SOLUTIONS

1. a = X³, b = 1, c = X² + X + 1 in Q[X].
   a and b are congruent if their difference is divisible by c. This is the case since X³ - 1 = (X - 1)(X² + X + 1).

   
   

2. a = X⁴ + X + 2, b = X + 3, c = X + 1 in Z/5Z[X].
   a - b is divisible by X + 1 if and only if 4 is a zero of it (mod 5). Substituting X = 4 (or X = -1) we see that this is indeed the case. Hence they are congruent modulo c.

   
   

3. a = (X³ + X + 1)⁵, b = (X² + 2X)⁵, c = X - 1 in Q[X].
   Again we need to check if 1 is a zero of a - b . Since 3⁵ - 3⁵ = 0, they are congruent modulo c.

   
   

Polynomial division gives:

1. a = X⁴ = (X² + X + 1)(X² - X) + X. Hence X is the representative of a of smallest degree.
2. a = X⁴ + X² + 1 = (X² + X + 1)² = d². Hence 0 is the representative of a mod d of smallest degree.

Representatives are those polynomials of degree less than the one with respect to which the residue class ring is formed. Hence, a complete set of representatives is as follows

1. For Z/2Z[X]/(X³ + 1): All polynomials of degree less than 3, that is {0, 1, X, X + 1, X², X² + 1, X² + X, X² + X + 1};
2. For Q[X]/(X - 1): All constants, that is, the set Q;
3. For R[X]/(2): All polynomials of degree less than 0, that is, {0}.

1. Since S is a vector space over Z/2Z of dimension 2, it has 4 elements. Notice that a³ = a^{2 + 1} = a (a + 1) = a² + a = 1. Thus 1, a, a² = a + 1 are distinct elements, so that, together with 0, they are all elements of S.
   

2. As aⁱ · a^j = a^{i + j}, the nonzero table entries can be read off from the results of the first part.
   

3. a¹⁷ = a^{15 + 2} = a² = a + 1.

1. To show that the map is well defined we have to prove that eval(f) is equal to eval(g) if f and g are congruent mod X - a. So let f and g be congruent. Then f - g is divisible by X - a. Hence f(a) - g(a) = 0. That is, eval(f) = eval(g).

2. A bijection is a mapping that is both injective and surjective. The latter condition is easy since for any element z R we have eval(z + (X - a)R[X]) = z.
   It remains to show that the mapping eval is injective. Let f, g R[X] be representatives of classes in R[X]/(X - a) and suppose that

   eval(f + (X - a)R[X]) = eval(g + (X - a)R[X]).

   Then f(a) = g(a), which means that f - g is a multiple of X - a, so f and g are congruent and f + (X - a)R[X]) = g + (X - a)R[X]). This shows that the mapping eval is injective.

We will only do the proofs for f₊ since the proofs for f_- are similar.

1. We show that f₊ is well defined. Let a and b be two polynomials that are congruent. We show that f₊(a ) = f₊(b). Since a and b are congruent, their difference is divisible by X²-2, that is a - b = (X² - 2) g for some g. Substituting 2 gives, a(2 )= b(2 ), i.e., f₊(a ) = f₊ (b).
2. We show that f₊ is injective. Let a and b be representatives of two classes. We may assume that they are of lowest degree. Suppose that f₊(a) = f₊(b). Thus, there are c, d, m, n Q with a = c + dX and b = m + nX. We find c + d2 = m + n2. In other words: c - m=(n - d)2. But the left-hand side is rational and the right-hand side is zero or irrational. Hence the only possible way they can be equal is when they are both zero. This implies a = b, and injectivity follows.
3. The image of f₊ is {c + d2 | c, d Q}. This is straightforward since f₊(c + dX) = c + d2.
4. We show the `multiplicative law' f₊(a)f₊(b) = f₊(ab) and leave the remaining parts to the reader.
   Taking c, d, m, n Q with a = c + dX and b = m + nX, we have f₊(a)f₊(b) = (c + d2) (m + n2) = cm + 2dn + (cn + dm)2 = (c + d2) (m + n2) = f₊(ab).

