SOLUTIONS

Solution of Exercise 1

1. The given set contains 0 and 1 of course. Starting with the elements x + y i 2 and u + v i 2, we find
   (x + y i 2) + (u + v i 2) = (x + u) + (y + v)i 2

   and

   (x + y i 2) · (u + v i 2) = (xu - 2yv) + (xv + yu)i 2,

   showing that the sum and product of two elements in the given set again belong to the same set. Since also -(x + y i 2) = - x - y i 2 belongs to the set, we conclude that we have subring.

2. If the given subset is a subring, then a² should also be in it, so we get a relation a² = x + ya for some integers x and y. This means that a is a zero of the polynomial X² - yX - x in Q[X]. But a is also a zero of X³ - 2. Since a is not a rational number, the polynomial X³ - 2 has no zero in Q and is therefore irreducible. This implies that the gcd of X² - yX - x and X³ - 2 must be 1. But the extended Euclidean algorithm shows that a is also a zero of the gcd. This is a contradiction and we conclude that the given subset is not a subring of C.

Solution of Exercise 2

To check that S is a subring we verify that 0, 1 S, and that, given x, y S, the elements x + y, -x, x * y are also in S.

• 0, 1 S: f(0) = 0 and f(1) = 1.
• Suppose x, y S. Then f(x + y) = f(x) + f(y) = x + y, f(-x) = -f(x) = -x and f(x * y) = f(x) * f(y) = x * y, showing that x + y, -x, x * y are also in S.

If f is complex conjugation, then we find the subring R of real numbers.

Solution of Exercise 3

First we check that f is a morphism:

• f(0) = a0a^-1 = 0,
• f(x + y) = a(x + y)a^-1 = (ax + ay)a^-1 = axa^-1 + aya^-1 = f(x) + f(y),
• f(1) = a1a^-1 = 1,
• f(x * y) = a(x * y)a^-1 = a(xa^-1ay)a^-1 = (axa^-1) * (aya^-1) = f(x) * f(y).

To show that f is an isomorphism one can show that f is bijective or give its inverse. We specify the inverse: The inverse of f is easily checked to be the map that sends x in R to a^-1xa.

If the ring is commutative f is just the identity map.

If the map g is a morphism, then g(1) = 1 implies a² = 1. In other words, a must be invertible and equal its own inverse. In that case the preceding shows that g is a morphism. In conclusion: g is a morphism if and only if a² = 1.

Solution of Exercise 4

1. The verification of the various requirements all follow the same pattern, so we confine ourselves to showing left distributivity, leaving the others to the reader. Suppose f, g, and h are all maps from S to R. To show that f * (g + h) = f * g + f * h we must verify that the left-hand side and right-hand side assume the same value for every s S:

   (f * (g + h))(s) = f(s) * (g + h)(s) = f(s) * (g(s) + h(s)) = f(s) * g(s) + f(s) * h(s) = (f * g)(s) + (g * h)(s) = (f * g + f * h)(s).

   Here, all equalities except the middle one follow from the definition of sum and product, whereas the middle one uses left distributivity in the ring R.

2. Surjectivity: If f is a map from Map(S, Z/2Z), then define A = { s S | f(s) = 1 } (= f^-1(1)). It is clear that f = f_A, showing that F maps A to f.
   Injectivity: If F(A) = F(B), i.e., f_A = f_B, then

   A = f_A^-1(1) = f_B^-1(1) = B.

   Note that G: Map(S, Z/2Z) -> P(S), given by G(f) = f^-1(1) is the inverse of F.

3. For subsets A and B of S, the `product' A * B should correspond to the product f_A * f_B, i.e., should be defined as (f_A * f_B)^-1(1). Since (f_A * f_B)(s) = 1 if and only if both f_A(s) = 1 and f_B(s) = 1, we find that

   A * B := A B.

