Responses to E-mail questions about Math 371- Fall 2014

INVARIANT FACTORS

QUESTION: "I'm looking at the definition of the isomorphism for a finitely generated module over a PID. I'm specifically wondering about the invariant factors - R^t + R/a1 + R/a2 + ... + R/an the book and my notes specify that a1 | a2 | ... | an. Are there other stipulations on the a's? I remember that for the fundamental theorem of finite abelian groups that all of the factors (the theorems somewhat resemble each other) all the ai's have a product of the initial group's order. Is there a condition like this for the ai's in a finitely generated module over a PID?"

ANSWER: The only conditions are that t >= 0, n >=0, the a_i are not units, and a_1|...|a_n. Keep in mind that the case t = 0 = n corresponds to the module {0}. Also, the invariant factors a_1,...,a_n are each unique only up to multiplying each of them by a possibly different unit. It is really just the ideals Ra_1,...,Ra_n which are uniquely determined by the isomorphism class of the module, along with the rank t of the module. The rank t is the maximal number of R-linearly independent elements in the module, and it is the dimension over the fraction field F of R of the F-vector space generated by the module in the way we discussed in class.

TESTING POLYNOMIALS FOR IRREDUCIBILITY

QUESTION: "I have a question about determining whether or not a polynomial is irreducible. The text references the Eisenstein test several times, but I don't think we discussed this test in class. Is there another way to determine whether or not a polynomial is irreducible? I know what to do for polynomials of degree 2 and 3, but I am unsure about degree 4. In particular, I am struggling with f(x) = x^4+2. "

ANSWER: I didn't discuss the Eisenstein criterion because of lack of time, so this will not be on the final exam. You could keep in mind that polynomials of degree 2 or 3 are irreducible if and only if they have no root in the base field. Also, if the base field is Q, all the roots in Q of a monic polynomial with coefficients in Z have to be in Z.

Here is an explicit approach to testing the irreducibility of monic polynomials f(x) in Q[x] such as f(x) = x^4 + 2. Suppose one can find the complex roots a_1,...,a_n of f(x), so that f(x) = (x-a_1)*...*(x-a_n) in C[x]. If f(x) is not irreducible then it has a monic factor g(x) in Q[x] which the product of (x - a_i) over some non-empty proper subset I of {1,...,n}. So to show that no such g(x) exists, one can simply work through the possibilities for I and check that the corresponding g(x) does not in fact have all its coefficients in Q[x].

Suppose, for instance that f(x) = x^4 + 2. The complex roots of f(x) are a_i = z^i * b where b is one choice of complex fourth root of 2 and z = \sqrt{-1}. We know that there is no fourth root of 2 in Q, so f(x) cannot have a linear factor in Q[x]. The only possible irreducible monic g(x) in Q[x] which can divide f(x) without being equal to f(x) are thus of degree 2. Such a g(x) must equal (x - a_i)*(x - a_j) for some 1 <= i , j <-= 4 with i <> j. This would require a_i*a_j to be a rational number. However, the complex absolute value of |z| is 1, so |z^i * b| = |b| = the real positive fourth root of 2. So |a_i * a_j| = |a_i|*|a_j| = the real positive square root of 2. Since a_i*a_j is supposed to be rational, we would conclude that there has to be a rational square root of 2 if g(x) exists, which is not possible. So f(x) is irreducible.

EUCLIDEAN RINGS, P.I.D.'s and U.F.D.'s

QUESTION: After doing a review of concepts with others in the class, I am still confused about the idea of a principal ideal domain. I have that a PID is an ideal generated by a single element. Does this mean that anything in the ideal has to be a multiple of g, g^2, g^3... if g is that single element? Additionally could you provide some examples of things that are a UFD but not a PID, or PID that is not a Euclidean domain?

ANSWER: An integral domain is a commutative ring R such that whenever b and c are elements of R such that bc = 0, then one of b or c have to be 0. For instance the integers Z are an integral domain. The ring Z/6 is not an integral domain, since b = (3 mod 6) and c = (2 mod 6) are not 0, but bc = (6 mod 6) = (0 mod 6).

An ideal I in a commutative ring R is an additive subgroup such that whenever i is in I and r is in R, the product ri is in I. For instance I = Z6 = {r6: r in Z} is an ideal inside R = Z.

One says that an ideal I in R is principal if I is the set Rd of all multiples of some element d of R. For instance I = Z6 is a principal ideal in R = Z on letting d = 6.

An integral domain R is a P.I.D. (principal ideal domain) if every ideal is principal. For instance, Z is a principal ideal domain and so is F[x] whenever F is a field. This is because both these rings are Euclidean domains, and all Euclidean domains are P.I.D.'s.

In class we talked about the following example of a non-principal ideal in a commutative ring. Let R = Z[x] be the ring of polynomials in one variable over the integers. Suppose I = R2 + Rx= {r2 + sx : r and s in R}. Then I is an ideal, since it is an additive subgroup closed under multiplication by elements of R. But I is not principal for the following reason. If I = Rd for some element d of R, then both 2 and x would have to be multiples of d. It is not hard to check that this forces d to be 1 or -1. However, then Rd= R while I it not all of R. This shows Z[x] is not a P.I.D..

