# 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 Mod-p 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 Mod-p 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 Mod-p 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 Mod-p 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 (x-c)(x-d),

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,
3-test 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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