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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 (x2+ax+1)(x2+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 x4 + 1 = x4 + (a + b)x3 + (ab + 2)x2 + (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, x2 + x + 1, and so x3 + 6 = (x + 6)(x2 + x + 1).
Note: x = 2 and x = 4 are also roots.

3. (a) x5 + 9x4 + 12x2 + 6: Eisenstein's Criterion with p = 3 does the job.

(b) x3-3x+3: If f(x) = x3-3x+2 then reduction mod p = 2 yields = x3+x+1 ∈ Z2[x].
Since this is degree 3 it's irreducible over Z2 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
x4 + 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) x3 + 2x2 + 4x + 1: If f(x) = x3 + 2x2 + 4x + 1 then reduction mod p = 3 yields
= x3 +2x2 +x+1 ∈ Z3[x]. Since this is degree 3 it's irreducible over Z3 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) x3 +3x2 +2. If f(x) = x3 +3x2 +2 then reduction mod p = 5 yields = x3 +3x2 +2.
Since this is degree 3 it's irreducible over Z5 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) x5+5x2+1: If f(x) = x5+5x2+1 then reduction mod p = 2 yields = x5+x2+1 ∈
Z2[x]. No roots in Z2 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 x2+x+1 long division shows us that does not factor into .
Therefore is irreducible over Z2 and the Mod-p Test then tells us that f(x) is
irreducible over Q.

4. Observe that there are p2 polynomials of the form x2 + ax + b in Zp since there are p choices
for each of a and b.

Now then, a polynomial which is reducible over Zp 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 Zp.

5. Look at x4 +x3 +x2 +1 ∈ Z3[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 Z3[x] there are only three irreducible
quadratic polynomials with leading coefficient 1. Note that we can ignore those with leading
coefficient 2 since if 2x2 + ... is a factor then so is 2(2x2 + ...) = x2 + ....

These three polynomials are x2 + 1, x2 + x + 2 and x2 + 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 Z3[x]/ < x4 + x3 + x2 + 1 > is a field. The elements in this field
have the form ax3 + bx2 + cx + d + < x4 + x3 + x2 + 1 > since higher powers simplify via x4 +
< x4 + x3 + x2 + 1 > = 2x3+2x2+2+ < x4 + x3 + x2 + 1 >. Since a, b, c, d ∈ Z3 there are 3·3·3·3 =
81 possible elements.

6. Look at x3+x+1 ∈ Z2[x]. This polynomial is irreducible over Z2 by the 2, 3-test since there are
no roots. Therefore the quotient ring Z2[x]/ < x3 + x + 1 > is a field. The elements in this field
have the form ax2+bx+c+< x3 + x + 1 > since higher powers simplify via x3+< x3 + x + 1 > =
-x - 1 + < x3 + x + 1 > = x + 1 +  < x3 + x + 1 >. Since a, b, c ∈ Z2 there are 2 · 2 · 2 = 8 possible
elements.

7. We know from problem 4 that there are polynomials of the form x2 + ax + b which are
irreducible over Zp. Pick any one, denote it x2 + a'x + b'. Then Zp[x]/  < x2 + a'x + b' > is a
field of with elements of the form αx + β +  < x2 + a'x + b' > with α, β ∈ Zp. Since there are p
choices for each of α, β there are p2 elements.

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