   The `sum' A + B should correspond to the sum f<sub>A</sub> + f<sub>B</sub>. This sum assumes the value 1 in s if and only if either f<sub>A</sub>(s) = 1 and f<sub>B</sub>(s) = 0. or f<sub>A</sub>(s) = 0 and f<sub>B</sub>(s) = 1. The corresponding set then consists of the union of A \ B and B \ A, where `\' denotes difference of sets. The resulting set can also be described as A B \ A B. So

   A + B := A B \ A B.

   The zero element of course is the empty set, the unit element the set S itself.

   Note: If you try to prove directly that P(S) is a ring (with the operations just described), you will run into substantial computations. Can you see why?

Solution of Exercise 5

1. Let f, g Q[X]. Then (f + g) = (f + g)(2) = f(2) + g(2) = (f) + (g). Likewise (f · g) = (f·g)(2) = f(2) · g(2) = (f) · (g). Of course, (0) = 0 and (1) = 1.
2. This is a straightforward verification based on the fact that reduction modulo n is a morphism of rings.

Solution of Exercise 6

1. (g o f)(0) = g(f(0)) = g(0) = 0; (g o f)(1) = g(f(1)) = g(1) = 1;
   (g o f)(x + y) = g(f(x + y)) = g(f(x) + f(y)) = g(f(x)) + g(f(y)) = (g o f)(x) + (g o f)(y);
   (g o f)(x * y) = g(f(x * y)) = g(f(x) * f(y)) = g(f(x)) * g(f(y)) = (g o f)(x) * (g o f)(y).
2. Ker(f) is contained in Ker(g o f). If f(x) = 0, then (g o f)(x) = g(f(x) = g(0) = 0.
   Ker(g o f) is contained in Ker(f). If (g o f)(x) = 0 then f(x) is in the kernel of g. But then f(x) = 0 since g is an isomorphism.
3. f(Ker(g o f)) is contained in Ker(g). Start with f(x), where x belongs to Ker(g o f). Since x satisfies g(f(x)) = 0, we find that f(x) belongs to Ker(g).
   Ker(g) is contained in f(Ker(g o f)). If x Ker(g), then f^-1(x) R and this element satisfies (g o f)(f^-1(x)) = g(x) = 0. So f^-1(x) belongs to Ker(g o f) showing that x = f(f^-1(x)) is an element of f(Ker(g o f)).

Solution of Exercise 7

• F is a morphism: F(0) = 0; F(1) = 1; F(f + g) = (f + g)(1/2) = f(1/2) + g(1/2) = F(f) + F(g); F(f * g) = (f * g)(1/2) = f(1/2) · g(1/2) = F(f) · F(g).
• The kernel of F. The polynomial p = 2X - 1 is obviously in the kernel. Let f be an element in the kernel and consider (in Q[X]) division with remainder: f = qp + r, with r a constant. Substituting 1/2, we find that r = 0. If

  q = a₀ + a₁X + ··· + a_nXⁿ,

  then working out the product q · (2X - 1) we see that -a₀ Z, 2a₀ - a₁ Z, 2a₁ - a₂ Z , etc., from which we conclude that a₀, a₁, ..., a_n Z. So the kernel consists precisely of all multiples of 2X - 1.

• The image of F. Given any polynomial f with integer coefficients, the rational number f(1/2) is of the form a/2^m, with a Z and m a nonnegative integer. All such rational numbers indeed occur: just use the polynomial aX^m. In conclusion, the image is precisely the set
  { a/2^m | a Z, m a nonnegative integer }.

Solution of Exercise 8

• F is a morphism: F(0) = 0; F(1) = 1; F(f + g) = (f + g)(2X + 3) = f(2X + 3) + g(2X + 3) = F(f) + F(g); F(f * g) = (f * g)(2X + 3) = f(2X + 3) · g(2X + 3) = F(f) · F(g).
• The inverse of F: The map G: Q[X] -> Q[X] defined by sending f to f(X/2 - 3/2) is easily checked to be the inverse.

