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MATH-4010: Abstract Algebra. Final examination, May 15, 2012.
- True or False. Show reasoning.
(a) p(x) and q(x) are polynomials in R[x], where R is a commutative ring with unity. Then p(x)q(x) = 0 implies that either p(x) = 0 or q(x) = 0.
Solution: False. R[x] will have zero divisors if R does. For example, in Z 24 [x], let p(x) = 4(x^2 + 5x + 1) and let q(x) = 6(x^3 − x + 7). Then p(x)q(x) = 0 because 4 ∗ 6 = 0 in Z 24. (b) p(x) and q(x) are polynomials in D[x], where D is an integral domain. Then p(x)q(x) = 0 implies that either p(x) = 0 or q(x) = 0.
Solution: True. If D is an integral domain, then so is D[x]. (c) The ring Z × 2 Z × 5 Z has unity.
Solution: False. If (a, 2 b, 5 c) is unity in this ring, then (a, 2 b, 5 c)(1, 2 , 5) = (1, 2 , 5), which forces a = 1 (OK), 4b = 2 and 25c = 5. The last two are impossible. (d) A subring of a ring without unity can have unity.
Solution: True. In the above case, take the subring {(a, 0 , 0) | a ∈ Z}. Then (1, 0 , 0) is unity in this subring. (e) Let n = ab, where a and b are both integers greater than 1. Then Zn can be a field for some values of a and b.
Solution: False. Zn is a field if and only if n is prime. You can also note that a 6 = 0, b 6 = 0 in this ring, and yet ab = 0. (f) Let Mn(F ) be the set of all invertible n × n matrices with elements in a field F. Then Mn(F ) is a ring under matrix addition and matrix multiplication.
Solution: False. It is a group under matrix multiplication, but it is not closed under addition. The sum of two invertible matrices is not necessarily invertible. In fact, A + −A = 0 for any A ∈ Mn(F ). (g) Q[x] is a PID (Principal Ideal Domain).
Solution: True. Q is a field, hence a PID and so Q[x] is also a PID. (h) Q[x, y] is a PID.
Solution: False. This is similar to the ideal generated by 5 and x in Z[x], where y takes the place of 5. (i) The congruence 6x ≡ 45 mod 63 has precisely three solutions between 1 and 62, inclusive.
Solution: True. Note that d = gcd(6, 63) = 3 and that 3|45. This means that a solution exists. Theory tells us that there are precisely d distinct solutions modulo 63.
(j) x^22 − 1 factors completely into a product of 22 polynomials of degree 1 in Z 23.
Solution: True. a^22 = 1 for all a 6 = 0 in this finite field, so
x^22 − 1 =
∏^22
i=
(x − i).
- In Z 11 [x], let f (x) = x^7 − x^6 + 6x^5 − 3 x^4 + x^3 + 3x + 2 and let g(x) = x^3 + 6x^2 + x + 1. Compute polynomials q(x) and r(x) such that f (x) = q(x)g(x)+r(x), where the degree of r(x) is less than 3.
Solution: This is a long division problem.
x^4 + 4x^3 + 3x^2 + 7x + 7
x^3+6x^2+x+1 | x^7 - x^6 + 6x^5 - 3x^4 + x^3 + 3x + 2
| x^7 + 6x^6 + x^5 + x^
4x^6 + 5x^5 + 7x^4 + x^3 + 3x + 2 4x^6 + 2x^5 + 4x^4 + 4x^
3x^5 + 3x^4 + 8x^3 + 3x + 2 3x^5 - 4x^4 + 3x^3 + 3x^
7x^4 + 5x^3 - 3x^2 + 3x + 2 7x^4 - 2x^3 + 7x^2 + 7x
7x^3 + x^2 - 4x + 2 7x^3 - 2x^2 + 7x + 7
3x^2 + 6
Thus, q(x) = x^4 + 4x^3 + 3x^2 + 7x + 7 and r(x) = 3x^2 + 6.
