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1 Homework 1
(1) Prove the ideal (3,x) is a maximal ideal in Z[x]. SOLUTION: Suppose we expand this ideal by including another generator polynomial, P ∈/ (3, x). Write P = n + x ∗ Q with n an integer not divisible by 3 (if 3|n then P ∈ (3, x) so we have not expanded the ideal) and Q is some polynomial. Then subtracting off a multiple of one generator from another does not change the ideal (analogous to row operations on a matrix not changing the row space), so in particular (3, x, P ) = (3, x, n). As n is not a multiple of 3, gcd(3, n) = 1, so 1 ∈ (3, x, P ) and thus (3, x, P ) is all of Z[x]. Thus (3,x) is maximal.
(2) Prove that (3) and (x) are prime ideals in Z[x]. SOLUTION: If P and Q are polynomials, then the constant term of P Q is the product of the constant terms of P and Q. Thus, if P Q ∈ (x) then the product of their constant terms is 0, and since Z is an integral domain, this means one of them has a constant term equal to 0, hence lies in (x). Thus (x) is prime. If neither P or Q is in (3), then each has at least one coefficient which is not a multiple of 3. Suppose pi and qj are such coefficients with i and j each minimal (i.e. come from the lowest degree term of P and Q such that the coefficient is not a multiple of 3). Consider the coefficient of xi+j^ in P Q, it is given by
[p 0 qi+j +... + pi− 1 qj+1] + [piqj ] + [pi+1qj− 1 +... + pi+j q 0 ]
As each of p 0 ,... , pi− 1 is divisible by 3 by assumption, the first piece is divisible by 3, and likewise each of q 0 ,... , qj− 1 is divisible by 3 so the third piece is also divisible by 3. But the middle term is not divisible by 3 since neither pi nor qj is divisible by 3, so the coefficient of xi+j^ in P Q is not divisible by 3, so P Q does not lie in (3). Thus (3) is prime.
(3) Prove that the kernel of ψ : Z[x] → R given by ψ(f ) = f (
(3)) is a principal ideal and find a generator for this ideal. Checking the additive and multiplicative properties for ker(ψ) is trivial, so it is an ideal. SOLUTION (I): Since every polynomial can be factored into linear factors over C, any P ∈ ker(ψ) has x −
3 as a factor. But this factor is not in Z[x], so P is also divisible by the other Z-conjugates of this factor, which in this case are just x +
- Thus P is divisible by [x − sqrt3][x +
3] = x^2 − 3 and hence this is a generator. SOLUTION (II): Alternatively, since the minimal degree polynomial in Z[x] which annihilates
3 is x^2 − 3 which has leading coefficient a unit in the co- efficient ring Z, by the division algorithm write an arbitrary P ∈ ker(ψ) as P (x) = (x^2 − 3)Q(x) + R(x) with R(x) = ax + b. Plugging in
3 into both sides of this expression, we see that R(
- = 0 so that a
3 + b = 0 for some
integers a and b. But this is clearly impossible unless a = b = 0, so R(x) = 0 and P is therefore divisible by x^2 −3 which is therefore a generator for the ideal.
(4) Determine whether the following are PIDs. If not, exhibit a nonprincipal ideal. (a) Z[t] SOLUTION: This is not a PID, this was shown in Problem (1). (b) R[x, y] SOLUTION: This is not a PID, the ideal (x,y) is not principal. (c) Z[
−5]
SOLUTION: Unlike the previous two examples, this ring is not a UFD. We use this fact to construct a nonprincipal ideal. Consider 6 = 2 · 3 = (1 +
−5), all four of these factors can be checked to be irre- ducible. Then ideals such as (2, 1 +
−5) obtained by taking one irreducible factor from each factorization are nonprincipal ideals in this ring.
(5) Let F be a field and F[[z]] the ring of power series. (a) Show that every element A =
i=0 aiz
i (^) such that a 0 6 = 0 is a unit in F[[z]]. SOLUTION: We want to show that there is B =
i=0 biz
i (^) such that AB =
1 + 0z + 0z^2 +... The coefficient b 0 must therefore be a− 0 1 which exists since a 0 is assumed to be a nonzero element in a field. Inductively, assume we have calculated b 0 ,... , bk− 1. The coefficient of zk^ in AB is given by:
0 = [zk]AB
∑^ k
i=
aibk−i
= a 0 bk +
∑^ k
i=
aibk−i
Therefore bk = −a− 01
∑k i=1 aibk−i^ which expresses^ bk^ in terms of things which are known, hence we can calculate all bk and thus B = A−^1 exists. (b) Show that every non-zero ideal of F[[z]] is of the form znF[[z]] for some n. SOLUTION: Let I = (f 1 , f 2 ,... , fj ) be an ideal in F[[z]]. As each fi has a leading nonzero coefficient we can write:
fi = ck(i)zk(i)^ + ck(i)+1zk(i)+1^ +... = zk(i)(ck(i) + ck(i)+1z +.. .) = zk(i) · ui
ui is a unit by (a). Thus I = (zk(1), zk(2),... , zk(j)) = (zmin(k(1),k(2),...,k(j))) since each zk(i)^ is clearly a multiple of this latter generator. (c) Show that F[[z]] is a PID. SOLUTION: This follows from (5b).
here that this ideal therefore contains a constant, hence is the whole ring. Thus (x, y^2 + 1) is maximal. Alternatively, look at the quotient ring R[x, y]/(x, y^2 + 1) ∼= R[y]/(y^2 + 1). This latter ring is isomorphic to C since it is obtained by adjoining an element y to R such that y^2 + 1 = 0; i.e. y behaves like
−1. Since C is a field, we conclude (x, y^2 + 1) is maximal. (b) Show (x^2 + y^2 + 1) is prime. SOLUTION: Specializing at y = 1 gives x^2 + 2 which is irreducible in R[x], thus x^2 + y^2 + 1 is irreducible in R[x, y]. Since R[x, y] is a UFD, principal ideals generated by irreducible elements are prime ideals, so (x^2 + y^2 + 1) is a prime ideal.
