# Abstract Nonsense

## UFDs (Pt. I)

Point of Post: In this post we discuss the most general “nice integral domain”, unique factorization domains.

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Motivation

So, up until this we point we have discussed the very nice integral domains which admit degree functions (i.e. Euclidean domains) and the, less nice still fantastic, integral domains whose ideals are all principal (i.e. PIDs). In this post we shall take another step down the niceness-ladder and discuss a proper subset of PIDs (and thus a “doubly proper” subset of Euclidean domains)–UFDs. Amusingly enough though, UFDs are perhaps the most useful of the three, because they pop up much more often than either Euclidean domains or PIDs. Perhaps the first indication of why this is so is captured by the nice interaction UFDs have with the construction $R\mapsto R[x]$. Indeed, we have seen $R[x]$ is a PID (let alone a Euclidean domain) if and only if $R$ is a field! So, in the vast majority of cases if one starts with a Euclidean domain or a PID $R$ and passes to the polynomial ring $R[x]$ one is not going to end up with a Euclidean domain or a PID. The classic example is that $\mathbb{Z}$ is about as nice of a Euclidean domain as you could want, yet $\mathbb{Z}[x]$ is not even a PID ($(p,x)$ for a prime $p$ is not principle)! That said, as we shall see the construction $R\mapsto R[x]$ does preserve UFDs in the sense that $R$ being a UFD implies that $R[x]$ is a UFD. So, for example, $\mathbb{Z}[x]$ while not a PID or a Euclidean domain is a UFD. Now, we have indicated before that polynomial rings play a pivotal role in not only algebra but algebra’s application to other fields of mathematics (e.g. algebraic geometry). With this in mind any properties that interact well with the construction $R\mapsto R[x]$ are definitely well-worth our time.

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So, I have yet to say actually what a UFD is. If you pulled some kid off the street and asked him what was one of the most startlingly useful and insightful ring theoretic properties of $\mathbb{Z}$ they might be apt to say “Oh! The fundamental theorem of arithmetic–every integer factors uniquely (up to differences in sign) into a product of primes!” (smart kid, huh?). Well, that kid would be right. Anyone who has done even the slightest bit of number theory is well-aware that the existence and uniqueness of integer factorization into primes is an invaluable tool. Often times easier in just a “makes things” simpler–how often are you trying to decide if some divisibility argument is true, and you just produce the prime factorization of things involved, and it becomes trivial. Well, this is what UFDs attempt to capture. We shall see that in any integral domain there are notions analogous to (integer) primality and UFDs are (roughly) those for which there is a notion of unique factorization into these prime elements. Indeed, this probably would have been clear if I had mentioned that UFD stands for Unique Factorization Domain.

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UFDs

We begin by defining the notions of primality that will be useful to us. Let $R$ be an integral domain, then a non-zero non-unit element $x\in R$ is said to be irreducible if whenever $x=ab$ either $a\in R^\times$ or $b\in R^\times$ (and so consequently the “other” element is an associate of $x$). Roughly this says that the only divisors of $x$ are units and associates. An element $x\in R$ is said to be prime if it generates a prime ideal (i.e. $(x)\in\text{Spec}(R)$)–note that this is equivalent to $x\mid ab$ implies $x\mid a$ or $x\mid b$.

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Now, wait a minute. These definitions seem to fly in the face of our time-honored and beloved definition of primality in $\mathbb{Z}$. Indeed, in $\mathbb{Z}$ we defined an element to be “prime” if it was irreducible and then it was a theorem (Euclid’s lemma) that a “prime” was…well…prime. So, what gives? First off, this makes us wonder if it’s even true if primality and irreducibility are, in general, different. Well, this is certainly true, let’s examine the classic example that $3$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$ but not prime. Indeed, we can recall that there is a certain field norm $N:\mathbb{Z}[\sqrt{-5}]\to\mathbb{Z}$ given by $N(a+b\sqrt{-5})=a^2+5b^2$ which is multiplicative. Now, if $3=(a+b\sqrt{-5})(c+d\sqrt{-5})$ then we can easily see that $9=(a^2+5b^2)(c^2+5d^2)$. Now, suppose that $3=a^2+5b^2$. Then, clearly $b=0$ and so $a^2=3$ which is ridiculous since $a\in\mathbb{Z}$. Thus, we can clearly conclude $a^2+5b^2\in\{1,9\}$. We assume that $a^2+5b^2=1$ (since by symmetry if $a^2+5b^2=9$ then $c^2+5d^2=1$ and the same logic will follow). We see then that evidently $b=0$ and $a=\pm1$ and so $a+b\sqrt{-5}=\pm 1\in\mathbb{Z}[\sqrt{-5}]^\times$. Thus, we see that every time we try to factor $3$ in $\mathbb{Z}[\sqrt{-5}]$ one of the factors must be a unit, in other words, $3$ is irreducible in $\mathbb{Z}[\sqrt{-5}]$. That said, a quick check shows that $9=(2+\sqrt{-5})(2-\sqrt{-5})$ so that $3\mid (2+\sqrt{-5})(2-\sqrt{-5})$. That said, if $2+\sqrt{-5}=3(a+b\sqrt{-5})=3a+3b\sqrt{-5}$ then we see by comparing coefficients that $3a=2$ which is preposterous and so $3\nmid 2+\sqrt{-5}$. Clearly though the same logic shows that $3\nmid2-\sqrt{-5}$. But, this implies that $3$ is not prime in $\mathbb{Z}[\sqrt{-5}]$ since we have found two elements of $\mathbb{Z}[\sqrt{-5}]$ not in $(3)$ whose product is in $(3)$. So, ok, the notions of primality and irreducibility are in general different. So, why do we “switch” the definitions in $\mathbb{Z}$? Well as we shall see there really is no foul since, in $\mathbb{Z}$ (or more generally in PIDs) primality and irreuducibility are the same. And considering that we usually first encounter primality when we are attempting to factorize integers and irreducibility tells us when we can “stop” (we’ve factored as much as we can) it makes sense that we choose irreducibility as our definition of “prime”.

