This section covers section 3.11 (“Polynomial Rings over Commutative Rings”).
Definition: Let be a commutative ring with a unit element. is defined, as in section 3.9, as the set of all symbols where is a non-negative integer and . The standard addition and multiplication make a ring. It will be shown in exercise 3.11.1 that so defined is a commutative ring with unit element.
Definition: Let be a commutative ring with unit element. By problem 3.11.1, is again a commutative ring with unit element. Therefore we may recursively define as .
Lemma 3.11.1: If is an integral domain, then is an integral domain.
Corollary: If is an integral domain, then is an integral domain for any positive integer .
Definition: Let be an integral domain. is a unique factorization domain (UFD) if (1) non-zero is either a unit or expressible as a product of a finite number of irreducible elements of , (2) the decomposition just mentioned is unique up to re-ordering and multiplication by units.
Lemma 3.11.2: If is a UFD and , then , i.e. their gcd exists in . Additionally, if then implies that .
Corollary: If is a UFD and is irreducible, then implies or .
Lemma 3.11.3: If is a UFD and are primitive, then is primitive.
Lemma 3.11.4 (Gauss Lemma): Let be a UFD and hence an integral domain. By lemma 3.11.1 and theorem 3.6.1, embeds in a field of fractions . If is primitive in , then it is irreducible in if and only if it is irreducible in .
Lemma 3.11.5: Let be a UFD and let be primitive. Then may be factored uniquely into irreducible elements in .
Theorem 3.11.1: If is a UFD, then so is .
Corollary: If is a UFD, then so is for any positive integer .
Corollary: If is a field, then 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.
This problem justifies the recursive definition of . It is trivial but tedious (and notationally cumbersome) to show that is a commutative ring, so I will take it for granted. The unit element is simply the constant polynomial given by for all .
What follows is a sketch of a proof. First consider a simple illustration using and . It is easy to see that the elements of the former all have the form
while elements of the latter have the form
Defining these polynomials as “formal symbols” as we do, it is not a priori obvious that monomials and , with , are in any sense equal. Still, it is tempting to simply move the indeterminates around.
In , the above comments suggest the mapping
Here, the coefficient on each monomial remains the same while the positions of the indeterminates are permuted. For simplicity of notation, we can imagine each of ranging from zero to infinity, although we note that only finitely many of the numbers in the array can be non-zero (a polynomial has finite degree) so there are no awkward questions of convergence, etc.
We assert that is an isomorphism, which is easy to believe. The only non-trivial thing to check is that respects ring multiplication. With correct bookkeeping, one sees that it does, but the details will be skipped here.
Let and let , where all the coefficients belong to and . Then the degree of is and the degree of is . The term of largest power in is , and its coefficient is because is an integral domain. Hence the degree of is .
Let be such that . We have by problem 3.11.3 that , which forces both and to have degree zero, i.e. to be elements of . But then so that is a unit. This is a slight abuse of notation, but the idea is clear that is a constant polynomial which we identify with its lone coefficient belonging to .
Because is a zero divisor, there exists such that . Then if , we have the product
where . Now this polynomial is identically zero, so each . For instance, we see that and that . Because , we can try multiplying the second equation by , in which case we see
This suggests an induction: we claim that for all and then we consider
By the induction hypothesis, all terms vanish except the first, leaving
which completes the induction.
Returning to the problem at hand, we see that is such that
If then we are done, because by the prohibition on nilpotent elements. The case of is somewhat trivial. If , we may rewrite as where the degree-zero coefficient of is non-zero, and then the above analysis goes through by using the coefficients of rather than those of .
Here is a clever and subtle proof of this result due to this math.stackexchange.com post.
Let non-zero be of minimal degree such that and write . Suppose that there does not exist any such having .
If all of the coefficients , annihilate , having , then we would certainly have
for every . However, this would lead to
contradicting our supposition that no such exists.
Therefore if we accept that no such exists, then some coefficient fails to annihilate . In particular, there is a largest index such that . Now,
because for . Expanding this product, it is seen that so that because we have previously determined that . We had supposed that was the non-zero polynomial of least degree which annihilated , but now we are forced to conclude that as well. This contradiction proves our result.
The above proof does not exhibit any concrete element of which sends to zero. A constructive proof shows that every coefficient of must indeed annihilate (contrast with the second paragraph) and has us conclude that
There is intuition to be gained here from writing
where , and then “Taylor expanding”:
Roughly speaking, this should only make sense when (1) 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 be a commutative, unital ring and let non-zero be nilpotent with . Then is a unit. Proof: Indeed, if we try the “Taylor expansion” as an inverse, we find
a telescoping sum. The series expansion for the inverse of 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 . Furthermore, if is any unit, we have is again a unit because is nilpotent.
Now suppose that is a unit in and are all nilpotent, say with nilpotency indices such that . We have that is a unit, and, by the lemma, is again a unit because is nilpotent, . The lemma applies because is a commutative, unital ring by exercise 3.11.1. Now repeating this stacking process more times, we conclude that is a unit.
On the other hand, suppose that were a unit and that
were such that . Writing out the product explicitly, we have that so that must be a unit. At the highest degree terms, we find
Multiplying the second equation by , the second term in the sum vanishes, and we are left with . This suggests that ever higher powers of should annihilate each successive coefficient of , counting downwards. Performing the relevant induction, we see that it is true and, as a result, annihilates every coefficient in . Now
shows that is nilpotent.
