391 Homework 8 solutions
• Exercises 2.3: 18. Show that Z[i] is an integral domain, describe its
field of fractions and find the units.
There are two ways to show it is an integral domain. The first is
to observe:
Any subring of a field is an integral domain.
(Proof: Suppose R is a subring of a field F . Take x, y ∈ R with
xy = 0. We need to show that one of x, y is zero. Suppose y 6= 0.
Then multiplying both sides by y −1 in the big field F , you get x = 0.
Done.)
The second way is to use some tricks. Suppose
wz = 0
for w, z ∈ Z[i]. Then wzwz = 0 hence wwzz = 0. Hence |w||z| = 0.
Hence either |w| = 0 or |z| = 0, i.e. either w = 0 or z = 0.
The field of fractions of Z[i] is isomorphic to Q[i], i.e. the complex
numbers of the form ab + dc i for a, b, c, d ∈ Z with b, d 6= 0.
Finally to find the units in Z[i], suppose w is a unit, so
wz = 1
for some z ∈ Z[i]. Then wzwz = 1 hence |w||z| = 1 hence ww =
|w|2 = 1. But if w = x + iy then |w|2 = x2 + y 2 = 1. That means
either x = ±1, y = 0 or x = 0, y = ±1. Hence the units are ±1, ±i.
• Exercises 2.4: 1(a)(b)(c).
(a) z 2 − (6 + i)z + (8 + 2i) = 0.
By the quadratic formula,
√
6 + i ± 1 + 4i
z=
.
2
You can simplify and put the answer in Cartesian form x + iy if you
like, but there is no need.
(b) z 2 − iz + 2 = 0.
By the quadratic formula,
√
i ± −9
i ± 3i
z=
=
.
2
2
so z = 2i or z = −i are the solutions.
(c) z 4 + 2z 2 + 4 = 0.
By the quadratic formula,
√
√
−2 ± −12
2
z =
= −1 ± 3i.
2
Let me simplify this time just for fun... first into polar form
z 2 = 2e±2π/3
1
2
Now square root once more...
√
z = ± 2e±π/3 .
If you like you can convert back into cartesian form, but there is no
need. If you do, you get
√
√
√ 1
3
1
3
z = ± 2( ± i
) = ± √ ± √ i.
2
2
2
2
(There are four different answers!)
• Exercises 3.1: 1(e), 2(a)(b)(c), 5, 6, 8, 9(a)(c), 10, 13, 14, 20(c).
1(e) f (x) = x7 + x6 + x4 + x + 1, g(x) = x3 + x + 1, F = Z2 .
Long division gives
x7 + x6 + x4 + x + 1 = (x3 + x + 1)(x4 + x3 + x2 + x) + 1.
VERY IMPORTANT: if you don’t know how I was able to just
write that answer down with no working, COME AND SEE ME
IN AN OFFICE HOUR!!!!!
2(a)(b)(c) (a) f (x) = x3 − 1, g(x) = x4 + x3 − x2 − 2x − 2, F = Q.
Long division gives...
x4 + x3 − x2 − 2x − 2 = (x3 − 1)(x + 1) − (x2 + x + 1)
x3 − 1 = (x2 + x + 1)(x − 1) + 0.
Hence the GCD is x2 + x + 1 and
x2 + x + 1 = (x4 + x3 − x2 − 2x − 2) − (x + 1)(x3 − 1).
√
√
(b) f (x) = x2 + (1 − 2)x − 2, g(x) = x2 − 2, F = R.
Long division and a little scratch work gives
√
√
√
√
x2 + (1 − 2)x − 2 = (x2 − 2).1 + (1 − 2)x + (2 − 2)
√
√
√
√
(x2 − 2) = ((1 − 2)x + (2 − 2))((−1 − 2)x + (−2 − 2)) + 0.
√
√
So the
GCD
is
(1
−
2)x
+
(2
−
2) made monic, which is
√
x − 2. Moreover,
√
√
√
1
1
√ (x2 + (1 − 2)x − 2) −
√ (x2 − 2).
x− 2=
2− 2
2− 2
(c) f (x) = x2 + 1, g(x) = x2 − i + 2, F = C.
Long division gives
x2 + 1 = (x2 − i + 2).1 + (i − 1).
Since i − 1 is a unit, the GCD is i = 1 made monic, which is 1.
Moreover,
1
1
1=
(x2 + 1) +
(x2 − i + 2).
i−1
1−i
3
5 Suppose deg(f (x)) = n and deg(g(x)) = m with m ≥ n. Prove
of give a counterexample...
(a) deg(f (x) + g(x)) = m.
FALSE. Take g(x) = −f (x) – then f (x) + g(x) = 0!!!
(b) deg(f (x)g(x)) = m + n.
FALSE. Take the coefficient ring to be Z4 and f (x) = g(x) =
2x + 1. Then f (x)g(x) = 4x2 + 4x + 1 = 1 since we can reduce
the coefficients mod 4. (BUT this is true if you assume that
f (x) and g(x) are not zero and the coefficient ring is an integral
domain).
6 Prove that if F is a field, f (x) ∈ F [x] and deg f (x) = n ≥ 1
then f (x) has at most n roots in F .
Proof. Proceed by induction on n. The case n = 1 is clear:
since f (x) is linear and F is a field it has exactly one root in
this case.
Induction step. Assume true if deg f (x) = k. Suppose deg f (x) =
k + 1. If f (x) has no roots in F , then we are certainly okay. So
assume that f (x) has at least one root c ∈ F . Then, (x − c) is
a factor, i.e.
f (x) = (x − c)g(x)
where deg g(x) = k. By induction g(x) has at most k roots. So
f (x) = (x − c)g(x) has at most (k + 1) roots.
