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Must PID contain 1? My concern arises when i consider the gcd of say a and b in the PID. Since it is a PID, it is generated by one element say k. k obviously $\in (k)$. However, if PID does not contain 1, then i can't write $k = k \cdot 1$, can 1? So is it true that then k must be equal to $k = k \cdot q$ for some $q \in$ PID. I find this kind of weird. Am I misunderstanding anything?

user10024395
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4 Answers4

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In most contexts involving domains, an identity is assumed to exist.

In principle you can define a PID without identity, but it is not likely you will encounter it in practice or text.

The main issue is that principal ideals are no longer simply multiples of their generator. In fact, $(x)=\{nx+ rx\mid n \in \Bbb Z, r\in R\}$, where $nx$ is special notation for "x added to itself n times" with the obvious accommodations for negative integers and zero.

You are completely correct in saying that $k\cdot 1$ does not represent a product of two ring elements, because the second symbol is it defined. Yet the ideal must contain $k$, by definition.

rschwieb
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  • so when we talk about gcd, is it implicitly implied that PID has 1? – user10024395 Apr 08 '15 at 02:37
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    @user136266 if you are taking a class, then yes, I'd say it's highly likely that assuming identity was intended. Do you have any reason to believe this might not be the case? – rschwieb Apr 08 '15 at 02:41
  • http://math.stackexchange.com/questions/20025/extending-ideals-from-principal-ideal-domains?lq=1 My concern arises from reading this post. Does the proof still work if the ring does not contain 1? – user10024395 Apr 08 '15 at 02:42
  • @user136266 The very first line uses the "with identity" description of the sum of two ideals, so of course it does not literally work without identity. – rschwieb Apr 08 '15 at 02:52
  • Sorry to bother, but I don't see any line that says "with identity". I use search function to search "with identity" but there is none – user10024395 Apr 08 '15 at 03:07
  • @user136266 I didn't claim that phrase appeared. I said that the description of the sum of two ideals presumes an identity. As I mentioned already, without identity you cannot say that (x)=xD. So (x)+(y) can't be described as xD + yD. – rschwieb Apr 08 '15 at 03:12
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From wikipedia:

"Integral domain" is defined almost universally as above, but there is some variation. This article follows the convention that rings have a 1, but some authors who do not follow this also do not require integral domains to have a 1

So authors who ask rings to have a $1$ ask their domains to have a $1$, but authors who don't ask for their rings to have a $1$ don't ask for their domains to have a $1$ and don't ask it of their PID's either.

I personally would suggest that you ask rings to have an identity and learn the theory there, I have heard from my proffesors that you can later study the subject without asking for rings to have $1$ without much difficulty (Authors who ask that their rings have a $1$ call the rings without $1$ Rngs since they lack the $i$ for identity)

Asinomás
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  • my concern arises when reading the solution of this post http://math.stackexchange.com/questions/20025/extending-ideals-from-principal-ideal-domains?lq=1. Is the result only true when PID has 1? – user10024395 Apr 08 '15 at 02:24
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A PID is in particular a domain, which is required to contain a $1$.

Nishant
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  • http://en.wikipedia.org/wiki/Integral_domain But wikipedia syas it is not required for integral domain to contain 1. – user10024395 Apr 08 '15 at 02:13
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    @user136266 this ultimately depends on your definition of a ring. Many authors nowadays require rings to have a unit since many useful applications of ring theory require units. – Cameron Williams Apr 08 '15 at 02:34
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    "Proof by definition" does not really acknowledge the OP's question here... I suppose in certain cases it is the right answer... – rschwieb Apr 08 '15 at 02:47
  • @user136266: Wikipedia cannot always be trusted. Many authors insist that already a ring is required to contain a $1$, and for them this question is moot. Many other authors don't. The readers must live with that. Even if Wikipedia, IMO unwisely, happens to take a side. – Jyrki Lahtonen Apr 08 '15 at 04:56
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May has or not. For Z: integers, we can say 2Z is a PID but it does not have 1.

PID means: every ideal of it is generated by only one element. And it does not have zero divisor. Domain part says that. R is a domain <=> every a,b element of R, if ab=0 => a=0 or b=0

Maja the bee
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