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Now that we have worked through that confusing point we can get back to the general theory. It’s clear that primality is stronger than irreducibility since if $x$ is prime and $x=ab$ we know that $x\mid a$ or $x\mid b$ and since $R$ is an integral domain this will clearly imply that the the other element of $a,b$ is a unit. But, as we showed above the converse is not true. More formally:

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Theorem: Let $R$ be an integral domain. Then, every prime element is irreducible but not conversely.

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With this in mind we are prepared to define UFDs. Namely a unique factorization domain, abbreviated UFD, is an integral domain $R$ where all irreducible elements are prime and every non-zero non-unit element of $R$ can be factored into the product of irreducibles.

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Now, how exactly does this definition imply that factorization is unique? Well, pretty much the same way it did in $\mathbb{Z}$:

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Theorem: Let $R$ be a UFD and let $r_1\cdots r_n=s_1\cdots s_m$ where $r_i,s_j$ are irreducible in $R$. Then, $n=m$ and there exists a bijection $\pi:[n]\to[n]$ such that $r_i$ is associate to $s_{\pi(i)}$.

Proof: Assume without loss of generality that $m\geqslant n$. Now, since $r_1$ is prime and $r_1\mid s_1\cdots s_m$ we have that $r_1\mid s_k$ for some $k\in[n]$, call $\pi(1)=k$. But, since $s_{\pi(1)}$ is a prime and $r_1$ is not a unit this implies that $r_1$ and $s_{\pi(1)}$ are associate, say $s_{\pi(1)}=u_1r_1$ for some unit $r_1$. We see then cancelling both sides that $\displaystyle r_2\cdots r_n=u_1 \prod_i s_{i}$ where the product is taken over $i\ne \pi(1)$. Now, since $r_2$ is prime and $\displaystyle r_2\mid u_1\prod_i s_{i}$ and $r_2\nmid u_1$ (since $u_1$ is a unit and $r_2$ is not) we see that $r_2\mid s_i$  for some $i\ne \pi(1)$, call $i=\pi(2)$. Using the same conclusion as before we know that $s_{\pi(2)}=u_2r_2$ for some $u_2\in R^\times$ and so $\displaystyle r_3\cdots r_n=u_1u_2\prod_{j}s_j$ where the product is taken over $j\ne \pi(1),\pi(2)$. Continuing in this way we find that we have an injection $\pi:[n]\to[m]$ such that $r_i$ is associate to $\pi(i)$, say by $s_{\pi(i)}=u_ir_i$, and such that $\displaystyle 1=u_1\cdots u_n \prod_{t}s_t$ where $t$ is taken over $[m]-\pi([n])$. Now, suppose that $[m]-\pi([n])$ was non-empty then the product $\displaystyle \prod_{t}s_t$ would be non-trivial, but this product is equal to the unit $u_1^{-1}\cdots u_n^{-1}$ and so, if $s_{t_0}$ is a factor in $\displaystyle \prod_{t}s_t$ we see that $s_{t_0}$ (being prime) divides one of $u_1^{-1},\cdots,u_n^{-1}$ which implies that $s_{t_0}$ is prime, which is a contradiction. Thus, $[m]=\pi([n])$ and the conclusion follows. $\blacksquare$

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Now, we have seen that PIDs possess greatest common divisors for any two elements, this is also true for UFDs. Moreover, it’s often easy (given their irreducible factorizations to find them):

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Theorem: Let $R$ be a UFD and let $x,y\in R$. We may write $x=up_1^{\alpha_1}\cdots p_n^{\alpha_n}$ and $y=vp_1^{\beta_1}\cdots p_n^{\beta_n}$ where $u,v\in R^\times$, $p_1,\cdots,p_n$ are irreducible in $R$, and $\alpha_i,\beta_j\in\mathbb{N}\cup\{0\}$. Then, $d=p_1^{\min\{\alpha_1,\beta_1\}}\cdots p_n^{\min\{\alpha_n,\beta_n\}}$ is a greatest common divisor for $x,y$.

Proof: Evidently $d\mid x,y$ and if $c\mid x,y$ it’s easy to see that $c=wp_1^{\gamma_1}\cdots p_n^{\gamma_n}$ where $w\in R^\times$, $\gamma_i\in\mathbb{N}\cup\{0\}$ and $\gamma_i\leqslant \min\{\alpha_1,\beta_1\}$ from where it easily follows that $c\mid d$. Thus, $d$ is a greatest common divisor. $\blacksquare$

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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.

October 22, 2011 -

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