(12345)(643125) = (56)(132)(4) so its order is the lest common
multiple of 1, 2, and 3, which is 6. The permutation is the product of
a 2 cycle (odd) and a 3-cycle (even) so it is an odd permutation.
2.
If a and b are real then N(a+bi) = a2+b2.
N(xy) = xy[`xy] = x[`x] y[`y] = N(x)N(y). We have
(a2+b2)(c2+d2)
= N(a+bi)N(c+di)
= N((a+bi)(c+di))
= N(ac-bd +i(ad+bc))
= (ac-bd)2+(ad+bc)2
Note that 13 = 32+22 and 101 = 102+12. So by the formula above
1313 = (3×10-2×1)2+(3×1+2×10)2 = 282+232.
Another solution is
1313 = (2×10-3×1)2+(2×1+3×10)2 = 172+322.
3.
First find an irreducible polynomial over the field with 13 elements.
x2-a will do for any a that is not a square; for example,
we could choose a = 2 and take the polynomial to be p(x) = x2-2.
Then F13[x]/(p(x)) is a field with 132 elements.
(Note that x2+1 does NOT work as it is reducible over F13.)
4.
(a)
Any polynomial of odd degree over the reals has a root
(as it has different signs for x large and positive and for x large and
negative). So any polynomial of odd degree at least 3 over the reals is
reducible.
(b)
The polynomial has degree at most 3, so it is irreducible if and only if
it has no roots over F2. There are only 2 elements 0 and 1 of F2
to check as roots. As f(0) and f(1) are both nonzero the polynomial
f(x) = x3+x+1 is irreducible.
(c)
This polynomial is reducible, as it is equal to (x2+x+1)2. To test
this polynomial, first check it has no roots, the check for
divisibility by polynomials of degree 2. This is not too hard as
x2+x+1 is the only irreducible polynomial of degree 2 over
F2. (There is also a fast way to see it is reducible that one or two
people used: recall that a2+b2+c2+¼ = (a+b+c+¼)2
whenever 2 = 0. So any sum of squares in a ring where 2=0 is also a
square. But x4+x2+1 = (x2)2+(x)2+12 is obviously a sum of
squares.)
5.
Use Euclid's algorithm:
x5+1
= (x2+1)(x3+x)+(x+1)
x2+1
= (x+1)(x+1)+0
So the greatest common divisor is x+1. The first line
gives x+1 = (x3+x)(x2+1)+1×(x5+1) so we can put
a(x) = x3+x, b(x) = 1.
Several people worked over the rationals rather than over F2.
In this case the greatest common divisor is 2 (or 1), a(x) = x4-x3-x2+x+1
and b(x) = 1-x.
File translated from TEX by TTH, version 2.53. On 4 Nov 1999, 15:21.