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MATH 403: Homework 12 Solutions
1. There are several ways to do some of these, so these are just examples.
Note: This is not obvious but not hard to attack. Since there are no roots in R
there
are no linear factors in R [x] and so it must factor into two quadratics both in
R[x]. We
don't know that it factors into something like (x^{2}+ax+1)(x^{2}+bx+1) but this is a
good
first start since any other options would have leading or constant coefficients
which were
not integers and that would be awful so we give this a shot.
Expanding the right gives us x^{4} + 1 = x^{4} + (a + b)x^{3} + (ab + 2)x^{2} + (a + b)x + 1
and
equating coefficients gives us a + b = 0 and ab = 2. Solving this gives us our
result and
we're darned lucky it worked, too!
(b) We know that x + 6 (aka x  1) is a factor since x = 6 (aka x = 1) is a
root . Long
division gives us the other, x^{2} + x + 1, and so x^{3} + 6 = (x + 6)(x^{2} + x + 1).
Note: x = 2 and x = 4 are also roots.
3. (a) x^{5} + 9x^{4} + 12x^{2} + 6: Eisenstein's Criterion with p = 3 does the job.
(b) x^{3}3x+3: If f(x) = x^{3}3x+2 then reduction mod p
= 2 yields = x^{3}+x+1 ∈
Z_{2}[x].
Since this is degree 3 it's irreducible over Z_{2} iff it has no roots and testing
x = 0, 1 tells
us it doesn't. The Modp Test then tells us that f(x) is irreducible over Q.
Note: Eisenstein works here too, with p = 3.
(c) In 2(a) we showed that in R[x]. Now then
suppose
x^{4} + 1 factored over Q. Since it has no linear factors (because no rational
roots) it must
factor into two quadratics in Q[x]. But these two quadratics are then also in
R[x] and
then we have two different factorizations over R which contradicts the fact that
R[x] is a
UFD.
Note: These factorizations are "different" in the sense that they're not just
rearranged or
replaced by associates in R[x] since the associates are obtained by multiplying
by nonzero
real numbers (units).
(d) x^{3} + 2x^{2} + 4x + 1: If f(x) = x^{3} + 2x^{2} + 4x + 1 then reduction mod p = 3
yields
= x^{3} +2x^{2} +x+1 ∈ Z_{3}[x]. Since this is degree 3 it's irreducible over Z_{3}
iff it has no
roots and testing x = 0, 1, 2 tells us it doesn't. The Modp Test then tells us
that f(x) is
irreducible over Q. Note that p = 2 does not work here.
(e) x^{3} +3x^{2} +2. If f(x) = x^{3} +3x^{2} +2 then reduction mod p = 5 yields = x^{3}
+3x^{2} +2.
Since this is degree 3 it's irreducible over Z_{5} iff it has no roots and testing
x = 0, 1, 2, 3, 4
tells us it doesn't. The Modp Test then tells us that f(x) is irreducible over
Q. Note
that p = 2, 3 do not work here.
(f) x^{5}+5x^{2}+1: If f(x) = x^{5}+5x^{2}+1 then reduction mod p = 2 yields =
x^{5}+x^{2}+1 ∈
Z_{2}[x]. No roots in Z_{2} tells us there are no linear terms but this may
factor into a degree
2 term and a degree 3 term. Let's check all possible degree 2 terms which don't
factor
themselves. There is only x^{2}+x+1 long division shows us that does not factor
into .
Therefore is irreducible over Z_{2} and the Modp Test then tells us that
f(x) is
irreducible over Q.
4. Observe that there are p^{2} polynomials of the form x^{2} +
ax + b in Z_{p} since there are p choices
for each of a and b.
Now then, a polynomial which is reducible over Z_{p} may be rewritten in
the form (xc)(xd),
where c and d are the roots. So how many of these are there? Well, there are p
choices if c = d
and if c ≠ d then there are choices; the division by 2 eliminates repeated pairs.
Therefore there are polynomials irreducible over Z_{p}.
5. Look at x^{4} +x^{3} +x^{2} +1 ∈ Z_{3}[x]. This polynomial has no linear terms since
there are no roots
(try x = 0, 1, 2.)
But maybe it has two irreducible quadratic factors ? In Z_{3}[x] there are only
three irreducible
quadratic polynomials with leading coefficient 1. Note that we can ignore those
with leading
coefficient 2 since if 2x^{2} + ... is a factor then so is 2(2x^{2} + ...) = x^{2} + ....
These three polynomials are x^{2} + 1, x^{2} + x + 2 and x^{2} + 2x + 2 (discovered by
brute force) so
we divide our polynomial by each of these in turn and we find a remainder each
time.
Therefore the quotient ring Z_{3}[x]/ <
x^{4} + x^{3} + x^{2} + 1 > is a field. The elements in this field
have the form ax^{3} + bx^{2} + cx + d + <
x^{4} + x^{3} + x^{2} + 1 > since higher powers simplify via x^{4} +
<
x^{4} + x^{3} + x^{2} + 1 > = 2x^{3}+2x^{2}+2+
<
x^{4} + x^{3} + x^{2} + 1 >. Since a, b, c, d ∈ Z_{3} there are 3·3·3·3 =
81 possible elements.
6. Look at x^{3}+x+1 ∈ Z_{2}[x]. This polynomial is irreducible over Z_{2} by the 2,
3test since there are
no roots. Therefore the quotient ring Z_{2}[x]/ <
x^{3} + x + 1 > is a field. The elements in this field
have the form ax^{2}+bx+c+<
x^{3} + x + 1 > since higher powers simplify via x^{3}+<
x^{3} + x + 1 > =
x  1 + <
x^{3} + x + 1 > = x + 1 + <
x^{3} + x + 1 >. Since a, b, c ∈ Z_{2} there are 2 · 2 · 2 = 8 possible
elements.
7. We know from problem 4 that there are
polynomials of the form x^{2} + ax + b which are
irreducible over Z_{p}. Pick any one, denote it x^{2} + a'x + b'. Then Z_{p}[x]/
<
x^{2} + a'x + b' > is a
field of with elements of the form αx + β + <
x^{2} + a'x + b' > with α, β ∈ Z_{p}. Since there are p
choices for each of α, β there are p^{2} elements.
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