In the residue class ring Q[X]/(X⁵), the element X⁵ = 0, so the product immediately reduces to (1 + X)(1 + X³)(1 + X⁴). Moreover, in expanding this product, the products X · 1 · X⁴, 1 · X³ · X⁴ and X · X³ · X⁴ vanish for the same reason. Therefore, the product consists of 2³ = 8 terms, 5 of which `survive', yielding the following representative of degree less than 5:

Reduce the equality modulo X² + 1. Since X² = -1 mod X² + 1 and therefore X³ = -X mod X² + 1 we find (-X + 1)f = -X - 1 mod X² + 1. If the remainder is bX + c, with b and c in Q, then -bX² + (b - c)X + c = -X - 1 mod X² + 1. Using X² = -1 mod X² + 1 once more, we find (b - c)X + b + c = -X - 1. Hence, b = -1, c = 0. The remainder is -X.

The proof is almost identical to the proof of the Chinese remainder theorem for integers:

Existence of f
Since c and d have gcd equal to 1, there are x and y with xc + yd = 1. Multiplying both sides with e = b - a yields xce + yde = b - a. Now consider f = a + xce = b - yde mod cd. This polynomial satisfies the relations.

Uniqueness of f
If f and g satisfy both relations, then f-g is a polynomial of degree less than nm and equal to 0 mod cd. However, there is only one such representative of the residue class of 0 mod cd, namely 0. Thus f = g.

• Use the extended Euclidean algorithm to solve x and y in the equation xc + yd = 1.
• f := a + xc(b-a) mod (cd).
• return f.

1. Using the extended Euclidean algorithm, we find that the inverse of (1 + X) modulo X⁴ is 1-X+X²-X³. Hence, the Taylor expansion is 1 - x + x² - x³.

2. Similarly, the inverse of (1 + X + X²) mod X⁴ is 1 - X + X³, so the Taylor expansion is 1 - x + x³.

3. The Taylor expansion of order 4 of cos(x) equals 1 - (1/2)x². The inverse of that polynomial mod x⁴ equals 1 + (1/2)x².

1. Reducing the relation af + bg = d mod p we find a'f' + b'g' = d' (where f' is obtained from f by reducing its coefficients mod p; similarly for g). Since gcd(a',b') divides the left-hand side of this relation it also divides the right-hand side.
2. In general, gcd(a,b) = gcd(X,X+p) = gcd(X,p).
   Since p is invertible in Q, this gives gcd(a,b) = 1 in Q[X]. On the other hand, in Z/pZ[X], we have gcd(a',b') = gcd(X,0) = X.

Since 1 + X and X² + 1 are relatively prime, the element a has an inverse, which we compute using the extended Euclidean algorithm:

The inverse is therefore X - 1 + (X² + 1). Multiplying through both sides of the equation and simplifying yields (modulo X² + 1)

Therefore z = 2X + (X² + 1).

1. The polynomial X³ + X + 1 has no roots in Q. So it is irreducible. (This argument suffices since the degree of the polynomial is 3.)

2. Since a(a² + 1) = -1, the inverse of a is -1-a².

3. By the extended Euclidean algorithm, we have (1/9)(X² - 2X + 5)(X + 2) + (-1/9)(X³ + X + 1) = 1. This shows that 5/9 - 2/9a + 1/9a² is the inverse of a.

4. Similarly to the previous part, we find 1/3(X² - 2X + 2)(X² + X + 1) +1/3(-X + 1)(X³ + X + 1) = 1, and so 2/3 - 2/3a + 1/3a² is the inverse of a² + a + 1.

1. False. Take f = X and d = X²
2. True. Every element of Z represents a different class.
3. True. Suppose a has an inverse, multiplying the equation ab = 0 by it from the left gives b = 0, contradicting the assumption that b is not zero. The proof that b is not invertible is similar.
4. False. Of course, a = 0 (and any choice for the other variables) gives a counterexample. But even if a is not zero it is false; for example, X + 1 is not equal to 1 but X(X + 1)= X·1 mod X².
5. True. Multiply with the inverse of a to prove it.
6. True. The identity (1 - a)(1 + a + a² + a³) = 1 - a⁴ = 1 shows that 1 - a is invertible.

Division with remainder shows that the map is surjective.
To show that the map is injective, consider two elements a and b of R. Their images are equal if they differ by a multiple of d. Since the degree of d is 1, this can only happen if this multiple is 0, i.e., if a = b.