Solution of Exercise 9

We have to show that R does not contain nonzero zero divisors. Let a R be nonzero and suppose ab = 0 for some b R. Then ab = a0. By the Cancellation Law it follows that b = 0.

Solution of Exercise 10

Suppose a/b and c/d are rational numbers. If the prime p is not a divisor of b and d, then p is not a divisor of the denominators of

a/b · c/d = (ac)/(bd) and a/b + c/d = (ad + bc) /bd.

(Here we must take care of course that we do not multiply numerator and denominator of a fraction by an integer having p as a factor.) Now consider a subring R generated by the elements a₁/b₁, ..., a_n/b_n and assume that a_i and b_i are relatively prime for i = 1, 2, ..., n. Choose a prime p that does not divide b₁, ..., b_n. By what we said above, p does not divide any denominator of a sum or product of elements from a₁/b₁, ..., a_n/b_n. In particular, 1/p does not belong to R. In other words: No finitely generated subring of Q equals Q, so that Q is not finitely generated.

Solution of Exercise 11

Put L = {x K | (x) = x}. Since (0) = 0 and (1) = 1, the elements 0 and 1 are in L. Let x, y L. Then (x · y) = (x) ·(y) = x · y so L is closed under multiplication.
Similarly, (x - y) = (x) - (y) = x - y, showing that x - y is in L.
Also (assuming x is not zero) (x) · (x^-1) = (x · x^-1) = (1) = 1, so (x^-1) = (x)^-1 = x^-1, which shows that L is closed under taking inverses.

Solution of Exercise 12

All we have to show is that Q + Qd contains 0 and 1 and is closed under +, -, * and / (all other properties like associativity, distributivity, etc., are inherited from R). Let x, y Q + Qd. Then there exist a, b, c, e Q such that x = a + bd and y = c + ed. Then x + y = a + bd + c + ed = (a + c) + (b + e)d and this is clearly an element of Q + Qd so Q + Qd is closed under addition. A similar computation shows that the product of x and y is also in Q + Qd. To show that x/y (for nonzero y) is in Q + Qd, one uses the well-known trick of multiplying numerator and denominator with c - ed.

The subfield coincides with Q if in the prime decomposition of d every prime occurs with even exponent.

Solution of Exercise 13

The map is well defined. If a/b = c/d, then ad = bc so that ab^-1 = cd^-1. So the definition of F(a/b) does not depend on the way we represented the fraction a/b.

The map is a morphism. F(0/1) = 0 · 1^-1 = 0; F(1/1) = 1 · 1^-1 = 1;

F(a/b + c/d) = F((ad + bc)/(bd)) = (ad + bc) · (bd)^-1 = ab^-1 + cd^-1 = F(a/b) + F(c/d);

F(a/b * c/d) = F((ac)/(bd)) = (ac)(bd)^-1 = (ab^-1)(cd^-1) = F(a/b) * F(c/d).

F is injective. If F(a/b) = 0, then ab^-1 = 0 so that a = 0 (since b and hence b^-1 are not equal to 0). But then a/b = 0/1.

F(Q(R)) is the smallest subfield of K containing R. If a/b is a nonzero element then a and b are both nonzero and F(b/a) is the inverse of F(a/b). Moreover, for each element a of R, F(a/1) = a. So F(Q(R)) is a field containing R. If L is a subfield of K containing R, then it contains all elements of the form ab^-1 with a, b R (and b nonzero). This means that L contains all elements F(a/b) with a/b Q(R). In conclusion, L contains F(Q(R)), so that the latter is indeed the smallest subfield of K containing the image of R.

Field of fractions of Z[X]. The domain Z[X] is a subring of the field Q(X). Since every fraction f/g in Q(X) can be rewritten as a fraction u/v of elements u and v in Z[X], we conclude that Q(X) itself is the smallest subfield of Q(X) containing Z[X]. Therefore, the field of fractions of Z[X] is (isomorphic) with Q(X).