3 Homework 3
(1) Show that the units in Z[i] are {± 1 , ±i}. SOLUTION: It is easy to check these are the only elements of Z[i] with norm 1, hence are the only units.
(2) Let f (x) =
∑d i=0 aix
i (^) ∈ R[x] such that ad ∈ R×. Show that for any
g(x) ∈ R[x], there are unique q(x) and r(x) such that g(x) = f (x)q(x) + r(x) and deg(r) < deg(f ). SOLUTION: Apply the division algorithm to divide g by f ; the successive terms of q(x) are calculated by dividing adxd^ into the highest remaining term at each step. As ad is a unit, this is always possible, hence the division algorithm gives a unique quotient. Uniqueness of r(x) follows from uniqueness of q(x).
(3) Let Xn be the n × n matrix whose (i, j)-entry is xi,j. Show det(Xn) is irreducible in C[xi,j ] 1 ≤i,j≤n. SOLUTION: Since the determinant is linear in rows, no monomial term of det(Xn) contains a subfactor of the form x^2 i,j or xi,j xi,k. Suppose det(Xn) = P Q. Write P = (x 1 , 1 a 1 , 1 + b 1 , 1 ) and Q = (x 1 , 1 c 1 , 1 + d 1 , 1 ) with a 1 , 1 , b 1 , 1 , c 1 , 1 and d 1 , 1 polynomials in variables which do not involve x 1 , 1. Since no terms of det(Xn) are divisible by x^21 , 1 , one of a 1 , 1 or c 1 , 1 is zero (but not both since there are terms of det(Xn) which contain x 1 , 1 ). WLOG P contains x 1 , 1 as part of its terms. Now, for each k, rewrite Q = (x 1 ,kc 1 ,k + d 1 ,k), then c 1 ,k = 0 for all k by the same logic as above. Hence, all x 1 ,k terms appear in P for all k. Using similar logic shows that all elements xj,k appear only in the same factor as x 1 ,k, but as this term appears in P , all variables only appear in P , hence Q is a unit and det(Xn) is irreducible.
(4a) Suppose s ∈ R is not nilpotent. Show that among all ideals which do not contain some power of s there is a maximal such ideal P containing all the others. SOLUTION: Letting P be the union of all such ideals, it is clear P does not contain any power of s, while any larger ideal containing P must contain some power of s (otherwise it would be contained in P by construction, a contradic- tion), so we can apply Zorn’s Lemma to conclude P is maximal. (b) Let I = {x ∈ R|x · sn^ ∈ P for some n}. Show I is an ideal not containing s. SOLUTION: If x, y ∈ I such that x·sm, y·sn^ ∈ P , then (x+y)·smax(m,n)^ ∈ P so x + y ∈ I. If r ∈ R is arbitrary, then xr · sm^ = x · sm^ · r ∈ P so xr ∈ I and I is an ideal. (c) Conclude I = P. SOLUTION: No power of s can be in I since otherwise one would have sm^ ∈ P for some m. Thus I ⊆ P. On the other hand, every element of P is clearly in I, so I = P. (d) Show P is prime.
It is easy to check that this gives an injective homomorphism of Z[
d] into M at 2 × 2 (R); so we have a matrix representation of this ring. Now the deter- minant of M (a + b
d) is a^2 − db^2 ; this number is defined to be the norm N (a + b
d). From properties of determinants, if a + b
d = [p + q
d] · [r + s
d] then N (a + b
d) = N (p +
d)N (r + s
d). Thus the norm is a multiplicative (but not additive) homomorphism Z[
d] → Z. An element of the ring such that N (a + b
d) = ±1 (i.e. the norm is a unit in the base ring Z) is a unit in Z[
d] since it can easily be checked that
its inverse is [a−b
√ d] N (a+b √ d) which is clearly an element of^ Z[
d]. This observation
further implies that if N (a+b
d) is prime in Z then a+b
d must be irreducible in Z[
d].
4.2 Specialization
The concept of specialization is useful in determining when multivariable poly- nomials factor and in showing that they do not factor. The idea is that replacing a variable by a number, we do not lose factorizability, (although we may gain factorizability in the specialization when none exists in the original polynomial). For example, suppose x^2 + y^2 factors as P (x, y)Q(x, y) in Q[x, y]. Then setting y = 1 for example, it must still factor: x^2 + 1 = P (x, 1)Q(x, 1). But as x^2 + 1 is irreducible in Q[x], we conclude the original polynomial was irreducible in Q[x, y] as well. On the other hand, since x^2 + 1 is reducible in C[x], we cannot conclude from specialization that x^2 + y^2 is irreducible in C[x, y], and indeed it is not: x^2 + y^2 = (x + iy)(x − iy). On the other hand, x^2 + y^2 + c is irreducible in C[x, y] for c 6 = 0, but any specialization of y always factors in C[x] (by the fundamental theorem of algebra), so factorizations in the specialization need not come from a factorization of the original polynomial.
