Abstract Nonsense

Crushing one theorem at a time

UFDs (Pt. II)

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

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What we’d now like to do is substantiate our claim that \left\{\text{PIDs}\right\}\supsetneq\left\{\text{UFDs}\right\}. The properness of any such containment is clear since as we have already mentioned \mathbb{Z}[x] is a UFD (still to be proven) which is not a PID. Thus, it clearly suffices to prove the containment. We do this in two steps corresponding to the two defining properties of UFDs. First we prove that irreducibility and primality are really the same thing in PIDs. Indeed:

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Theorem: Let R be a PID and x\in R be irreducible, then x is prime.

Proof: Consider the ideal (x). We claim that (x)\in\text{MaxSpec}(R) from where it will follows that (x)\in\text{Spec}(R) and so x is prime. To see this suppose that \mathfrak{a}\supseteq(x), since R is a PID we know that \mathfrak{a}=(a) for some a\in R and so x=ab for some b\in R. Since x is irreducible this implies that either a is associate to x, in which case \mathfrak{a}=(a)=(x), or a is a unit in which case \mathfrak{a}=(a)=R. Thus, no proper ideal properly contains (x) and since x\notin R^\times we know that (x)\ne R and so (x)\in\text{MaxSpec}(R) as claimed. \blacksquare

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What we now show is that every element of a PID can be factored into a product of irreducibles. To do this we shall first need a lemma which says that every ascending chain of ideals in a PID eventually terminates (this is actually equivalent to the more general notion of a Noetherian ring (one in which every ideal is finitely generated)). Indeed:

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Lemma: Let R be a PID, then if \mathfrak{a}_1\subseteq\mathfrak{a}_2\subseteq\cdots is a ascending sequence of ideals in R then \mathfrak{a}_i is eventually constant (i.e. there exists N\in\mathbb{N} such that n,m\geqslant N implies \mathfrak{a}_n=\mathfrak{a}_m).

Proof: Since R is a PID we know that \mathfrak{a}_j=(a_j) for some a_j\in R. Let \displaystyle \mathfrak{a}=\bigcup_{j=1}^{\infty}(a_j), we know that, being the union over a chain of ideals, \mathfrak{a} is an ideal and so by assumption that R is a PID we have that \mathfrak{a}=(a) for some a\in R. Now, since a is the union over the chain we have that a\in(a_k) for some k\in\mathbb{N} and so \mathfrak{a}=(a)\subseteq(a_k) from where it readily follows that (a_k)=\mathfrak{a}. Now, if m>k we know that (a_m)\subseteq\mathfrak{a}\subseteq (a_k) and since (a_k)\subseteq (a_m) this implies that (a_k)=(a_m). Thus, the chain is eventually constant. \blacksquare

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From this we can prove that every element of a PID can be factored into irreducibles:

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Theorem: Let R be a PID and x\in R be non-zero and non-unit. Then, there exists irreducibles r_1,\cdots,r_n\in R such that x=r_1\cdots r_n.

Proof: If x is irreducible, we’re done, so assume not. Then, x=r_1r_2 where r_1,r_2\in R. If r_1 and r_2 are irreducible, then we’re done. So, assume not, without loss of generality say that r_1 is not irreducible. Write r_1=r_{2,1}r_{2,2}. We see then that x=r_{2,1}r_{2,2}r_2. If r_{2,1},r_{2,2}r_2 are all irreducible we’re done, so assume not, say r_{2,1}=r_{3,1}r_{3,2}. Continuing in this way we arrive at a chain of ideals (x)\subseteq (r_1)\subseteq (r_{2,1})\subseteq (r_{3,1})\subseteq\cdots. By the lemma we have to have that this chain is eventually constant, say (r_{n,1})=(r_{n+1,1}) this contradicts that r_{n,1} is not irreducible. Thus, r_{n,1} is irreducible. Continuing this process factors x into irreducibles as desired. \blacksquare

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Combining these two gives:

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Theorem: Let R be a PID, then R is a UFD, but not conversely.

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and a corollary:

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Corollary: Let R be a Euclidean domain, then R is a UFD, but not conversely.

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With this we can finish a proof from our last post on PIDs

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Theorem: Let (R,d) be a Euclidean domain with d satisfying the d-inequality and let x,y\in R. Then, d\in R is a greatest common divisor of x,y if and only if d\mid x,y and if c\mid x,y then d(c)\leqslant d(d).

Proof: As remarked earlier it’s clear that if d is a greatest common divisor then these properties hold. So, conversely suppose that d\mid x,y and has maximal degree among all such common divisors. If d is not a greatest common then there exists a c\mid x,y such that c\nmid d. Now, since R is a UFD we can factor x and y as up_1^{\alpha_1}\cdots p_n^{\alpha_n} and v p_1^{\beta_1}\cdots p_n^{\beta_n} where u,v\in R^\times and p_j is irreducible. Now, since d,c\mid x,y we know that c=wp_1^{\gamma_1}\cdots p_n^{\gamma_n} and d=ep_1^{\delta_1}\cdots p_n^{\delta_n} where w,e\in R^\times and \gamma_i,\delta_i\leqslant\min\{\alpha_i,\beta_i\}. Since c\nmid d we must have that \gamma_{i_0}>\delta_{i_0} for some i_0 and so d'=u p_1^{\delta_1}\cdots p_{i_0}^{\gamma_{i_0}}\cdots p_n^{\delta_n} is also a divisor of x,y and since it’s equal to d multiplied by an irreducible (and so non-unit) we know (since d satisfies the d-inequality) that d(d')>d(d) which is a contradiction. Thus, c\mid d. Since c was arbitrary the conclusion follows. \blacksquare

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[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.


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

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