1
$\begingroup$

I'm getting frustrated with the definition of an integral domain. I'm trying to prove the Gaussian integers are an integral domain, having just proven they're a subring of the complex numbers.

However, I understand the part about it being a commutative ring where it has no zero divisors, but what on Earth does it mean to say 1 can't equal 0... When is that ever the case!?

I got this definition from the web:

An integral domain is a commutative ring with an identity (1 =\= 0) with no zero-divisors. That is ab = 0 implies a = 0 or b = 0.

So please explain to me carefully what they mean

Thanks

$\endgroup$
8
  • 1
    $\begingroup$ As the two answers have pointed out, the set $\{ 0 \}$ is actually a commutative ring with a multiplicative identity, and this is something you can easily check. In this case (and this case only), $1 = 0$ here since the element is both the additive identity and the multiplicative identity. The question is: is this an integral domain? According to your definition, it's not since we need a multiplicative identity that's different from the additive identity. In every other ring aside from this ring with only one element, the additive identity $0$ is different from the multiplicative... $\endgroup$
    – layman
    Feb 18, 2015 at 1:07
  • $\begingroup$ ...identity $0$. Do you know why this is? If not, here is why: Suppose $R$ is a ring with more than one element. Then it must have $0$ since it is a ring. And we know $0* r = 0$ for all $r \in R$. But it must also have a multiplicative identity $1$ since it is a ring, and $1 * r = r$ for all $r \in R$. But then since $R$ has more than one element, there is an element $r \neq 0$. Then $1*r = r \neq 0$ which means $1 \neq 0$ because $0 * r = 0$. So it must be that $1 \neq 0$ if $R$ has more than one element. $\endgroup$
    – layman
    Feb 18, 2015 at 1:10
  • 2
    $\begingroup$ I genuinely feel you are thinking when we write $1$ of the number 1. But that's not the case. We are writing $1$ as a symbol for the multiplicative identity. So when we write $1 = 0$, we aren't saying the number $1$ equals the number $0$, we are saying the multiplicative identity of the ring is the same as the additive identity. $\endgroup$
    – layman
    Feb 18, 2015 at 1:14
  • 1
    $\begingroup$ Thank the heavens... you're right, I was thinking they were numbers... not symbols... That helps so much. I'm trying to catch up here so like I said, I'm needing everything to be explained... SO 1 denotes the multiplicative identity, and 0 the additive. So in your set {0}, is it true that 1=0 since the multiplicative identity is 0 and so is the additive? $\endgroup$
    – Douglas
    Feb 18, 2015 at 1:18
  • 1
    $\begingroup$ And thank you for putting up with my lack of understanding and persisting with me. You've saved me from going insane thinking 1=0 (the numbers). Thumbs up returned. $\endgroup$
    – Douglas
    Feb 18, 2015 at 1:23

3 Answers 3

Reset to default
2
$\begingroup$

Note that 1 is the multiplicative identity and 0 times anything is 0. Thus if $1=0$ we have $0x=0=x=1x$ for all $x$, so the ring has exactly one element, and that is 0.

Even though this satisfies the other conditions of an integral domain because if $ab=0$ then $a=0$ and $b=0$, it is explicitly excluded from the definition.

When we say "1" and "0" in ring theory, we are not speaking of numbers necessarily. It is somewhat an abuse of notation. "1" means "the multiplicative identity of the ring we are considering" and "0" refers to the additive identity. You are right in saying that the numbers 1 and 0 are not equal. It may be more comfortable for you to read "1=0" as "the multiplicative identity is equal to the additive identity."

$\endgroup$
4
  • $\begingroup$ I think the point of the OP's question is, if we do only have one element, namely, $R = \{ 0 \}$, is this an integral domain? $\endgroup$
    – layman
    Feb 18, 2015 at 1:03
  • $\begingroup$ I'm just trying to understand the first sentence of the definition... please explain it really simply because it's going over my head. when you say 'if 1=0'... that can't ever happen so what are you saying? $\endgroup$
    – Douglas
    Feb 18, 2015 at 1:06
  • $\begingroup$ Please see my edit. $\endgroup$
    – Matt Samuel
    Feb 18, 2015 at 1:10
  • $\begingroup$ @Douglas The element "$1$" is the element of the ring such that $k\cdot 1 = k$, even if our ring is of colors, or children's names, or loops in a space and has nothing to do with our usual concept of numbers. I.e. $1$ is the multiplicative identity. The element "$0$" is our element in the ring such that $k+0 = k$, I.e. it is our additive identity. There is nothing preventing the multiplicative identity to also be the additive identity, as shown here, however if that happens then the entire ring is made up of only a single element. $\endgroup$
    – JMoravitz
    Feb 18, 2015 at 1:10
0
$\begingroup$

It may help to realise that what we are talking about here is abstract algebra. This means that we are concentrating on the properties satisfied by "objects" and "operations" rather than the specific details of those objects and operations.

So here is an example: I am going to talk about the number $0$ and no other numbers at all, ever! (Very boring? - yes definitely.) Which of the axioms of an integral domain does this satisfy. Well for example, it satisfies the associative law since the only possible sums of three numbers we have are $$0+(0+0)\quad\hbox{and}\quad (0+0)+0\ ,$$ and these are definitely equal. Another: is there a number $u$ such that $ux$ is always equal to $x$? Well, the only possible value of $x$ is $0$ - that's the only number we're talking about, remember? So all we need is $u0=0$, and that's true when $u=0$. So $0$ is the multiplicative identity, and we usually write the multiplicative identity as $1$. Remember, that doesn't mean it is $1$ in the usual sense - as I said above, we are not really paying attention to the specific details of our numbers. So in this case it is reasonable to say that $0=1$.

But as mentioned above, this is a really boring example and is not typical of other integral domains. That's why we choose to specifically exclude it in the definition.

$\endgroup$
1
  • $\begingroup$ Further, the zero ring can be very confusing, since $\,0^{-1} = 0 = 1,\,$ and $\, 0 = 0/0 = 1,\,$ etc. Though this is sometimes useful (e.g. see the infamous thread on $\,0/0\,$ in A Rudin proof), and the related discussion here), algebraists have lost so much hair explaining $\,0^{-1}\,$ in the zero ring, that they simply outlaw the zero ring! More seriously, see here and here for why $\{0\}$ is not a field/domain. $\endgroup$
    – Bill Dubuque
    Feb 18, 2015 at 1:58
0
$\begingroup$

The detail that $1\neq0$ is typically stated as "an integral domain is a nonzero commutative ring..." by most accounts. Indeed, there is exactly one ring with this property: http://en.wikipedia.org/wiki/Zero_ring

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.