- Compute 4139^1648 mod 41 without using a calculator.
Solution: Note that 4139 = 100 ∗ 41 + 39 ≡ 39 ≡ − 2 mod 41, and that 1648 = 410 ∗ 40 + 8. Fermat’s little theorem states that ap−^1 ≡ 1 mod p if p is prime and a is not a multiple of p. Then
41391648 mod 41 ≡ [(−2)^4 0]^410 ∗ (−2)^8 mod 41 ≡ 1410 (−2)^8 mod 41 ≡ 44 mod 41 ≡ 64 ∗ 4 mod 41 ≡ 23 ∗ 4 mod 41 ≡ 46 ∗ 2 mod 41 ≡ 5 ∗ 2 mod 41 ≡ 10 mod 41.
pq^ = Ip(1) + qIp(q) = p + qIp(q).
Thus qIp(q) = pq^ − p.
Then
I 3 (15) =
315 − 3 − (3^5 − 3) − (3^3 − 3)
It is not necessary to get the final exact answer if no calculator is available.
- Let F = Z 13. Show that x^2 − 2 and x^2 − 11 are both irreducible in F [x]. Let α be a zero of x^2 − 2 and let E = F (α). E is an extension field of F and its elements are of the form a + bα, where a and b are in F. Show that x^2 − 11 is reducible in E by computing the square root of 11. That is, find a + bα such that (a + bα)^2 = 11 in E.
Solution: Compute the squares of all the non-zero elements of F = Z 13 , noting that for any prime p, a^2 = (p − a)^2 mod p. a 1 2 3 4 5 6 7 8 9 10 11 12 a^2 1 4 9 3 12 10 10 12 3 9 4 Note that 2 and 11 are missing in the second row. This means that both 2 and 11 have no square root in F. This is equivalent to saying that x^2 − 2 and x^2 − 11 are both irreducible in F [x]. Let E = F (α), where α is a zero of x^2 − 2. Suppose that (a + αb)^2 = 11 in E. Then a^2 +2b^2 = 11 and 2abα = 0. If b = 0, then we end up with a^2 = 11, which is impossible, so a must be 0. Thus,
b^2 = (2)−^1 ∗ 11 = 7 ∗ 11 = 12.
Note from the table that b = 5 works, so (5α)^2 = 11. (Check: (5α)^2 = 25 ∗ 2 = 50 = 3 ∗ 13 + 11) Thus (x^2 − 11) = (x − 5 α)(x + 5α) in E. We know that
11 ∈/ Q(
2). That is, x^2 − 11 remains irreducible in Q(
2). Note how the situation differs in the finite field Z 13. In fact, if α is a zero of any particular irreducible polynomial of degree two in Zp, then all polynomials of degree 2 in Zp(α) are reducible! To see why this is true, let a generate the multiplicative group of non-zero elements of Zp. We know that this group is cyclic. It is also clear that x^2 − a is irreducible, because if a = b^2 , then powers of b would generate 2(p − 1) distinct elements, which is impossible. Let α be a zero of x^2 − a in E = Zp(α). Any non-zero c ∈ Zp equals ai^ for some i such that 1 ≤ i < p. Let i = 2k + r, where r = 0 or 1. Then c = (ak)^2 or c = (αak)^2. In either case, c is the square of an element in E.
- In the field F = Z 7 [x]/〈p(x)〉, where p(x) is an irreducible polynomial of degree 3, let 〈α〉 = {α, α^2 ,... } denote the cyclic subgroup generated by any α 6 = 0. Compute the maximum order of such a subgroup.
Solution: Any finite field has pn^ elements, where p is a prime and n is a positive integer. Such fields exist for all primes and for all n. The non-zero elements comprise a cyclic group under multiplication, and the order of this group is pn^ − 1. For the given problem, p = 7 and n = 3, so the maximum order is 7^3 − 1 = 342.
