Math 306, Spring 2017
Homework 5 — Solutions
Due: April 19
(1) Suppose that α ∈ Q̄. Prove that minα (x) is irreducible in Q[x]. [Note: to say that a polynomial
f (x) ∈ Q[x] is reducible in Q[x] means that there exists some factorization f (x) = g(x)h(x) with
0 < deg(g(x)), deg(h(x)) < deg(f (x)). To say that f (x) is irreducible means that f (x) is not reducible.]
Solution. Suppose, on the contrary, that minα (x) = g(x)h(x) for some g(x), h(x) ∈ Q[x] with
0 < deg(g(x)), deg(h(x)) < deg(minα (x)). Evaluating at α we’re left with 0 = minα (α) = g(α)h(α).
Now since this is an expression in C, and since C has no zero divisors (since it’s a field), we must
therefore have either g(α) = 0 or h(α) = 0. But then we’d have that α is the root of some nonzero
rational polynomial of degree less than deg(minα (x)), contradicting the minimality of minα (x).
(2) Suppose that α ∈ Q̄, and let deg(minα (x)) = d. Prove that {1, α, · · · , αd−1 } is a basis for the Q-vector
space
Q[α] = {q0 + q1 α + q2 α2 + · · · + qn αn : n ∈ N and q0 , q1 , · · · , qn ∈ Q}.
[Note that this means that dimQ Q[α] = d, and since we argue that Q[α] = Q(α) on this handout, it
follows that dimQ (Q(α)) = [Q(α) : Q] = deg(minα (x)).]
Solution. We start with a lemma: any element of the form αi for i ∈ N is contained in the Q-span
of {1, α, · · · , αd−1 }. The result is clearly true for i ≤ d − 1. Suppose, by induction, that we have the
result for all j < i. Note in particular that αi−1 is then in the Q-span of {1, α, · · · , αd−1 }: there exist
c0 , · · · , cd−1 ∈ Q with
αi−1 = c0 + c1 α + · · · + cd−1 αd−1 .
Furthermore, if we write minα (x) = b0 + b1 x + · · · + bd−1 xd−1 + xd (with each bi ∈ Q), then we can use
the equation 0 = minα (α) to write
αd = −b0 − b1 α − · · · − bd−1 αd−1 .
Using the previous two equations, we then have
αi = ααi−1
= α(c0 + c1 α + · · · + cd−1 αd−1 )
= c0 α + c1 α2 + · · · + cd−1 αd
= (c0 α + c1 α2 + · · · + cd−2 αd−1 ) + cd−1 αd
= (c0 α + c1 α2 + · · · + cd−2 αd−1 ) + cd−1 −b0 − b1 α − · · · − bd−1 αd−1
= −b0 + (c0 − b1 )α + (c1 − b2 )α2 + · · · + (cd−2 − bd−1 )αd−1 + cd−1 αd−1 .
This proves our lemma.
Now we prove that {1, α, · · · , αd−1 } is a Q-basis. First, we’ll show the set spans Q[α]. If we take a
given z ∈ Q[α], then we can write
z = q0 + q1 α + · · · + qn αn
for some n ∈ N and all qi ∈ Q. Our lemma tells us that each for each 1 ≤ i ≤ n we have αi =
c0,i + c1,i α + · · · + cd−1,i αd−1 for some cj,i ∈ Q. Plugging this into our expression for z gives
z = q0 + q1 α + · · · + qn αn
= q0 + q1 c0,1 + c1,1 α + · · · + cd−1,1 αd−1 + · · · + qn c0,n + c1,n α + · · · + cd−1,n αd−1
= (q0 + q1 c0,1 + · · · + qn c0,n ) + (q1 c1,1 + · · · + qn c1,n ) α + · · · + (q1 cd−1,1 + · · · + qn cd−1,n ) αd−1 .
Since Q is closed under addition and multiplication, each coefficient is in Q, and therefore z is in the
Q-span of {1, α, · · · , αd−1 }.
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Math 306, Spring 2017
Homework 5 — Solutions
Due: April 19
Now to show independence, suppose that we had some relation
q0 + q1 α + · · · + qd−1 αd−1 = 0,
where each of the qi is rational. Let i be as large as possible so that qi 6= 0. If we assume that
{1, α, · · · , αd−1 } is dependent (for the sake of contradiction), then such a qi exists. We can then divide
the relation through by qi to get
q0
q1
qi−1 i−1
0=
+ α + ··· +
α
+ αi .
qi
qi
qi
i−1
+ xi is a monic polynomial in Q[x] that has α
It then follows that f (x) = qq0i + qq1i x + · · · + qi−1
qi x
as a root and which satisfies deg(f (x)) = i < d = deg(minα (x)). This contradicts the assumption that
minα (x) is the minimal polynomial for α.
(3) Suppose that F and K are fields, with F ⊆ K. Suppose further that W is a vector space over K; this
means that there are functions ⊕ : W × W → W and : K × W → W which satisfy the usual vector
space axioms [rules like k (w1 ⊕ w2 ) = (k w1 ) ⊕ (k w2 ) for all k ∈ K and w1 , w2 ∈ W , or that
w1 ⊕ w2 = w2 ⊕ w1 for all w1 , w2 ∈ W ]. In 206, you would have called ⊕ ”vector addition” and ”scaling.”
