Cevaplar

2012-10-10T20:26:24+03:00

Section 4.3.
2. Suppose g(x) = x
n + cn1x
n1 + + c1x + c0 and h(x) = x
n + dn1x
n1 + + d1x + d0 were
both (monic) associates of f(x). Then g(x) = uf(x) and h(x) = vf(x) for some nonzero u; v 2 F.
Thus u
1
g(x) = f(x) = v
1
h(x), which means that u
1
x
n + u
1
cn1x
n1 + + u
1
c1x + u
1
c0 =
v
1
x
n + v
1
dn1x
n1 + + v
1
d1x + v
1
d0. Then the leading coecients are equal, so u
1 = v
1
,
and thus u = v. Then g(x) = h(x) = uf(x) = vf(x).
4. Let f(x) = cnx
n + + c1x + c0, with cn = 0, be a nonzero polynomial in 6 Zp[x]. Now, for
each 0 < r (p 1), r is a unit, so rf(x) is an associate of f(x). It suces to show that if
0 < r1; r2 (p 1) and r1 =6 r2, then r1f(x) =6 r2f(x). So suppose 0 < r1; r2 (p 1) and r1 =6 r2.
Then r1f(x) = r1cnx
n + + r1c1x + r1c0 and r2f(x) = r2cnx
n + + r2c1x + r2c0. Since F is
a eld, cn = 0, and 6 r1 =6 r2, we know that r1cn =6 r2cn, and thus r1f(x) =6 r2f(x). Therefore,
1f(x); 2f(x); 3f(x); dots;(p 1)f(x) are exactly the (p 1) associates of f(x).
6. If x
2 + 1 were not irreducible in Q[x], then by Theorem 4.10, x
2 + 1 = (ax + b)(cx + d) for some
a; b; c; d 2 Q. Thus, x
2 + 1 = acx
2 + (bc + ad)x + bd, which implies that ac = 1, bc + ad = 0, and
bd = 1. So in particular, none of a; c; b or d is zero. There are a number of contradictions one could
reach. Here is one: ac = 1 implies that a =
1
c
(recall, we are in Q, so
1
c makes sense). And bd = 1
implies that b and d are either both positive, or both negative. So, substituting for a in bc + ad = 0,
we get bc +
1
c
d = 0. Then, multiplying by c gives bc
2 + d = 0, which is equivalent to c
2 =
d
b
. Since
b and d have the same sign,
d
b
is negative. But c is rational, so c
2
cannot equal a negative number.
This is a contradiction, so x
2 + 1 must be irreducible in Q[x].
10.
(a) x
2 3 is irreducible in both Q[x]. It is not irreducible in R[x], because x
2 3 = (x +
p
3)(x
p
3).
(b) x
2 + x 2 is not irreducible in Z3[x] or in Z7[x] because x
2 + x 2 = (x + 2)(x 1). Notice,
however, that (x + 2)(x 1) = (x + 2)(x + 2) in Z3[x], but (x + 2)(x 1) = (x + 2)(x + 6) in Z7[x].

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2012-10-10T20:26:27+03:00

cevabı -20 olacak........

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