This page covers section 3.9 (“Polynomial Rings”). Throughout, is a field.
Topics covered: 3.9
is the set of polynomials with coefficients from the field .
Addition, multiplication and equivalence of polynomials are defined in the obvious ways.
Definition: The degree of non-zero , , is the coefficient of the highest non-zero power of in . The degree is not defined for the zero polynomial.
Lemma 3.9.1: If are non-zero, then .
Corollary: is an integral domain.
Lemma 3.9.2: Given , , there exist such that with either or . This is the Euclidean algorithm.
Theorem 3.9.1: is a Euclidean ring with used as norm.
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 3.9.1: Find the greatest common divisor in of
(a) and .
(b) and .
(a) Unlike the exercise on Gaussian integer divisors, polynomial long division takes away any interesting guess work in computing the greatest common divisor (we use the algorithm in problem 3.7.4). The numbers for this particular example work out horrendously, but we find (in the notation of 3.7.4) that
which is a unit in , so and are coprime in .
(b) In stark contrast to the hideous arithmetic of part (a), we find that actually divides . Specifically,
so their greatest common divisor is .
(a) Show that is irreducible over .
(b) Show that is irreducible over .
(c) Show that is irreducible over .
(d) Show that is reducible over .
(a) Suppose that with , neither being a constant. Then we must have . Write and , where can each take only the values or . Expanding the product, we have which implies , and implies . Then the cross term is , but the coefficient of is supposed to be . Thus can not be factored unless one of the factors is a unit, so it is irreducible.
(b) As before, let us try to write . We find by expanding that and . In fact, we can factor out and to get . Equivalently, just call . Then we have
which has us conclude that . But then
However, because is a prime, this is not possible (problem 3.8.4) and therefore is irreducible over . Note that this argument fails if we work over other fields such as . In that case, we have and , so
(c) If is reducible, then it must have at least one linear factor by simple considerations of degree. We can therefore write a proposed factorization as
The coefficients give conditions , and . Combining all three, we find that
By brute force we see that this can never happen, so is irreducible over .
Alternately, as suggested in this math.stackexchange.com post, we note that is a cyclic group of order so that . Then if is expressible as a cube, it must be the case that . However, we can easily show that : first switch back to computing because is smaller and easier to square by hand. We have and then and . Finally, plugging back in, we get
This shows that is not the cube of any number in , so the polynomial is irreducible.
(d) The same reasoning as above brings us to the condition . This can be simplified slightly by applying Fermat’s little theorem. We can see that
From here it can easily be brute forced by hand, and we find that so and constitutes a solution. Specifically,
over . It is immaterial whether factors further (it doesn’t), because the object was simply to show that is reducible.
Herstein 3.9.3: Let be two fields and suppose are relatively prime. Show that they are relatively prime in .
Because and are relatively prime in , there exist such that
for all . The elements and also belong to , so this identity holds there, too.
Herstein 3.9.4: Let .
(a) Prove that is irreducible over and explicitly prove that is a field with elements.
(b) Prove that is irreducible over and explicitly prove that is a field with elements.
(c) Prove that .
(a) is irreducible over by precisely the same argument as in problem 3.9.2b. Denote the ideal by . We have defined
and we see by the Euclidean algorithm that any representative may be written as
with or by the Euclidean algorithm. Hence . We can express , where each have possible values. Therefore we can enumerate elements in . It is conceivable that there is some overcounting, but if
then must divide . This is only possible if both coefficients are zero, so and and we must have no duplicates. Therefore there are elements in .
To show that is a field, we take the properties of a ring for granted and would just like to show that every element has a multiplicative inverse. That is, we would like to explicitly produce an inverse of the arbitrary element . If and , then the inverse is ; if and , the inverse is because
Therefore we are left to solve
for and , with . Comparing coefficients quickly indicates the forms and . Then
It will always be possible to solve this if the latter term in parentheses is non-zero, because can be chosen as its inverse. Suppose that, for some , . Then or . This we know to be impossible by problem 3.8.4. Therefore if we take
then the inverse of in is given by and, consequently, is a field.