Solution of Exercise 14

The relation eaq cannot be transitive since otherwise it would follow from (1,1) eaq (2,2) and (2,2) eaq (0,2) that (1,1) eaq (0,2) which is clearly not the case.

Solution of Exercise 15

1. 0 · 2 = 0, 1 · 2 = 2, 2 · 2 = 4, 3 · 2 = 0, 4 · 2 = 2, 5 · 2 = 4, so ({2}) = {0, 2, 4}.
2. Since 1 = -1 · 2 + 1 · 3 ({2, 3}), it follows that r = r · 1 ({2, 3}) for every r R. So ({2, 3}) = R.
3. ({2}) = {0, 2, 4, 6}.
4. If f(X) = 2g(X) + Xh(X) for some g, h Z[X], then f(0) is even. Conversely, if f(0) is even, then f(X) = 2g(X) + Xh(X) if we take g(X) = f(0)/2 and h(X) = (f(X) - f(0))/X. Hence,
   ({2, X}) = {f Z[X] | f(0) is even}.

Solution of Exercise 16

If f(X,Y) ({X+Y, X-Y}), then there exist polynomials g and h in Q[X,Y] such that f(X,Y) = g(X,Y)(X+Y) + h(X,Y)(X-Y). Hence, f(X,Y) = (g(X,Y)+h(X,Y))X + (g(X,Y)-h(X,Y))Y ({X,Y}). Conversely, if f ({X,Y}), then there exist polynomials g and h in Q[X,Y] such that f = g·X + h·Y. Then f = ((g+h)/2)·(X+Y) + ((g-h)/2)·(X-Y) ({X+Y, X-Y}).

This does not hold in Z[X,Y]: Suppose X ({X+Y, X-Y}). Then there exist polynomials g and h in Z[X,Y] such that X = (X+Y)g + (X-Y)h. Setting X = Y = 1 shows that 1 = 2 g(1,1), which is impossible in Z.

Solution of Exercise 17

1. V is not an ideal: start with 2 + 2i in V. If V were an ideal then (2 + 2i)i should also be in V. But -2 + 2i is evidently not in V.
2. This is an ideal. Of course, 0 is in the ideal; if f and g are in the ideal, then f(0) = f(3) = g(0) = g(3) = 0, and consequently (f + g)(0) = (f + g)(3) = 0 so that f + g is in the ideal. Similarly, if f is in the ideal and g is an arbitrary element, then (fg)(0) = (fg)(3) = 0 since f(0) = f(3) = 0. We conclude that fg is in the ideal.

Solution of Exercise 18

1. This is a prime ideal. Suppose fg ({p}), i.e., all coefficients of fg are divisible by p, and write
   f = a₀ + a₁X + ··· + a_mX^m, g = b₀ + b₁X + ··· + b_nXⁿ.

   We want to show that f or g is in the ideal, i.e., that all coefficients of at least one of the two polynomials are divisible by p. If p does not divide all coefficients of f nor all of g, then there is a smallest index r such that a_r is not divisible by p and a smallest index s such that b_s is not divisible by p. It is easily seen that this implies that the coefficient of X^{r + s} in fg is not divisible by p. Contradiction.
   The ideal is not maximal: ({p,X}) is an ideal that strictly contains ({p}), but is still not the whole ring.

2. This ideal is not prime (and therefore not maximal): 5 = (1 + 2i)(1 - 2i) is in the ideal, but both 1 + 2i and 1 - 2i are not.
3. This ideal is maximal (and therefore prime). If f is an element in the ring that is not contained in the ideal, then we argue that ({2, X² + X + 1, f}) = Z[X]. To show this, it suffices to show that ({2, X² + X + 1, f}) contains the element 1.

