Topics in Algebra, Chapter 3.11, Part 2
This section covers section 3.11 ("Polynomial Rings over Commutative Rings"). This page is split into parts so as to not get excessively long. **See also**: [part 1](/journal/topics-in-algebra-chapter-3-section-11a) (3.11.1 through 3.11.8). The problems below are paraphrased from/inspired by those given in Topics in Algebra by Herstein. The solutions are my own unless otherwise noted. I will generally try, in my solutions, to stick to the development in the text. This means that problems will not be solved using ideas and theorems presented further on in the book. ### Herstein 3.11.9 (Lemma 3.11.2): Let $R$ be a unique factorization domain and let $a,b,c\in R$. ### (a) Show that $a$ and $b$ have a greatest common divisor in $R$. ### (b) If $(a,b)=1$, then show that $a\mid bc$ implies that $a\mid c$. ### (c) If $a$ is irreducible and $a\mid bc$, then prove that $a\mid b$ or $a\mid c$. **(a)** The idea here is very simple: construct the greatest common divisor as the product of all the primes which $a$ and $b$ share. Formally, suppose that $a=u\pi_1\cdots\pi_m$ and $b=v\rho_1\cdots\rho_n$ where $u,v\in R$ are units and the $\pi$'s and the $\rho$'s are all irreducible elements of $R$, not necessarily distinct. Because $R$ is a UFD, these decompositions of $a$ and $b$ are finite and unique up to associates. Out of those decompositions, run through the list of $\pi_1,\ldots,\pi_m$ and pick out those $\pi_i$ which are associated to some $\rho_j$. This must be done without repetition, so if such a $\rho_j$ is found, it is removed from consideration for further $\pi$'s (for instance, consider $a=3\cdot 3$ and $b=3$; we would pick out $3$ only once). Now, define $d\in R$ to be the product of those $\pi_i$ which were picked out by this algorithm. We claim that $d$ is the greatest common divisor of $a$ and $b$. It is clear that $d\mid a$ because it was constructed out of a subset of factors of $a$. We also have that $d\mid b$ because each of those factors was associated to a factor of $b$, and we were mindful of repetitions. Now suppose that $c\mid a$ and $c\mid b$ and perhaps $c\nmid d$. Then there would exist an irreducible $\sigma\in R$ with $\sigma\mid c$ and $\sigma\nmid d$. Of course, we must have $\sigma\mid a$ and $\sigma\mid b$, but then $\sigma\nmid d$ contradicts the construction of $d$. Therefore it must be true that $c\mid d$ and as a result $d$ is the greatest common divisor. **(b)** Let $a=u\pi_1\cdots\pi_l$, $b=v\rho_1\cdots\rho_m$ and $c=w\sigma_1\cdots\sigma_n$ where $u,v,w\in R$ are units and the $\pi$'s, $\rho$'s and $\sigma$'s are all irreducible elements of $R$. For each $\pi_i$, $a\mid bc$ implies that $\pi_i$ is associated to some $\rho_j$ or $\sigma_j$. Because $(a,b)=1$, we know that $\pi_i$ is not associated to any $\rho_j$. Hence it is associated to some $\sigma$. This argument holds for every $\pi_i$, and thus $a\mid c$. **(c)** If $(a,b)=1$ then $a\mid c$ by part (b). The only other choice for $(a,b)$ is for $(a,b)=a$, because $a$ is irreducible. In that case, we have $a\mid b$ because $b$ will always be divisible by $(a,b)$. ### Herstein 3.11.10: Let $R$ be a unique factorization domain and let $f\in R[x]$. ### (a) Prove that we may express $f(x)=a g(x)$ where $a\in R$ and $g\in R[x]$ is primitive. ### (b) Prove that the decomposition of part (a) is unique up to associates. **(a)** Suppose $f(x)=a_0+a_1 x+\cdots+a_n x^n$. By trivial extension of exercise 3.11.9, the greatest common divisor $(a_0,\ldots,a_n)$ exists in $R$. Then we may take $a=(a_0,\ldots,a_n)$, the content of $f$, and the coefficients of $g$ are those of $f$ with the common irreducible divisors pulled out. By definition of gcd, we know that $g$ is primitive. **(b)** The corollary to lemma 3.11.3 states that, with $R$ a UFD, the content of a product of polynomials in $R[x]$ is the product of the contents of the polynomials, modulo multiplication by units. Suppose $f(x)=ag(x)=bh(x)$ where $a,b\in R$ and $g,h\in R[x]$ are primitive. Applying the statement above, we