(b) Suppose that factors in as
with . Then we find and (ignoring the trivial possibilities that or ) and, plugging in and extracting a factor of , we see that
The condition on the coefficient is then which may be restated as for some . If we can show that there is no solution to this last equation, then we have shown that is irreducible over . Multiply through by to get or . Now, modulo , we have that
and we know that is not possible. Therefore the polynomial is irreducible over .
Letting , the Euclidean algorithm argument from part (a) shows that there are exactly elements in . Again, we would like to explicitly produce an inverse of in . If and , then the inverse is simply . If and , then we have
Write this as
Then we have and modulo . The latter equation is solved by multiplying through by . Thus
Finally, we come to the case of . If , then the same trick of adding and subtracting terms gives
Then and so
Can ? Multiply through by to put it in the form
where . As shown above, this is equivalent to which is not possible modulo . Hence we can always solve for
so the inverse is constructed and is a field.
(c) We would like to construct an explicit isomorphism . An important first question is whether such a can refer to the coefficients and of in a well-defined way. The answer is yes: if , then is divisible by . This is only possible if both coefficients are zero, so and .
A natural guess for the isomorphism is the simple . A homomorphism must satisfy for . However, we can see, for example, that
so this map fails to be a homomorphism.
If we work from the restriction that be a homomorphism, we know that
Then the map will be fully determined by computing its action on and . We know by problem 3.4.20 that if is an isomorphism then it maps onto and thus . From this, it easily follows that for any . For , we have
This implies that must be an element of which squares to . Such an element is , because
So we posit that defined by
is an isomorphism. The only non-trivial property to check is that respects products. Letting , we have
Therefore respects the ring product. It is very easy to check that it respects ring addition and that it is injective, and so we have proven that .
Herstein 3.9.5: Letting , show that as fields.
First we show that is irreducible over . Suppose that . Then, comparing coefficients, and imply that , and . Then the middle coefficient says . This gives which in turn implies that , a contradiction.
Therefore is irreducible over and lemma 3.9.6 says that is maximal in . Theorem 3.5.1 then says that is a field much like the one studied in problem 3.9.4a. The elements look like with . The obvious choice for a map to is
which is well-defined by the comments made in 3.9.4c. One has to be a little careful here: (with the and flipped from the definition above) fails to respect the ring multiplication. With the definition of , we have
The additive property is very easy to show, so is a homomorphism. It is also trivially surjective because is the image of , and trivially injective because implies or and . Therefore, .
Herstein 3.9.6*: Let and define the formal derivative of polynomial in the usual way. Prove that is divisible by the square of a non-constant polynomial if and only if and have a greatest common divisor of positive degree.
Note that, over , the degree of a polynomial’s derivative is exactly one less than the degree of the polynomial. In other fields, this may not be the case. For instance, the derivative of over is zero.
First suppose that is divisible by the square of a non-constant polynomial , writing
Then the derivative is so that . As is a common divisor, it also divides the greatest common divisor , and so as desired.
On the other hand, suppose that we know and have a non-trivial greatest common divisor . Factor into primes as
and take any of the primes, say . We have that is a common divisor of and , so write and . We can also compute the derivative by the chain rule, which gives . Equating the two representations gives
Because is prime, it divides either or . The latter is impossible, because is strictly smaller than . Therefore and .
Herstein 3.9.7: Let with prime, and let be irreducible and have degree . Prove that is a field with elements.
By lemma 3.9.6, is maximal and, by theorem 3.5.1, is a field. A generic element in is where . By the division algorithm, we can write
where or . Therefore, and so we will always be able to identify an equivalence class by a polynomial of degree smaller than . The representative of any equivalence class can be written
where for each . With choices for each of the coefficients, there are at most equivalence classes comprising . However, there are also no duplicates. Indeed, if , then is divisible by . Because each of and has degree smaller than , so does their difference. The only way for the difference to be divisible by is if it is zero, i.e. if . Therefore each of the polynomials described above represents a unique equivalence class, and has elements.