8 Let F be a field. Prove that if f (x) ∈ F [x] is a polynomial of
degree 2 or 3 then f (x) is irreducible in F [x] if and only if f (x)
has no root in F .
Proof. I’ll show f (x) is reducible in F [x] if and only if f (x) has
a root.
(⇒). Suppose f (x) is reducible, then f (x) = g(x)h(x) where –
because f (x) is of degree 2 or 3 – g(x) must be a linear factor.
But linear factors always have roots. So f (x) has a root too.
(⇐). Suppose f (c) = 0. Then (x − c) is a linear factor of f (x),
so f (x) is reducible.
9(a)(c) (a) Show that unique factorization fails horribly in R[x] when
R is not an integral domain.
Consider x2 − 1 over Z8 . You can factor it in two different
ways: (x − 1)(x + 1) or (x − 3)(x + 3). So unique factorization
fails when the coefficient ring is not an integral domain.
(c) How many roots does f (x) = 2x − 4 in Z6 [x] have?
Search... x = 2, 5 are both roots, so it has two roots!
10 Decide whether each of the following is irreducible.
(a) f (x) = x2 + 1 over Z5 . REDUCIBLE: x = 2 is a zero.
(b) f (x) = x2 + 1 over Z7 . IRREDUCIBLE: it doesn’t have any
zeros so its irreducible (as it is of degree 2 or 3).
(c) f (x) = x2 + 1 over Z1 9.
4
IRREDUCIBLE: it doesn’t have any zeros (note it is enough to
check x = 0, 1, 2, . . . , 9 to see this!).
(d) f (x) = x3 − 9 over Z1 1.
x = 4 gives 55 = 0 so it is REDUCIBLE.
(e) f (x) = x3 + x + 1 over Z2 .
IRREDUCIBLE as no zeros (and it is of degree 2 or 3).
(f) f (x) = x4 + x2 + 1 over Z2 .
Since it has no zeros it cannot have a LINEAR factor. But it
could still possible factor as a product of two irreducible quadratics. There is only one irreducible quadratic over Z2 , namely,
x2 + x + 1. So if x4 + x2 + 1 factors it must factor as (x2 + x +
1)(x2 + x + 1) – and it does! So its REDUCIBLE.
13 List all the irreducible polynomials in Z2 [x] of degree ≤ 4. Factor x7 + 1 as a product of irreducibles in Z2 [x].
Linear polynomials are always irreducible, so x, x + 1 count.
Then for quadratics there is only x2 +x+1 – the other 3 possible
quadratics all have zeros.
For cubics there are 8 possibilities at first: x3 , x3 +1, x3 +x, x3 +
x + 1, x3 + x2 , x3 + x2 + 1, x3 + x2 + x, x3 + x2 + x + 1. Obviously
we must have +1 at the end else x = 0 is a root. Also there
must be an odd number of non-zero terms else x = 1 is a root.
So this leaves just x3 + x + 1 and x3 + x2 + 1. Since these have
no roots and are cubics they ARE irreducible.
For quartics, similar reasoning narrows the candidates down to
x4 + x + 1, x4 + x2 + 1, x4 + x3 + 1, x4 + x3 + x2 + x + 1. Now
which of these are irreducible? They have no linear factors, but
they could factor as a product of two irreducible quadratics, i.e.
as (x2 + x + 1)2 . That rules out x4 + x2 + 1. The remaining
three MUST be irreducible.
Answer: x, x + 1, x2 + x + 1, x3 + x + 1, x3 + x2 + 1, x4 + x +
1, x4 + x3 + 1, x4 + x3 + x2 + x + 1.
14 For each of the following numbers c find an irreducible polynomial in √
Q[x] that has c as a root.
(a) 1 + √3.
c = 1 + 3, so (c − 1)2 = 3 so c is a √
root of x2 − 2x − 2. That
is irreducible over Q as its zeros 1 ± 3 are not rational...
(b) 2 + 21/3 .
c is a solution of (x−2)3 = 2, hence a root of the cubic x3 −6x2 +
12x − 10. To see that the latter is irreducible, use Eisenstein’s
criterion with p = 2 (see Theorem 3.5 on p.109).
(c) 2 + i.
It is a solution of (x − 2)2 = −1, so a root of x2 − 4x + 5. That
is irreducible
over Q as it has no zeros.
p
√
(d) 1 + 3.
5
√
c2 = 1 + 3 so (c2 − 1)2 = 3. So c is root of x4 − 2x2 − 2. To
see that really is irreducible, use Eisenstein with p = 2.
20(c) Solve
f (x) ≡ x2 + 1
(mod x3 + x + 2),
f (x) ≡ 2x + 1
(mod x2 + x + 2)
in Z3 [x].
First we must find the GCD...
x3 + x + 2 = (x2 + x + 2)(x + 2) + 1
So the GCD is 1 (so we can use the Chinese Remainder Theorem directly) and
1 = (x3 + x + 2) − (x + 2)(x2 + x + 2).
Now to solve the equation, one solution is
f (x) = (2x + 1)(x3 + x + 2) − (x2 + 1)(x + 2)(x2 + x + 2).
The general solution is therefore
f (x) ≡ (2x+1)(x3 +x+2)−(x2 +1)(x+2)(x2 +x+2)
(mod (x3 +x+2)(x2 +x+2)).