not sure why you are bringing it up. Also, I'm confused by your usage of $2_R$. – J126 Dec 07 '10 at 02:14
  • thankyou qwirk. thats exactly what i wnated. so basically the first element of a ring is the zero element? – Tyler Hilton Dec 07 '10 at 02:15
  • @Joe, i wanted to know if the 0 element was the 0 from integers. I think thats where im confused. But since Qwirk said that ${2}$ has a zero element (namely 2),it makes more sense – Tyler Hilton Dec 07 '10 at 02:17
  • Okay. Glad it's cleared up. – J126 Dec 07 '10 at 02:19
  • @Tyler: Not the 'first' element (how do we say what the first element is?). But the definition of a ring requires that we have an element such that $a+0_R=0_R+a=a$ for all $a$ in the ring. Similarly we must have a multiplicative identity. Hence for a ring with a single element, they must be equal (see Qiaochu's response) – Juan S Dec 07 '10 at 02:20
  • just to finally clear everything up. so an ID is pretty much saying that the zero element can not be an identity element. An identity element is something where $a \cdot x = a$ – Tyler Hilton Dec 07 '10 at 02:20
  • I would say the additive identity (the 'zero') cannot be the multiplicative identity (the 'one') – Juan S Dec 07 '10 at 02:26
  • @Tyler: that is part of the definition of an integral domain, but there is also a second (much more important) part, which is described in the answer. – Qiaochu Yuan Dec 07 '10 at 02:27
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    @Tyler: I agree with Qiaochu: the "really important part" (if one part may be said to be more important than others in a definition) is that in an integral domain, if $ab=0$, then $a=0$ or $b=0$. The requirement that the multiplicative identity should be different from the additive one (that the ring cannot consists only of one element) is more technical, and is akin (in fact, closely related to) the part of the definition of "prime number" that excludes $1$ from the class of primes by fiat. It's important, but you really want to remember the second part of the definition. – Arturo Magidin Dec 07 '10 at 03:41
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An integral domain is a ring with no zero divisors, i.e. $\rm\ xy = 0\ \Rightarrow\ x=0\ \ or\ \ y=0\:.\:$ Additionally it is a widespread convention to disallow as a domain the trivial one-element ring (or, equivalently, the ring with $\: 1 = 0\:$). It is the nonexistence of zero-divisors that is the important hypothesis in the definition. The exclusion of the trivial ring is merely a convention that proves convenient in many contexts (see my post here for some reasons why the convention proves convenient).

One may think of a domain as a ring-theoretic analog of a field, since a ring is a domain iff it is a subring of some field. Indeed, a field clearly has no zero-divisors so ditto for all of its subrings. Conversely, any domain may be enlarged to its field of fractions by adjoining an inverse for every nonzero element. Although domains needn't contain inverses for every $\rm\ c\ne 0\:,\:$ they do retain a trace of this property. Namely, nonzero elements $\rm\:c\:$ are cancellable: $\rm\ c\ x = c\ y\ \Rightarrow\ x = y\:.\ $ This enables transfer of many (but not all) properties of fields to domains. Here is one useful example: $\ $ a polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in the ring $\rm\:D\ $ iff $\rm\ D\:$ is a domain. For the simple proof see my post here, where I illustrate it constructively in $\rm\ \mathbb Z/m\ $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd\:$. The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m\:,\:$ e.g. a square root of $1$ that is not $\:\pm 1$.

Community
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Bill Dubuque
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Some examples might be in place:

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 latter denotes a polynomial ring with countably many indeterminates.

NON-INTEGRAL DOMAINS: (i.e. every ring in which you can multiply a non-zero number to get zero)

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

Fredrik Meyer
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