have that $a$ is associated to $b$ (and both are associates of the content of $f$). Writing $b=au$ with unit $u\in R$, we see that $$0=ag(x)-bh(x)=a\big(g(x)-uh(x)\big)$$ also implies that $g\sim h$. Therefore the decomposition of part (a) is unique up to associates. ### Herstein 3.11.11: Let $R$ be an integral domain, $F$ its field of fractions, and $f\in F[x]$. Show that $f(x)$ may be decomposed as $g(x)/a$ where $a\in R$ and $g\in R[x]$. Again this is a fairly easy idea just couched in some abstract language. The idea is to clear the denominators. Write $$f(x)=\frac{r_0}{s_0}+\frac{r_1}{s_1}x+\cdots+\frac{r_n}{s_n}x^n$$ with $r_i,s_i\in R$ and $s_i$ restricted to non-zero values. The fraction notation is shorthand for equivalence classes in $F$ where, as we recall, $a/b=c/d$ if and only if $ad-bc=0$. The clear thing to consider is $$p(x)=\frac{1}{s_0\cdots s_n}\Big(r_0 s_1\cdots s_n+s_0 r_1 s_2\cdots s_n x+\cdots+s_0\cdots s_{n-1}x^n\Big),$$ which is in the form $p(x)=g(x)/a$ with $a\in R$ and $g(x)\in R[x]$. We must establish that $p(x)=f(x)$ in $F[x]$. This is easily worked out for each coefficient individually. For instance, for the zeroth coefficient, we would like to show that $$\frac{r_0}{s_0}=\frac{r_0 s_1\cdots s_n}{s_0\cdots s_n},$$ and of course $$r_0\Big(s_0\cdots s_n\Big)-s_0\Big(r_0 s_1\cdots s_n\Big)=0$$ so we are correct. All other coefficients follow in the same manner, and we see that the result has been established. ### Herstein 3.11.12 (converse of Lemma 3.11.4): Let $R$ be a unique factorization domain and $F$ its field of fractions. Let $f\in R[x]$ be primitive and, when considered in $F[x]$, irreducible. Show that $f$ is also irreducible in $R[x]$. This does not seem to require that $f$ be primitive. Suppose that $f$ were reducible in $R[x]$, say $f(x)=g(x)h(x)$ with $g,h\in R[x]$ of positive degree. Then such a factorization also exists in $F[x]$ because $g$ and $h$ can be considered as elements of $F[x]$ if we replace each coefficient $a$ by the fraction $a/1$ (the unit element $1$ exists because $R$ is a UFD, but it is not essential: the fraction $ar/r$ for arbitrary non-zero $r\in R$ would suffice). So if $f$ is reducible in $R[x]$ then it must be reducible in $F[x]$, and the contrapositive is the statement that we want. ### Herstein 3.11.13 (corollary to Theorem 3.11.1): Let $F$ be a field. Show that $F[x_1,\ldots,x_n]$ is a unique factorization domain. Theorem 3.11.1 and its corollary state that if $R$ is a UFD then $R[x_1,\ldots,x_n]$ is also a UFD. Therefore we must show here that a field $F$ is indeed a UFD. But this is vacuously true of a field. We recall the definition: "Let $R$ be an integral domain. $R$ is a **unique factorization domain** (UFD) if (1) non-zero $r\in R$ is either a unit or expressible as a product of a finite number of irreducible elements of $R$, (2) the decomposition just mentioned is unique up to re-ordering and multiplication by units." A field is an integral domain by exercise 3.2.12. Every non-zero element in a field is invertible under multiplication, i.e. is a unit. The condition (2) is trivial in this case. Therefore $F$ is a UFD and the result holds by Theorem 3.11.1. ### Herstein 3.11.14: Let $R$ be a principal ideal domain. Show that $R$ is a unique factorization domain. Here's a sketch of a proof which was guided by [this PDF](http://www.maths.nuigalway.ie/MA416/section4-2.pdf). One really has to wonder if there's some elusive result in the text which simplifies the exercise, because it seems that this should really be a starred problem. To show that $R$ is a UFD we must prove two facts: **(1)** Every element $r\in R$ can be factored into a finite number of irreducible elements of $R$, **(2)** The factorization from (1) is unique up to reordering and multiplication by units. **(1)** Let $r\in R$ be such that it cannot be factored into finitely many irreducible elements. This means that, no matter how $r$ has been factorized, it can be broken down still further into non-trivial (i.e. non-unit) factors. Now, this also