1. g(a) = 0 means that g(X) = 0 mod f. That is just saying that f divides g.
2. The class a satisfies a² + a + 1 = 0. Now subsitute in X⁶ + X³ + 1:
   a⁶ + a³ + 1 = a³(a³ + 1) + 1 = a³(a + 1)(a² + a + 1) + 1 = 1.

   So f does not divide g.

1. Let f be an arbitrary polynomial of degree m. If m < n, then it obviously has a representative mod d of degree less than n.
   We proceed by induction on m. So suppose that the assertion holds for all polynomials of degree less than m > n - 1 and let f be a polynomial of degree m. If its leading coefficient is a then f - a dX^{m - n} is congruent to f mod d, but is of degree less than m. Hence, by induction, it has a representative of degree less than n. This then, is also a representative of f mod d.

2. If X has a representative of degree 0, then there exists a c Z/4Z such that 2X divides X - c. However all coefficients of a multiple of 2X are 0 or 2, while the coefficient of X in X - c is 1. This is a contradiction. Hence X does not have representative of degree less than 1.

1. The polynomial d has no roots. Thus, if it is reducible, then it is the product of two irreducible polynomials of degree 2. However, the only irreducible polynomial of degree 2 is p = X² + X + 1. But d does not equal p².

2. The field S = Z/2Z[X]/(d) is a vector space of dimension 4 over Z/2Z. Therefore, S has 2⁴ = 16 elements.
   Let a be the element X + (d) of S. We write down some powers of a:

   a0 = 1	a1 = a
   a2 = a2	a3 = a3
   a4 = a + 1	a5 = a2 + a
   a6 = a3 +a2	a7 = a3 + a + 1
   a8 = a2 + 1	a9 = a3 + a
   a10 = a2 + a + 1	a11 = a3 + a2 + a
   a12 = a3 + a2 + a + 1	a13 = a3 + a2 + 1
   a14 = a3 + 1	a15 = 1

   We have found all 15 nonzero elements of S as powers of a. From this it is easy to write down an addition and multiplication table.

3. A subfield of order 4 contains 0, 1 and at least one element of the form aⁱ (with i between 0 and 15). Then it will also contain a²ⁱ, a³ⁱ, etc. If the subfield is to have only 4 elements, then we conclude from the table that i = 5 or 10. It is easily checked that {0, 1, a⁵, a¹⁰} is indeed a subfield.

1. Since X³ + X + 1 has degree 3, it suffices to check that X³ + X + 1 has no zeros in Z/2Z (reducibility of X³ + X + 1 is equivalent to having a degree one factor and that is equivalent to having a zero). Substituting 0 and 1 in X³ + X + 1 leads to the value 1. So X³ + X + 1 is an irreducible polynomial in Z/2Z[X].
   The number of elements in the quotient ring is 2³ = 8.
2. Division with remainder: X⁷ + 1 = (X³ + X + 1)(X⁴ + X² + X + 1). Since a³ + a + 1 = 0, this implies that a⁷ = 1.
   It suffices to show that 1, a, a², ..., a⁶ are all distinct. Suppose aⁱ = a^j for 0 i < j 6. Cancelling powers of a we reduce to considering the relation 1 = a^j for 0 < j < 7. If a^j - 1 = 0, then X^j - 1 = 0 mod X³ + X + 1 so that X³ + X + 1 | X^j -1. This is clearly impossible if j 3. If j is at least 4, then, by multiplying both sides of 1 = a^j with a^{7 - j}, we find again a relation 1 = aⁱ with 0 < i 3 (here we use that a⁷ = 1); this is impossible.
3. a² is a zero: (a²)³ + a² + 1 = (a³ + a)² + 1 = 1 + 1 = 0.
   a⁴ is a zero: (a⁴)³ + a⁴ + 1 = (a³ + a)⁴ + 1 = 1 + 1 = 0.
4. The zeros of X³ + X² + 1 are the elements a³, a⁵, a⁶ as is easily verified.
   Note that 1, a, a², a³, a⁴, a⁵, a⁶ are precisely the zeros of X⁷ + 1 and that X⁷ + 1 = (X + 1)(X³ + X² + 1)(X³ + X + 1).