Finally, we consider the polynomial
which is a unit by the lemma, because is a unit and was just shown to be nilpotent. Repeating the argument from above, we conclude that is nilpotent. By induction, the same may be said of each coefficient with .
Some nudges toward this solution were provided by this excellent site. There is also a very well known, elegant proof that unit implies nilpotent coefficients. It can be found in this math.stackexchange.com 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 is a non-trivial statement, equivalent to the axiom of choice.
We try the most natural thing, which is to look at the ideal generated by and . This hasn’t been defined in the text, but we take a stab, writing
It is trivial to see that is indeed an ideal of .
Let be arbitrary. Because has no zero divisors and is built on top of and , we must have . 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 were to generate the ideal, we would also have and , which is only possible if is a constant, having degree zero. Therefore the ideal is not principal and is not a principal ideal domain.
(a) The idea here is very simple: construct the greatest common divisor as the product of all the primes which and share. Formally, suppose that and where are units and the ’s and the ’s are all irreducible elements of , not necessarily distinct. Because is a UFD, these decompositions of and are finite and unique up to associates. Out of those decompositions, run through the list of and pick out those which are associated to some . This must be done without repetition, so if such a is found, it is removed from consideration for further ’s (for instance, consider and ; we would pick out only once). Now, define to be the product of those which were picked out by this algorithm. We claim that is the greatest common divisor of and .
It is clear that because it was constructed out of a subset of factors of . We also have that because each of those factors was associated to a factor of , and we were mindful of repetitions. Now suppose that and and perhaps . Then there would exist an irreducible with and . Of course, we must have and , but then contradicts the construction of . Therefore it must be true that and as a result is the greatest common divisor.
(b) Let , and where are units and the ’s, ’s and ’s are all irreducible elements of . For each , implies that is associated to some or . Because , we know that is not associated to any . Hence it is associated to some . This argument holds for every , and thus .
© If then by part (b). The only other choice for is for , because is irreducible. In that case, we have because will always be divisible by .
(a) Suppose . By trivial extension of exercise 3.11.9, the greatest common divisor exists in . Then we may take , the content of , and the coefficients of are those of with the common irreducible divisors pulled out. By definition of gcd, we know that is primitive.
(b) The corollary to lemma 3.11.3 states that, with a UFD, the content of a product of polynomials in is the product of the contents of the polynomials, modulo multiplication by units.
Suppose where and are primitive. Applying the statement above, we have that is associated to (and both are associates of the content of ). Writing with unit , we see that
also implies that . Therefore the decomposition of part (a) is unique up to associates.
Again this is a fairly easy idea just couched in some abstract language. The idea is to clear the denominators. Write
with and restricted to non-zero values. The fraction notation is shorthand for equivalence classes in where, as we recall, if and only if .
The clear thing to consider is
which is in the form with and . We must establish that in . This is easily worked out for each coefficient individually. For instance, for the zeroth coefficient, we would like to show that
and of course
so we are correct. All other coefficients follow in the same manner, and we see that the result has been established.
This does not seem to require that be primitive. Suppose that were reducible in , say with of positive degree. Then such a factorization also exists in because and can be considered as elements of if we replace each coefficient by the fraction (the unit element exists because is a UFD, but it is not essential: the fraction for arbitrary non-zero would suffice). So if is reducible in then it must be reducible in , and the contrapositive is the statement that we want.
Theorem 3.11.1 and its corollary state that if is a UFD then is also a UFD. Therefore we must show here that a field 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∈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 is a UFD and the result holds by Theorem 3.11.1.
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 is a UFD we must prove two facts: (1) Every element can be factored into a finite number of irreducible elements of , (2) The factorization from (1) is unique up to reordering and multiplication by units.
(1) Let be such that it cannot be factored into finitely many irreducible elements. This means that, no matter how 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 of (that is, not an associate of ) with the same “infinite divisibility” property: if had a factorization where none of the factors had the property, then itself would be expressible as a finite product of irreducibles, a contradiction. By the same argument, has a proper factor with the infinite divisibility property, and so forth. Now consider the chain of ideals
where each inclusion is proper. It is easy to show that the set
is also an ideal of : first note that if then for some . If and , then for some and . If , with and , then, because of the inclusion chain, both and belong to so that .
Finally, we employ the fact that is a PID. Because is an ideal, it is principal, say with . Because , we know that for some . But then if , is supposed to properly contain , which is impossible because is at least as large as all of the 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 dividing must divide one of or . The result holds for Euclidean rings (lemma 3.7.6) and for UFDs (exercise 3.11.9c), but not, a priori, for PIDs.
With a PID, have a greatest common divisor. Consider the ideal , which must be principal, say for . Clearly is a divisor of both (with and ) and (with and ). It is also the greatest such divisor: let be such that . Then if and , then . Therefore is the greatest common divisor of and , and .
If is irreducible and and , we would like to show that or . If , then . Therefore there exist with and consequently . Because , we have that divides the left hand side, but of course then it also dividies the right hand side: . Therefore if , then . Otherwise . 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 be two factorizations of into irreducibles. Because divides the right hand side, it must be associate to one of the . Because we are working in a domain, we can cancel from both sides, leaving a unit on the right hand side but one less irreducible. Continuing this process, we end up with 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 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.
By corollary 1 to theorem 3.11.1, we have that, when is a UFD, is a UFD. The fact that is a UFD is the statement of the fundamental theorem of arithmetic (theorem 1.3.1). Therefore the result follows.