means that there must be some proper factor $a_0$ of $r$ (that is, $a_0$ not an associate of $r$) with the same "infinite divisibility" property: if $r$ had a factorization where none of the factors had the property, then $r$ itself would be expressible as a finite product of irreducibles, a contradiction. By the same argument, $a_0$ has a proper factor $a_1$ with the infinite divisibility property, and so forth. Now consider the chain of ideals $$(a_0)\subset(a_1)\subset(a_2)\subset\cdots$$ where each inclusion is proper. It is easy to show that the set $$U=\bigcup_{i=0}^\infty\ (a_i)$$ is also an ideal of $R$: first note that if $x\in U$ then $x\in (a_i)$ for some $i$. If $s\in R$ and $x\in U$, then $x\in(a_i)$ for some $i$ and $sx\in (a_i)\subset U$. If $x,y\in U$, with $x\in(a_i)$ and $y\in(a_j)$, then, because of the inclusion chain, both $x$ and $y$ belong to $(a_{{\rm max}(i,j)})$ so that $x+y\in(a_{{\rm max}(i,j)})\subset U$. Finally, we employ the fact that $R$ is a PID. Because $U$ is an ideal, it is principal, say $U=(\alpha)$ with $\alpha\in R$. Because $\alpha\in U$, we know that $\alpha\in(a_i)$ for some $i$. But then if $j\ge i$, $(a_j)$ is supposed to properly contain $U$, which is impossible because $U$ is at least as large as all of the $a$ ideals. Therefore no element of a PID can have this infinite divisibility property: it can be factored into finitely many irreducible elements. **(2)** Here we point out that the proof of theorem 3.7.2, uniqueness of prime factorization in Euclidean rings, carries over almost identically to PIDs. The only difficulty is that we have no result in PIDs which states that a prime element of $R$ dividing $ab$ must divide one of $a$ or $b$. The result holds for Euclidean rings (lemma 3.7.6) and for UFDs (exercise 3.11.9c), but not, a priori, for PIDs. With $R$ a PID, $a,b\in R$ have a greatest common divisor. Consider the ideal $I=\{ax+by\mid x,y\in R\}$, which must be principal, say $I=(d)$ for $d\in R$. Clearly $d$ is a divisor of both $a$ (with $x=1$ and $y=0$) and $b$ (with $x=0$ and $y=1$). It is also the greatest such divisor: let $x_0,y_0\in R$ be such that $d=ax_0+by_0$. Then if $c\mid a$ and $c\mid b$, then $c\mid (ax_0+by_0)=d$. Therefore $d$ is the greatest common divisor of $a$ and $b$, and $d\in R$. If $\pi\in R$ is irreducible and $a,b\in R$ and $\pi\mid ab$, we would like to show that $\pi\mid a$ or $\pi\mid b$. If $\pi\nmid a$, then $(\pi,a)=1$. Therefore there exist $x,y\in R$ with $\pi x+ay=1$ and consequently $\pi xb+aby=b$. Because $\pi\mid ab$, we have that $\pi$ divides the left hand side, but of course then it also dividies the right hand side: $\pi\mid b$. Therefore if $\pi\nmid a$, then $\pi\mid b$. Otherwise $\pi\mid a$. This proves the result, which is a generalization of lemma 3.7.6. Now it would be possible to quote the proof of theorem 3.7.2 verbatim here. Instead, we will give the rough sketch of the argument: let $r=\pi_1\cdots\pi_m=\rho_1\cdots\rho_n$ be two factorizations of $r$ into irreducibles. Because $\pi_1$ divides the right hand side, it must be associate to one of the $\rho_i$. Because we are working in a domain, we can cancel $\pi_1$ from both sides, leaving a unit on the right hand side but one less irreducible. Continuing this process, we end up with $1$ on the left hand side and, if there are any non-units remaining on the right hand side, then we have a contradiction. Hence we must have $n=m$ and all of the irreducibles in the two factorizations pair off as associates. This proves that the factorization is unique up to rearranging and multiplication by units. Taking (1) and (2) together, we see that any principal ideal domain must be a unique factorization domain. ### Herstein 3.11.15: Prove that $\mathbb{Z}[x_1,\ldots,x_n]$ is a unique factorization domain. By corollary 1 to theorem 3.11.1, we have that, when $R$ is a UFD, $R[x_1,\ldots,x_n]$ is a UFD. The fact that $\mathbb{Z}$ is a UFD is the statement of the fundamental theorem of arithmetic (theorem 1.3.1). Therefore the result follows.