   By applying division with remainder to f and X² + X + 1 we see that we may as well assume f to be of degree at most 1. By using the element 2 in the ideal, we may as well assume f to have its coefficients in {0,1}. This leaves us with 4 cases to consider, 3 of which lead to the desired conclusion, one of which is against our assumption on f:

   • 0, but then f is an element of ({2, X² + X + 1}) against our assumption;
   • 1, and we are done;
   • X, but then again 1 = (X² + X + 1) - X(X + 1) is in the ideal and we are done;
   • X + 1, but then again 1 = (X² + X + 1) - X(X + 1) is in the ideal and we are done.

   Strictly speaking we also have to show that the ideal ({2, X² + X + 1, f}) itself is not the whole ring. If ({2, X² + X + 1, f}) = Z[X], then there exist polynomials g and h with integer coefficients such that 2g + (X² + X + 1)h = 1. It follows that the leading coefficient of h is even. Going down by steps of 1, we easily see that all coefficients of h must be even. But the constant term of h must also be odd, leading to a contradiction.

   Another approach in solving this problem is to consider the quotient ring Z[X]/({2, X² + X + 1}). It consists of 4 elements and it suffices to show that this ring is a field.

Solution of Exercise 19

1. Let (i₁, j₁), (i₂, j₂) I × J. Then (i₁, j₁) + (i₂, j₂) = (i₁ + i₂, j₁ + j₂) I×J for i₁ + i₂ I (since I is an ideal) and j₁ + j₂ J (since J is an ideal). Let (i,j) I × J and let (r,s) R × S. Then (r,s) · (i,j) = (ri,sj) I × J for ri I (since I is an ideal) and sj J (since J is an ideal).
2. Let j₁, j₂ J. Since J = (I), there exist i₁, i₂ I such that (i₁) = j₁ and (i₂) = j₂. Hence, j₁ + j₂ = (i₁) + (i₂) = (i₁ + i₂) (I) = J. Let s S and let j J. Since is surjective, there exists r R such that s = (r). Since J = (I), there exists an i I such that j = (i). Hence, sj = (r)(i) = (ri) (I) = J. This shows that J is an ideal in S. Of course, J is not empty.
   If r₁, r₂ R satisfy r₁ + I = r₂ + I, then r₁ - r₂ I. It follows that (r₁) - (r₂) = (r₁ - r₂) J, i.e., (r₁) + J = (r₂) + J. This shows that is well defined. The remaining properties of a morphism are easy to verify.
   is not necessarily injective. For example, take R = S = Z, the identity map, I = ({0}), J = ({2}). Then is simply the natural map `reduction mod 2' from Z to Z/2Z. This map is not injective.

Solution of Exercise 20

The map : Z/2Z[X] -> Z/2Z[X] defined by (f(X)) = f(X + 1) is an automorphism of the ring Z/2Z[X]. For example, (f(X)) + (g(X)) = f(X + 1) + g(X + 1) = (f + g)(X + 1) = ((f + g)(X)). Notice that is clearly surjective since every polynomial can also be written as a polynomial in X + 1.

Of course, maps the ideal (X³ + X + 1) onto the ideal ((X + 1)³ + (X+1) + 1) = (X³ + X² + 1). Hence, induces a morphism

: Z/2Z[X]/(X³ + X + 1) -> Z/2Z[X]/(X³ + X² + 1).

Now it is easily checked that induces an isomorphism.

Solution of Exercise 21

1. Since X⁴ + X² + 1 is not equal to 0 for X = 0 and X = 1, X⁴ + X² + 1 has no factor of degree one, and therefore also no factor of degree three. The only irreducible polynomial of degree two in Z/2Z[X] is X² + X + 1 but (X² + X + 1)² = X⁴ + X² + 1 f. Using the characterization of maximal ideals in a polynomial ring we deduce that (f) is a maximal ideal. By the Remainder Theorem each element of S has a unique representation as a polynomial in of degree < 4.
2. By the above, two elements are the same if their a + b + c² + d³ representations are identical.
3. The nonzero elements form a (multiplicative) group of order 15. By Lagrange's Theorem, this implies that the order of is a divisor of 15, so 1, 3, 5, or 15. The cases 1 and 3 are ruled because of item 1. Since ⁵ = ³ + 1 (again by item 1), we conclude that the order is 15. This implies that the elements , ², ..., ¹⁵ are all distinct: If ^k = ^m for k, m in {1, 2, ..., 15}, then 15 divides m - k, so that m = k.

