Topics in Algebra, Chapter 3.11

This section covers section 3.11 (“Polynomial Rings over Commutative Rings”).

Topics covered: 3.11

  • Definition: Let R be a commutative ring with a unit element. R[x] is defined, as in section 3.9, as the set of all symbols a0+a1x++amxm where m is a non-negative integer and a0,,amR. The standard addition and multiplication make R[x] a ring. It will be shown in exercise 3.11.1 that R[x] so defined is a commutative ring with unit element.

  • Definition: Let R be a commutative ring with unit element. By problem 3.11.1, R[x] is again a commutative ring with unit element. Therefore we may recursively define R[x1,,xn] as R[x1,,xn1][xn].

  • Lemma 3.11.1: If R is an integral domain, then R[x] is an integral domain.

  • Corollary: If R is an integral domain, then R[x1,,xn] is an integral domain for any positive integer n.

  • Definition: Let R be an integral domain. R is a unique factorization domain (UFD) if (1) non-zero rR 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.

  • Lemma 3.11.2: If R is a UFD and a,bR, then (a,b)R, i.e. their gcd exists in R. Additionally, if (a,b)=1 then abc implies that ac.

  • Corollary: If R is a UFD and aR is irreducible, then abc implies ab or ac.

  • Lemma 3.11.3: If R is a UFD and f,gR[x] are primitive, then fgR[x] is primitive.

  • Lemma 3.11.4 (Gauss Lemma): Let R be a UFD and hence an integral domain. By lemma 3.11.1 and theorem 3.6.1, R[x] embeds in a field of fractions F[x]. If fR[x] is primitive in R[x], then it is irreducible in R[x] if and only if it is irreducible in F[x].

  • Lemma 3.11.5: Let R be a UFD and let pR[x] be primitive. Then f may be factored uniquely into irreducible elements in R[x].

  • Theorem 3.11.1: If R is a UFD, then so is R[x].

  • Corollary: If R is a UFD, then so is R[x1,,xn] for any positive integer n.

  • Corollary: If F is a field, then F[x1,,xn] is a UFD.

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 3.11.1: Let R be a commutative, unital ring. Prove that R[x] is a commutative unital ring.

This problem justifies the recursive definition of R[x1,,xn]. It is trivial but tedious (and notationally cumbersome) to show that R[x] is a commutative ring, so I will take it for granted. The unit element is simply the constant polynomial fR[x] given by f(x)=1R for all x.

Herstein 3.11.2: Prove that R[x1,,xn]R[xσ(1),,xσ(n)] where σSn permutes the indices.

What follows is a sketch of a proof. First consider a simple illustration using R[x,y] and R[y,x]. It is easy to see that the elements of the former all have the form aijxiyj

while elements of the latter have the form aijyixj.

Defining these polynomials as “formal symbols” as we do, it is not a priori obvious that monomials rxiyj and ryjxi, with rR, are in any sense equal. Still, it is tempting to simply move the indeterminates around.

In R[x1,,xn], the above comments suggest the mapping ϕ:R[x1,,xn]R[xσ(1),,xσ(n)]

given by ϕ(ai1inx1i1xnin)=ai1inxσ(1)iσ(1)xσ(n)iσ(n).

Here, the coefficient a on each monomial remains the same while the positions of the indeterminates are permuted. For simplicity of notation, we can imagine each of i1,,in ranging from zero to infinity, although we note that only finitely many of the numbers in the a array can be non-zero (a polynomial has finite degree) so there are no awkward questions of convergence, etc.

We assert that ϕ is an isomorphism, which is easy to believe. The only non-trivial thing to check is that ϕ respects ring multiplication. With correct bookkeeping, one sees that it does, but the details will be skipped here.

Herstein 3.11.3: Let R be an integral domain and let f,gR[x]. Prove that deg(fg)=deg(f)+deg(g).

Let f=r0+r1x++rnxn and let g=s0+s1x++smxm, where all the coefficients belong to R and rn,sm0. Then the degree of f is n and the degree of g is m. The term of largest power in fg is rnsmxm+n, and its coefficient is rnsm0 because R is an integral domain. Hence the degree of fg is m+n.

