Download Solutions to Second Midterm

MATH 100B
Elham Izadi
February 21 2014: Second midterm
Please prove all your assertions and do not forget quantifiers when needed. State all the definitions, propositions, theorems, lemmas that you use precisely. Good luck!
(1) (42 points) Define the following objects
(a) (20 points) the field of fractions of an integral domain (define the field of fractions
as a set and say how the binary operations are defined and what the additive and
multiplicative identities are)
(b) (12 points) the minimal polynomial of an algebraic element of a field extension
(c) (10 points) an element of R constructible over a subfield F of R
Solution. (a) the field of fractions of an integral domain: For an integral domain D, the
field of fractions Q(D) is defined as the set of equivalence classes
in D ⇥ (D \ {0})
for the equivalence relation
(a, b) ⇠ (c, d) , ad
bc = 0.
Denoting the equivalence class of (a, b) by ab , addition is defined as
a c
ad + bc
+ :=
,
b d
bd
with zero 01
and multiplication is defined as
a c
ac
· :=
b d
bd
with identity 01 .
(b) the minimal polynomial of an algebraic element ↵ 2 L
of a field extension L K
is the monic
polynomial f (x) 2 K[x]
of smallest degree
such that f (↵) = 0.
(c) an element u of R is called constructible over a subfield F of R if it is possible to
construct a segment of length |u|
starting from points with coordinates in F
and using only a straight edge
and compass.
⇤
(2) (42 points) Prove that the regular heptagon is not constructible. Hint: letting ⇣ := e2⇡i/7 ,
show that ⇣ + ⇣ 1 is a root of the polynomial x3 + x2 2x 1. Deduce that ⇣ + ⇣ 1 is not
constructible and hence cos( 2⇡
) is not constructible.
7
Solution. We have the equation ⇣ 7 = 1.
So we can write
0 = ⇣7
1 = (⇣
1)
6
X
⇣ i.
i=0
Therefore
6
X
⇣ i = 0.
i=0
Substituting ⇣ + ⇣
1
for x in the polynomial x3 + x2
(⇣ + ⇣
1 3
) + (⇣ + ⇣
1 2
)
2(⇣ + ⇣
1
)
2x
1=
1 and simplifying, we obtain
3
X
⇣ i.
i= 3
Multiplying the above equation by ⇣ 3 , we obtain
6
X
⇣i
i=0
1 3
1 2
which is zero. So (⇣ + ⇣ ) + (⇣ + ⇣ )
2(⇣ + ⇣ 1 ) 1 = 0.
This polynomial is irreducible over Q because it has no rational roots:
if pq is a rational root, then p| 1 and q|1.
So the only possible roots are ±1.
Substituting ±1 into the polynomial we see that they are not roots.
Therefore x3 + x2 2x 1 is the minimal polynomial of ⇣ + ⇣
and [Q(⇣ + ⇣ 1 ) : Q] = 3.
Since this is not a power of 2, ⇣ + ⇣ 1 is not constructible.
Since 2 cos( 2⇡
)=⇣ +⇣
z
1
, cos( 2⇡
) is not constructible either.
7
1
⇤
(3) (40 points) Assume R is a commutative ring and that R is either finite or a finite dimensional
vector space over a field. Prove that every prime ideal in R is maximal.
Solution. Let I ⇢ R be a prime ideal and let D = R/I be the quotient of R by I.
Then, since I is prime, D is an integral domain.
To prove that I is maximal, we prove that D is a field.
To prove this, we need to prove that every nonzero element of D is invertible.
Let a 2 D be a nonzero element. Consider the map
a
: D ! D
x 7 ! ax.
Then a is a homomorphism of additive groups,
and, when D is a finite dimensional vector space over a field K, a is K-linear.
If x 2 ker a , then ax = 0 and, since a 6= 0 and D is an integral domain, we have x = 0.
So the kernel of a is trivial and a is injective.
Since D is either finite or a finite dimensional vector space over a field K (and a is K-linear), it
follows that a is bijective.
Hence a is surjective
and there exists b 2 D such that a (b) = 1, i.e., ab = 1 and a is invertible.
⇤