Abstract Nonsense

Crushing one theorem at a time

PIDs (Pt. II)


Point of Post: This is a continuation of this post.

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We now would like to discuss the notion of greatest common divisors in PIDs. Recall that the notation a\mid b, a divides b, for a,b in some ring R is that there exists some c\in R such that b=ac. Note that this is equivalent to saying that (a)\supseteq (b). Now, note that if R is an integral domain and a\mid b and b\mid a, say a=ub and b=u'a we see then that a=ub=uu'a and so by cancellation uu'=1 so that u,u'\in R^\times.  We call two elements related in this way, a\mid b and b\mid a, associated and we call them associates. It’s evident that being associate is an equivalence relation on R. We then call an element d\in R a greatest common divisor of a,b\in R if d\mid a, d\mid b and whenever c\mid a and c\mid b then c\mid d. An emphasis was placed on the article ‘a’ because greatest common divisors are not unique. For example, in \mathbb{Z} a greatest common divisor of 2 and 3 is 1, but so is -1. In general, greatest common divisors are unique up to associates:

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Theorem: Let R be an integral domain and let a,b\in R. Then, if d is a greatest common divisor of a,b then so is ud for any u\in R^\times. Moreover, every greatest common divisor is of this form.

Proof: Suppose first that ud is an associate of d, then evidently du\mid a,b and if c\mid a,b then c\mid d so that c\mid ud and so ud is a greatest common divisor of a,b. Conversely, suppose that x is a greatest common divisor of a,b. Since x\mid a,b and d is a greatest common divisor we have that x\mid d. But, since x is also a greatest common divisor of a,b applying the same argument shows that d\mid x. From previous observation we have that d and x are associates. \blacksquare

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So, how exactly do greatest common divisors factor into PIDs? Well, given two elements a,b\in R we know that the ideal (a,b) generated by the two is principal, and so equal to (d) for some d\in R. Well, unsurprisingly, as is easily seen in \mathbb{R}, one has that d is a greatest common divisor of a,b. Indeed:

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Theorem: Let R be an integral domain and suppose that d\in R are such that (d)=(a,b). Then, d is a greatest common divisor of a,b.

Proof: Since a,b\in (d) we have by definition that d\mid a,b. Suppose now that c\mid a,b then we know that (c)\supseteq(a,b) and so (c)\supseteq(d). But, by definition this implies that c\mid d. \blacksquare

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So, think about how this jives with our usual notion of greatest common divisors in \mathbb{Z}. We usually call the greatest common divisor of a,b\in\mathbb{Z} the largest (positive) common divisor of a,b. In other words \text{gcd}(a,b)=\max\left\{x\in\mathbb{N}:x\mid a,b\right\}. Now,  from the previous theorem we know that if d=\text{gcd}(a,b) and c\mid a,b then (c)\supseteq (d) and so, in particular, we have that d\in(c). But, by previous theorem we know (c) is generated by an element of least degree in (c). In particular, d(c)\leqslant d(x) for every x\in (c) and so, in particular, d(c)\leqslant d(d). From this it’s not hard to conclude that, agreeing with our idea of greatest common divisors in \mathbb{Z}, one has the following:

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Theorem: Let (R,d) be a Euclidean domain with d satisfying the d-inequality. Then if a,b\in R one has that d is a greatest common divisor of a,b if and only if d\mid a,b and whenever c\mid a,b one has that d(c)\leqslant d(d).

Proof: Suppose first that d is a greatest common divisor of a,b and let c\mid a,b. As mentioned in the previous paragraph we have that (c)\supseteq (d) and so d\in (c) and so d(c)\leqslant d(d). We shall prove the other direction in our next post. \blacksquare

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Regardless, another common fact (known colloquially as Bezout’s identity) which holds true in \mathbb{Z} holds more generally in PIDs:

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Theorem: Let R be a PID. Then, if a,b\in R one has that 1 is a greatest common divisor of a,b if and only if ax+by=1 for some x,y\in R.

Proof: Suppose first that a,b have 1 as a greatest common divisor. We know then that (a,b)=(1)=R and so, in fact, for any r\in R there exists x,y\in R such that ax+by=r. Conversely, suppose that ax+by=1 for some x,y\in R then evidently (a,b)=R=(1) and so 1 is a greatest common divisor of a,b. \blacksquare

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If a,b\in R have 1 as a greatest common divisor we say that a,b are coprime. In fact, we see that the above theorem may be stated as follows:

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Theorem: Let R be a PID and a,b\in R. Then, a,b are coprime if and only if (a),(b) are comaximal

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And so the CRT tells us that if a,b are coprime then R/(ab)=R/((a)(b))\cong R/(a)\times R/(b).

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As a last structural result we mention the following:

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Theorem: Let R be a PID, then R/\mathfrak{a} is a PID if and only if \mathfrak{a}\in\text{Spec}(R).

Proof: The necessity for \mathfrak{a} being prime is clear since we need R/\mathfrak{a} to be an integral domain. Suppose now that \mathfrak{a} is prime, then R/\mathfrak{a} is an integral domain. Now, if \mathfrak{b} is an ideal in R/\mathfrak{a} then we know by the lattice isomorphism theorem that \mathfrak{b}=\mathfrak{c}/\mathfrak{a} for some \mathfrak{c}\supseteq\mathfrak{a}. But, by assumption that R is a PID we know that \mathfrak{c}=(c) for some c\in R and it’s not hard to see then that \mathfrak{b}=\mathfrak{c}/\mathfrak{a}=\left(c+\mathfrak{a}\right). \blacksquare

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

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October 22, 2011 - Posted by | Algebra, Ring Theory | , , , , , , , ,

2 Comments »

  1. […] we have seen that PIDs possess greatest common divisors for any two elements, this is also true for UFDs. […]

    Pingback by UFDs (Pt. I) « Abstract Nonsense | October 22, 2011 | Reply

  2. […] this we can finish a proof from our last post on […]

    Pingback by UFDs (Pt. II) « Abstract Nonsense | October 22, 2011 | Reply


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