Herstein 3.11.4: Let R be a unital integral domain and let uR[x] be a unit. Show that u is also a unit in R.

Let vR[x] be such that uv=1R[x]. We have by problem 3.11.3 that 0=deg(uv)=deg(u)+deg(v), which forces both u and v to have degree zero, i.e. to be elements of R. But then uv=1R so that uR is a unit. This is a slight abuse of notation, but the idea is clear that uR[x] is a constant polynomial which we identify with its lone coefficient belonging to R.

Herstein 3.11.5: An element x in a ring is nilpotent if there exists positive nZ such that xn=0. Let R be a commutative ring with no non-zero nilpotent elements and let fR[x] be a zero divisor given by f(x)=a0+a1x++amxm. Prove there exists non-zero bR such that ba0=ba1==bam=0.

Because f is a zero divisor, there exists gR[x] such that fg=0. Then if g(x)=b0+b1x++bnxn, we have the product (fg)(x)=k=0m+nckxk

where ck=akb0+ak1b1++a0bk. Now this polynomial is identically zero, so each ck=0. For instance, we see that c0=a0b0=0 and that c1=a0b1+a1b0=0. Because a0b0=0, we can try multiplying the second equation by b0, in which case we see 0=a0b0b1+a1b02=a1b02.

This suggests an induction: we claim that b0+1a=0 for all <k and then we consider 0=b0kck+1=b0k+1ak+b0kak1b1+b0kak2b2++b0ka0bk.

By the induction hypothesis, all terms vanish except the first, leaving 0=b0k+1ak

which completes the induction.

Returning to the problem at hand, we see that b=b0m+1 is such that ba0=ba1==bam=0.

If b00 then we are done, because b=b0m+10 by the prohibition on nilpotent elements. The case of b0=0 is somewhat trivial. If b0=0, we may rewrite g as g(x)=xrh(x) where the degree-zero coefficient of h(x) is non-zero, and then the above analysis goes through by using the coefficients of h rather than those of g.

Herstein 3.11.6*: Let R be a commutative ring and let fR[x] be a zero divisor given by f(x)=a0+a1x++amxm. Prove there exists non-zero bR such that ba0=ba1==bam=0. This is the same as problem 3.11.5 without the nilpotency condition.

Here is a clever and subtle proof of this result due to this post.

Let non-zero gR[x] be of minimal degree such that fg=0 and write g=b0+b1x++bnxn. Suppose that there does not exist any such b having ba0==bam=0.

If all of the coefficients ai, i=0,,m annihilate g, having aig=0, then we would certainly have aibn=0

for every i. However, this would lead to bnf=bna0++bnamxm=0,

contradicting our supposition that no such b=bn exists.

Therefore if we accept that no such b exists, then some coefficient fails to annihilate g. In particular, there is a largest index i such that aig0. Now, fg=(a0++aixi++amxm)(b0++bnxn)=(a0++aixi)(b0++bnxn)=0 \begin{aligned} fg&=\Big(a_0+\cdots+a_i x^i+\cdots+a_m x^m\Big)\Big(b_0+\cdots+b_n x^n\Big)
=&\Big(a_0+\cdots+a_i x^i\Big)\Big(b_0+\cdots+b_n x^n\Big)=0 \end{aligned}

because ajg=0 for j>i. Expanding this product, it is seen that aibn=0 so that deg(aig)<n because we have previously determined that aig0. We had supposed that g was the non-zero polynomial of least degree which annihilated f, but now we are forced to conclude that f(aig)=0 as well. This contradiction proves our result.

The above proof does not exhibit any concrete element of R which sends f to zero. A constructive proof shows that every coefficient of f must indeed annihilate g (contrast with the second paragraph) and has us conclude that bnf=0.

Herstein 3.11.7*: Let R be a commutative, unital ring. Show that fR[x] given by f(x)=a0+a1x++anxn is a unit in R[x] if and only if a0 is a unit in R and a1,,an are nilpotent in R.

There is intuition to be gained here from writing 1f=1a0(1+f),

where f=(fa0)/a0, and then “Taylor expanding”: 1f=1a0(1f+f2).