Notice that since F ⊆ K, the elements of F are themselves scalars on W , and so we can view W as a
vector space over F as well. It’s also true that we can view K as a vector space over F , since K comes
endowed with its own additive structure, and multiplying elements of K by elements of F satisfy all the
necessary rules of a vector space.
(a) Prove that dimF (W ) = dimF (K) dimK (W ) by arguing that if {w1 , · · · , wn } is a basis for W over K,
and if {k1 , · · · , k` } is a basis for K over F , then {k1 w1 , · · · , k` w1 , k1 w2 , · · · , k` w2 , · · · , k1 wn , · · · , k` wn }
is a basis for W over F .
Solution. First, we argue that the given set spans W (under F coefficients). Let w ∈ W be
given. Since {w1 , · · · , wn } is a basis for W over K, we have elements a1 , · · · , an ∈ K so that
w = a1 w1 + · · · + an wn . Furthermore, since {k1 , · · · , k` } is a basis for K over F , we know for each
ai there exist bi,1 , · · · , bi,` ∈ F so that ai = bi,1 k1 + · · · + bi,` k` . Plugging these equations into our
expression for w yields
w = a1 w1 + · · · + an wn
= (b1,1 k1 + · · · + b1,` k` ) w1 + · · · + (bn,1 k1 + · · · + bn,` k` ) wn
=
n X
`
X
bi,j kj wi .
i=1 j=1
Hence we have that w is in the F -span of {k1 w1 , · · · , k` w1 , k1 w2 , · · · , k` w2 , · · · , k1 wn , · · · , k` wn }.
Now we argue the set is linearly independent over F . Suppose that we have some coefficients
fi,j ∈ F (with 1 ≤ i ≤ n and 1 ≤ j ≤ `) so that
0=
n X
`
X
fi,j kj wi .
i=1 j=1
We can rewrite this as
0 = (f1,1 k1 + · · · + f1,` k` ) w1 + · · · + (fn,1 k1 + · · · + fn,` k` ) wn ,
where the coefficient in front of each wi is an element of K since the fi,j and kj are elements of K
(which is closed under addition and multiplication). But we know that {w1 , · · · , wn } is a basis for
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Math 306, Spring 2017
Homework 5 — Solutions
Due: April 19
W over K, and hence is independent. So this is only possible if — for all 1 ≤ i ≤ n — we have
fi,1 k1 + · · · + fi,` k` = 0.
But note that each of these equations is an F -relation of the elements {k1 , · · · , k` }, which we’re
assuming is a basis over F . Since they are therefore linearly independent over F , we must have
that each of these coefficients is 0. Hence we have that all fi,j = 0, and so the only F -relation
amongst {k1 w1 , · · · , k` w1 , k1 w2 , · · · , k` w2 , · · · , k1 wn , · · · , k` wn } is the trivial relation. So the set
is linearly independent.
(b) Prove that if F ⊂ K ⊂ L are all fields, then [L : F ] = [L : K][K : F ].
Solution. This follows from part (a) by taking L for the vector space W from the question.
(The only thing to do would be to argue that L is a vector space over K, which would amount to
identifying a “vector addition” on the set L and some “scalar multiplication” by K which satisfies
the appropriate rules. In this case, one defines “vector addition” to be the addition we get from
the field structure of L, and the “scalar multiplication” is simply the multiplication we get from
the field structure of L (which is possible since K ⊆ L). One then argues that the appropriate
properties are satisfied by citing the corresponding field-theoretic axiom. For instance, we have
k (l1 ⊕ l2 ) = (k l1 ) ⊕ (k l2 ) because we know that multiplication distributes across addition
in a field. It’s perhaps also worth mentioning that the field structure of L also gives us a way to
talk about multiplying the “vectors” of L, and hence we get that L is a vector space endowed with
“vector multiplication.” This kind of object is called an algebra.)
(4) (?) Let √
K be a quadratic number field, and define the discriminant of K in the following way: if
K = Q( d) where d 6= 1 and d is squarefree, then we define
d
if d ≡ 1 (mod 4)
dK =
4d if d ≡ 2, 3 (mod 4)
h √ i
Prove that OK = Z dk +2 dk .
Solution. We give a proof by cases. First, suppose that√d ≡ 2, 3 (mod 4), so that dK = 4d. We know
from a result in class that in this case, we have OK = Z[ d]. So we simply need to show that
"
√ #
√
4d + 4d
Z[ d] ⊆ Z
.
2
√
√
Observe that for any a + b d ∈ Z[ d] we have
√
√
a + b d = a − 2bd + 2bd + b d = a − 2bd + b(2d +
√
d) = a − 2bd + b
√ !
4d + 4d
.
2
h √ i
√
Since a − 2bd, b ∈ Z, we have Z[ d] ⊆ Z 4d+2 4d . For the reverse containment, we simply run this
process backwards:
√ !
√
√
4d + 4d
a+b
= a + b2d + b d ∈ Z[ d].