With some more work, one can show Z[x] is in fact a U.F.D.. We did not cover the background necessary to show that, so giving an example of a U.F.D. which is not a P.I.D. won't be a question on the exam. But Z[x] is such an example.

The question of constructing a P.I.D. which is not Euclidean with respect to any norm also leads off in a direction we did not cover in the course, so I won't say more about this. The main thing to keep in mind that a Euclidean ring is a P.I.D., and being a P.I.D. is sufficient to show a ring is a U.F.D..

JORDAN FORM

QUESTION: In question 6 of Homework 7, James determined the dimension of the 1-eigenspace of A. Why do we need to do this? How do we determine the sizes of the Jordan Blocks?

ANSWER: If J(r,n) is an n by n Jordan block with eigenvalue r and I_n is the n by n identify matrix, then T(r,n) = J(r,n) - r*I_n is an upper triangular matrix of size n by n which has 1's just above the diagonal. Suppose we are working over a field F. Then we can multiply column vectors of size n on the left by T(r,n), giving an action of T(r,n) on F^n. Fix a positive integer c. Then the set B(c) of vectors v in F^n such that T(r,n)^c * v = 0 is a vector space of dimension c if 1 \le c < n and of dimension n if c >= n. Suppose now that g is an F-linear transformation of some vector space F^m, and that all the eigenvalues of g lie in F, so that g can be put in Jordan canonical form. From the dimensions of the spaces of vectors v in F^m which are sent to 0 by (g-r)^c for each eigenvalue r and each integer c > 1, one can deduce the shape of the Jordan blocks. The most important case to keep in mind is just the difference between a diagonalizable matrix, in which all the blocks are 1 by 1, and the case when some block is larger than 1. All the blocks are 1 by 1 if an only if for all eigenvalues r, the vectors v such that (g - r) * v = 0 are the same as the vectors v such that (g-r)^2 *v = 0. To see this, just think about writing down a diagonal matrix and then calculating g - r*Identity.

DISCRIMINANTS

QUESTION: I'm reading through the discussion on discriminants in the textbook and I am a bit confused about the use of the discriminant in finding the splitting field of the irreducible polynomial f(x), specifically a cubic. On page 613, "the splitting field for the irreducible cubic f(x) is obtained by adjoining sqrt(D) and a root of f(x) to F." Is it strictly easier to obtain the splitting field by adjoining the square root of the discriminant in place of adjoining all the roots of the polynomial? To be clear, I do understand that the determinant tells us what the Galois group will be isomorphic to, I am just wondering about that line where the splitting field is mentioned and whether it is easier to use the discriminant to find the splitting field.

ANSWER: The following general fact is true for irreducible separable monic polynomials f(x) in F[x]. Let E be the field generated over F by the roots of f(x). Then E/F is a Galois extension. The action of G = Gal(E/F) on the roots {r_1,...,r_n} of f(x) gives an embedding of G into S_n. The alternating group A_n is a subgroup of index 2 in S_n for n >= 2. The group H = A_n \cap G is a subgroup of index either 1 or 2 in G. The fixed field E^H of the action of H on E is the field F(\sqrt{disc(f)}) generated over F by either square root of the discriminant of f = f(x). One can see this by using the fact that an automorphism \sigma in G fixes \sqrt{disc(f)} if and only if \sigma lies in A_n. When n = 3, either G = A_3 or G = S_3. If G = A_3, then the discriminant is a square in F. Otherwise G = S_3, then E has degree 6 = #S_3 over F, and E must the the field generated by on root f(x) together with the quadratic extension F(\sqrt{disc(f)}). This is a special situation for cubic separable irreducible polynomials.

IDEALS IN RINGS

QUESTION: I have a relatively simple question - how to find ideals in a ring. In class, you gave the example of getting all the ideals of Z/4: they are {0}, {0,2}. and the whole ring, but I may have missed the explanation on how you found ideals in the first place. Is finding ideals of a ring Z mod n just finding all the factors that divide n? For example, ideals of Z/12 would be Z/2, Z/3, Z/4, Z/6, Z/12?

And another question: for a ring Z/n, would the maximal ideals be all the Z/m where m is a prime number that divides n? For ex., Z/2 is a max ideal of Z/4 since 2|4 and 2 is prime.

ANSWER: Left ideals are additive subgroups of a ring which are closed under left multiplication by arbitrary elements of the ring. So in the ring R = Z/12, you are looking for subgroups, not quotient groups. Each subgroup in this case will be an ideal since it will be closed under left multiplation by integers. But for other rings, the requirement of being closed under left multiplication is not automatically satisfied by every additive subgroup. The ideals in R = Z/12 are the R itself, {0}, R2, R3, R4 and R6. As for the second question, keep in mind that we are looking for subgroups, not quotient groups. It is true that the maximal ideals of the ring A = Z/n for n > 1 are of the form Ap when p is a prime divisor of n.

Last updated: 8/24/14