Roughly speaking, this should only make sense when (1) a0 is invertible, and (2) the series terminates due to nilpotency. Now we seek to formalize this argument.

One way of doing so is to postulate this lemma: Let S be a commutative, unital ring and let non-zero nS be nilpotent with nk=0. Then 1+n is a unit. Proof: Indeed, if we try the “Taylor expansion” as an inverse, we find (1+n)(1n+n2+(1)k1nk1)=1+(1)k1nk=1,

a telescoping sum. The series expansion for the inverse of 1+n may make sense in a formal context even if it does not terminate, but in our case we don’t need to worry – it terminates thanks to the nilpotency of n. Furthermore, if uS is any unit, we have u+n=u(1+u1n) is again a unit because u1n is nilpotent.

Now suppose that a0 is a unit in R and a1,,an are all nilpotent, say with nilpotency indices ki such that aiki=0. We have that f0=a0 is a unit, and, by the lemma, f1=a0+a1x is again a unit because a1x is nilpotent, (a1x)k1=a1k1xk1=0. The lemma applies because R[x] is a commutative, unital ring by exercise 3.11.1. Now repeating this stacking process n1 more times, we conclude that f=fn is a unit.

On the other hand, suppose that f were a unit and that g(x)=b0+b1x++bnxn

were such that fg=1. Writing out the product explicitly, we have that a0b0=1 so that a0 must be a unit. At the highest degree terms, we find anbm=0

and anbm1+an1bm=0.

Multiplying the second equation by an, the second term in the sum vanishes, and we are left with an2bm1=0. This suggests that ever higher powers of an should annihilate each successive coefficient of g, counting downwards. Performing the relevant induction, we see that it is true and, as a result, anm+1 annihilates every coefficient in g. Now anm+1=anm+1(fg)=f(anm+1g)=0

shows that an is nilpotent.

Finally, we consider the polynomial fanxn

which is a unit by the lemma, because f is a unit and anxn was just shown to be nilpotent. Repeating the argument from above, we conclude that an1 is nilpotent. By induction, the same may be said of each coefficient ai with i>1.

Some nudges toward this solution were provided by this excellent site. There is also a very well known, elegant proof that unit f implies nilpotent coefficients. It can be found in this post. However, it is clearly not the solution Herstein had in mind, because prime ideals have not been introduced at this point in the text. Also, to say that a prime ideal exists in R is a non-trivial statement, equivalent to the axiom of choice.

Herstein 3.11.8: Let F be a field. Prove that F[x1,x2] is not a principal ideal domain.

We try the most natural thing, which is to look at the ideal generated by x1 and x2. This hasn’t been defined in the text, but we take a stab, writing I=(x1,x2)={ax1+bx2a,bF[x1,x2]}.

It is trivial to see that I is indeed an ideal of F[x1,x2].

Let dI be arbitrary. Because F has no zero divisors and d is built on top of x1 and x2, we must have deg(d)1. Here we define the degree of a multivariate monomial as the sum of the powers on its indeterminates, and the degree of a multivariate polynomial as the largest out of the degrees of its constituent monomials. If d were to generate the ideal, we would also have dx1 and dx2, which is only possible if d is a constant, having degree zero. Therefore the ideal I is not principal and F[x1,x2] is not a principal ideal domain.

Herstein 3.11.9 (Lemma 3.11.2): Let R be a unique factorization domain and let a,b,cR.

(a) Show that a and b have a greatest common divisor in R.

(b) If (a,b)=1, then show that abc implies that ac.

© If a is irreducible and abc, then prove that ab or ac.

(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=uπ1πm and b=vρ1ρn where u,vR are units and the π’s and the ρ’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 π1,,πm and pick out those πi which are associated to some ρj. This must be done without repetition, so if such a ρj is found, it is removed from consideration for further π’s (for instance, consider a=33 and b=3; we would pick out 3 only once). Now, define dR to be the product of those πi which were picked out by this algorithm. We claim that d is the greatest common divisor of a and b.