   For each i = 0, 1, 2, ..., 14, compute the a + b + c² + d³ representation of ⁱ (this gives the so-called logarithm table of the field). Notice that we may assume that i > j and since ⁱ + ^j = ^j · (^i-j + 1), we may also assume j = 0.
   Look up the representation of ⁱ.
   Add 1.
   Look up the value of k satisfying ⁱ + 1 = ^k.

4. ⁰ = 1,
   ⁵ = ⁴ = ( + 1) = ² + ,
   ¹⁰ = (⁵)² = (² + )² = ⁴ + ² = ² + + 1.
5. Do item 3) for an arbitrary polynomial f instead of f = X⁴+X+1.

Solution of Exercise 22

1. F is a morphism: F(0) = (0,0,0); F(1) = (1,1,1);
   F(f + g) = ((f + g)(-1), (f + g)(0), (f + g)(1)) = (f(-1) + g(-1), f(0) + g(0), f(1) + g(1)) = (f(-1), f(0), f(1)) + (g(-1), g(0), g(1)) = F(f) + F(g);
   F(f · g) = ((f · g)(-1), (f · g)(0), (f · g)(1)) = (f(-1) · g(-1), f(0) · g(0), f(1) · g(1)) = (f(-1), f(0), f(1)) * (g(-1), g(0), g(1)) = F(f) * F(g).
2. The kernel consists of the polynomials f such that f(-1) = f(0) = f(1) = 0. But this means exactly that we are dealing with the polynomials which are divisible by X - 1, X and X + 1. Since these are relatively prime, we conclude:
   Ker(f) = {(X³ - X) · h | h Q[X]}.
3. If g and h satisfy g(-1) = h(-1) = r, g(0) = h(0) = s, g(1) = h(1) = t, then g - h is in the kernel of F. So given one polynomial, g say, satisfying g(-1) = r, g(0) = s and g(1) = t, we have
   g + Ker(F) = {f Q[X] | f(-1) = r, f(0) = s and f(1) = t}.

Solution of Exercise 23

There are 4 · 5 = 20 substitutions. But the substitutions aX + b and -aX - b yield the same polynomial. So from f we find 10 irreducible polynomials by these substitutions. They all have leading coefficient 1 or -1, since the square of a nonzero element in Z/5Z is always 1 or -1. If a substitution yields a polynomial h, then it is easy to check that it is impossible to get -h by a substitution (show that (aX + b)² - 2 = -((cX + d)² - 2) has no solutions). The number of irreducible polynomials with leading coefficient 1 or -1 is 20 (there are 50 polynomials of degree 2 with leading coefficient 1 or -1; 2 · 5 = 10 among them are reducible and have a single zero with multiplicity 2; 2 · 10 = 20 among them are reducible and have two distinct zeros). In conclusion, we find all irreducible polynomials, up to a sign.

Another approach is to start with an irreducible polynomial of degree 2 and leading coefficient 1, say, and then to complete the square. You obtain (X + c)² - 2 or (X + c)² + 2. In the first case you are done, in the second case you consider (2X + 2c)² - 2 = -((X + c)² + 2).

Now given a field Z/5Z[X]/(h) with 25 elements, we may just as well assume that h has leading coefficient 1 or -1 (this doesn't change the ideal). We choose the sign so that h is of the form F_a,b(f) for some a and b. Then F_a,b induces an isomorphism between F and Z/5Z[X]/(h).

Solution of Exercise 24

For all x and y in R we have

x + y = (x + y)² = x² + xy + yx + y² = x + y + xy + yx.

Hence xy = -yx.

But we also have 1 = (-1)² = -1 and we find xy = yx.

We can conclude that R is commutative.