2
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Math 306, Spring 2017
Homework 5 — Solutions
Due: April 19
Now suppose we are in the case d ≡ 1 (mod 4), so that dK = d. In this case we know OK = Z
So we prove the equality of sets
"
"
√ #
√ #
d+ d
1+ d
Z
=Z
.
2
2
Let a + b
a+b
h
√ i
1+ d
.
2
√ d+ d
2
be given. Then we have
√ !
√ !
d+ d
d−1+1+ d
d−1
=a+b
=a+b
+b
2
2
4
√ !
1+ d
.
2
Now
that since d ≡ 1 (mod 4) we have (d − 1)/4 ∈ Z,and hence the quantity above is in
h observe
√ i
√
1+ d
1+ d
Z
.
For
the
reverse
direction,
observe
that
for
a
+
b
and d = 4k + 1 we have
2
2
!
!
√
√
√ !
1+ d
−4k + 4k + 1 + d
d+ d
a+b
=a+b
= a − 2b + b
.
2
2
2
√
(5) Let K be an imaginary quadratic number field, say √
K = Q( d) for d < 0 and
√ squarefree; assume
furthermore that d 6≡ 1 (mod 4). For an element a + b d ∈ OK , define N (a + b d) = a2 − db2 . Recall
that an element α ∈ OK is called a unit in OK if there exists some γ ∈ OK so that αγ = 1.
(a) Prove that N is multiplicative: if α1 , α2 ∈ OK , then N (α1 )N (α2 ) = N (α1 α2 ).
Solution.
One can approach this in two ways. The first is the direct approach: suppose that αi =
√
ai + bi d, and we’ll compute both N (α1 α2 ) and N (α1 )N (α2 ). On the one hand, we immediately
have
N (α1 )N (α2 ) = a21 − db21 a22 − db22 = (a12 a22 + d2 b21 b22 ) − d(b21 a22 + b22 a21 ).
On the other hand, we can compute that
√
√
√
α1 α2 = a1 a2 + a1 b2 d + a2 b1 d + b1 b2 d = (a1 a2 + db1 b2 ) + (a1 b2 + a2 b1 ) d.
Note that the latter expression is of the appropriate form to evaluate N (α1 α2 ), and using the
provided formula we get
2
N (α1 α2 ) = (a1 a2 + db1 b2 ) − d(a1 b2 + a2 b1 )2
= a21 a22 + 2a1 a2 db1 bd + d2 b21 b22 − da21 b22 − 2da1 b2 a2 b1 − da22 b21
= a21 a22 + d2 b21 b22 − d a21 b22 + a22 b21 .
This is precisely the value we computed for N (α1 )N (α2 ), so we get the desired result.
Here’s a second way to approach this problem if√you remember some basic facts about complex
numbers. Note that N (α) = αᾱ, where ᾱ = a − b d is the (so-called) complex conjugate of α. The
relevant fact to remember about complex conjugation is that respects multiplication: if z1 , z2 ∈ C
are given, then z1 z2 = z̄1 z̄2 . With this in mind, we get that
N (α1 α2 ) = α1 α2 α1 α2 = α1 α2 ᾱ1 ᾱ2 = α1 ᾱ1 α2 ᾱ2 = N (α1 )N (α2 ).
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Math 306, Spring 2017
Homework 5 — Solutions
Due: April 19
(b) Prove that α ∈ OK is a unit if and only if N (α) = 1.
√
√
Solution. Since we assume d ≡ 1 (mod 4), we must have OK = Z[ d]. Let α = a + b d. If
N (α) = 1, then we have a2 − db2 = 1. Observe that
√
√
(a + b d)(a − b d) = a2 − b2 d = 1.
√
√
√
Hence a − b d is a multiplicative
inverse of a + b d. Furthermore we have that a − b d ∈ OK
√
since it is conspicuously in Z[ d].
Conversely, if α is a unit then there exists some γ ∈ OK so that αγ = 1. By applying norms we
find that N (α)N (γ) = 1. But observe that since the form x2 − dy 2 is a positive definite quadratic
form, and since N (α), N (γ) ∈ Z, this is possible if and only if N (α) = 1.
(c) Prove that if d 6= −1, then the only units of OK are ±1.
Solution.√ Let α ∈ OK be a unit,
√ and suppose αγ = 1. By hypothesis we are in the case where
OK = Z[ d], so that α = a + b d for some a, b ∈ Z. If N (α) = 1, then we have a2 − db2 = 1.
Observe that if we have d ≤ −2, then if b 6= 0 we arrive at a2 − db2 ≥ −db2 ≥ −d ≥ 2. Hence we
must be in the case where b = 0, so that a2 = 1. But then a = ±1. (
[Alternatively, one can use the fact that if we view x2 − dy 2 as a quadratic form in the usual
“ax2 + bxy + cy 2 ” notation, then our assumption gives us that c > a. By Sheridan’s proof of the
fact that each equivalence class of quadratic forms (under proper equivalence) contains a unique
reduced form, we know that the minimum value attained by the quadratic form is a (i.e., 1), and
furthermore that this minimum value is achieved only for (x, y) = ±(1, 0).]
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