It is clear that da because it was constructed out of a subset of factors of a. We also have that db because each of those factors was associated to a factor of b, and we were mindful of repetitions. Now suppose that ca and cb and perhaps cd. Then there would exist an irreducible σR with σc and σd. Of course, we must have σa and σb, but then σd contradicts the construction of d. Therefore it must be true that cd and as a result d is the greatest common divisor.

(b) Let a=uπ1πl, b=vρ1ρm and c=wσ1σn where u,v,wR are units and the π’s, ρ’s and σ’s are all irreducible elements of R. For each πi, abc implies that πi is associated to some ρj or σj. Because (a,b)=1, we know that πi is not associated to any ρj. Hence it is associated to some σ. This argument holds for every πi, and thus ac.

© If (a,b)=1 then ac by part (b). The only other choice for (a,b) is for (a,b)=a, because a is irreducible. In that case, we have ab because b will always be divisible by (a,b).

Herstein 3.11.10: Let R be a unique factorization domain and let fR[x].

(a) Prove that we may express f(x)=ag(x) where aR and gR[x] is primitive.

(b) Prove that the decomposition of part (a) is unique up to associates.

(a) Suppose f(x)=a0+a1x++anxn. By trivial extension of exercise 3.11.9, the greatest common divisor (a0,,an) exists in R. Then we may take a=(a0,,an), 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,bR and g,hR[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 uR, we see that 0=ag(x)bh(x)=a(g(x)uh(x))

also implies that gh. 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 fF[x]. Show that f(x) may be decomposed as g(x)/a where aR and gR[x].

Again this is a fairly easy idea just couched in some abstract language. The idea is to clear the denominators. Write f(x)=r0s0+r1s1x++rnsnxn

with ri,siR and si 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 adbc=0.

The clear thing to consider is p(x)=1s0sn(r0s1sn+s0r1s2snx++s0sn1xn),

which is in the form p(x)=g(x)/a with aR and g(x)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 r0s0=r0s1sns0sn,

and of course r0(s0sn)s0(r0s1sn)=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 fR[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,hR[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 rR 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[x1,,xn] is a unique factorization domain.

Theorem 3.11.1 and its corollary state that if R is a UFD then R[x1,,xn] 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 RR be an integral domain. RR is a unique factorization domain (UFD) if (1) non-zero rRrR is either a unit or expressible as a product of a finite number of irreducible elements of RR, (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. 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 rR 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 rR 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 a0 of r (that is, a0 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, a0 has a proper factor a1 with the infinite divisibility property, and so forth. Now consider the chain of ideals (a0)(a1)(a2)

where each inclusion is proper. It is easy to show that the set U=i=0 (ai)

is also an ideal of R: first note that if xU then x(ai) for some i. If sR and xU, then x(ai) for some i and sx(ai)U. If x,yU, with x(ai) and y(aj), then, because of the inclusion chain, both x and y belong to (amax(i,j)) so that x+y(amax(i,j))U.

Finally, we employ the fact that R is a PID. Because U is an ideal, it is principal, say U=(α) with αR. Because αU, we know that α(ai) for some i. But then if ji, (aj) 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,bR have a greatest common divisor. Consider the ideal I={ax+byx,yR}, which must be principal, say I=(d) for dR. 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 x0,y0R be such that d=ax0+by0. Then if ca and cb, then c(ax0+by0)=d. Therefore d is the greatest common divisor of a and b, and dR.

If πR is irreducible and a,bR and πab, we would like to show that πa or πb. If πa, then (π,a)=1. Therefore there exist x,yR with πx+ay=1 and consequently πxb+aby=b. Because πab, we have that π divides the left hand side, but of course then it also dividies the right hand side: πb. Therefore if πa, then πb. Otherwise πa. 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=π1πm=ρ1ρn be two factorizations of r into irreducibles. Because π1 divides the right hand side, it must be associate to one of the ρi. Because we are working in a domain, we can cancel π1 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 Z[x1,,xn] is a unique factorization domain.

By corollary 1 to theorem 3.11.1, we have that, when R is a UFD, R[x1,,xn] is a UFD. The fact that Z is a UFD is the statement of the fundamental theorem of arithmetic (theorem 1.3.1). Therefore the result follows.