Section 13.2 and 13.3.
Homework #9
Masaya Sato
Sec 13.2 3. Determine the minimal polynomial over Q for the element 1 + i. Solution: Observe that 1 + i is a root of the quadratic polynomial x2 + 2x + 2. By the the 2 minimal al Rational Root Theorem , the polynomial is irreducible over Q. So x + 2x + 2 is the minim
polynomial.
√3 and of 1 + √2 + √4. √ √ First consider the extension Q(2 + 3) of Q. Observ Observee that 2 + 3 is a root of 3
4. Determine the degree over Q of 2 + Solution:
3
the polynomial polynomial x2
− 4x + 1,
and x2 4x + 1 is irreducible over Q by the Rational Root Theorem or Gauss’s Lemma . So the polynomial is the minimal polynomial and therefore [ Q(2 + 3) : Q] = 2. Next consider the extension Q(1 + 2 + 4) of Q. Observe that Q(1 + 2 + 4) = Q( 4) since 1 and 2 are generated by 4. And 4 is a root of the polynomial x3 4, which is irreducible over Q by Rational Root Theorem or Gauss’s Lemma . So
−
√
√ √ √ √ 3
√ 3
3
3
[Q(1 +
√ 3
3
3
2+
√ 3
√ √ −
√
3
3
√ 3
4) : Q] = [Q( 4) : Q] = 3.
5. Let F = Q(i). Prove that x3
3
− 2 and x − 3 are irreducible over F . F . Proof. First suppose by contradiction that x − 2 is reducible over F . F . Then at least one root is in F since x − 2 decomposes into two polynomials, i.e., x − 2 = (x − α)q(x), √ where α ∈ F and q(x) is some monic quadratic polynomial. Now let ζ = + i. Then the roots of x − 2 are √ √ √ 3
3
3
1 2
3
3
2,
3
2ζ , and
3
3 2
2ζ ,
where ζ denotes the complex conjugate of ζ . Observe Observe that every every element in F is of the form a + bi, bi, Howeve verr none of the roots is of the form. form. Theref Therefore ore x3 2 is a and b are elements in Q. Howe irreducible over F . F . 3 Similarly, Similarly, if x if x 3 is reducible over F , F , then the roots are 3, 3ζ , and 3ζ . However, none of the roots is of the form a + bi, bi, where a and b are in Q. Therefore x3 3 is irreducible over F . F .
√ √ 3
−
10. Determine the degree of the extension Q(
3
−
√ − 3
√
3 + 2 2) over Q.
√
Solution: Observe first that 3 + 2 2 is a root of the polynomial p(x) = x4 6x2 + 1. p(x) is irreducible over Q by Rational Root Theorem or Gauss sLemma . So p(x) is the minimal
√
polynomial and therefore [ Q( 3 + 2 2) : Q] = 4.
Abstract Abstract Algebra Algebra by Dummit and Foote 1
−
Section 13.2 and 13.3.
Homework #9
Masaya Sato
12. Suppose the degree of the extension K/F is a prime p. Show that any subfield E of K
containing F is either K or F . F . Proof. By assumption the degree [K [ K : F ] F ] = p, where p is some prime. Since E is a subfield of K containing F , F , i.e. F E K ,
≤ ≤
][E : F ]. p = [K : F ] F ] = [K : E ][E F ]. Then consider the following two cases. Case 1: [K : E ] = 1 and [E [E : F ] F ] = p. Then E = K . [E : F ] Case 2 : [K : E ] = p and [E F ] = 1. Then E = F . F .
∈ Q for i = 1, 2, . . . , n.n. Prove that √2 ∈/ F . F . positive integer Proof. Observe first that the degree√of extension F /Q is 2 for some k. √ √ Then suppose by contradiction that 2 ∈ F . F . Any field containing 2 includes Q( 2) as a 13. Suppose F = Q(α1 , α2 , . . . , αn ) where αi2
3
k
3
3
3
subfield. Therefore
√
√
3
√
3
3
[F : Q] = [F : Q( 2)][Q( 2) : Q] = 3[F 3[F : Q( 2)], and thus [F [F : Q] is divisible by 3. This contradicts that [F [ F : Q] = 2k . Hence
√2 ∈/ F . F . 3
[ F ((α) : F ] 14. Prove that if [F F ] is odd then F ( F (α) = F ( F (α2 ). Proof. Since [F [F ((α) : F ] F ] is finite, α is algebraic over F . F . This implies that α2 = αα is algebraic
over F . [ F ((α2 ) : F ] F . So the simple extension [F F ] is finite. Therefore [F ( )][F ((α2 ) : F ] F (α) : F ] F ] = [F ( F (α) : F ( F (α2 )][F F ]
2
2
[F ((α) : F ( )][F ((α ) : F ] ⇒ 2k + 1 = [F F (α )][F F ] Obse Observ rvee that that [F [F ((α) : F ( F (α )] ≤ 2. If [F ( F (α) : F ( F (α )] = 2,
2 2 for some positive integer k . then this contradicts that [F [ F ((α) : F ] Ther eref efor oree [F ( F ] = 2k + 1. Th F (α) : F ( F (α2 )] = 1 and thus F ( F (α) = F ( F (α2 ).
16. Let K/F be an algebraic extension and let R be a ring contained in K and containing
F . F . Show that R is a subfield of K containing F . F . Proof. For every nonzero α
∈ R α is algebraic over F since α ∈ K . So α is a root of some
irreducible polynomial p(x) of the form
p(x) = an xn +
···+ a x+a , 1
0
where a0 , a1 , . . . , an F . F . If a0 = 0, then p(x) = x since otherwise p(x) would be reducible. Moreover 0 = p(α) = α. So suppose that a0 = 0. Then
∈
0 = p(α) = an αn +
· · · + a α, 1
and thus α−1 =
1
n 1
−(a )− (a α − + · · · + a ) ∈ R 0
n
1
is a multiplicative inverse of α. Therefore R is a field.
Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 13.2 and 13.3.
Homework #9
Masaya Sato
19. Let K be an extension of F of degree n.
(a) For any α K prove that α acting by left multiplication on K is an F -linear F -linear transformation of K .
∈
(b) Prove Prove that that K is isomorphic to a subfield of the ring of n n matrices over over F , F , so the ring of n n matrices over F contains an isomorphic copy of every extension of F of degree n.
≤
×
×
Proof. (a) Let f α : K
→ K be a map defined by f α (x) = αx. αx.
Then for all x and y in K f α (x + y) = α(x + y) = αx + αy = f α (x) + f α (y), and for every λ
∈ F f α (λx) λx) = α(λx) λx) = (αλ αλ))x = (λα λα))x = λ(αx) αx) = λf α (x).
Therefore the map f λ , which corresponds to left multiplication on K by α, is an F -linear F -linear transformation on K . (b) For α and β in K assume that f α (x) = f β K . Then β (x) for every x
∈
f α (x) = f β β (x)
⇒ αx = βx
and thus α = β by setting x = 1. So the action of an element in K by left multiplication is an injective F -linear F -linear transformation. Moreover for all α and β in K and x K
∈
f α+β (x) = (α + β )x = αx + βx = f α (x) + f β β (x), and (f α f β )(x). f αβ βx ) = f α (f β αβ (x) = (αβ )x = α(βx) β (x)) = (f β )(x Therefore K is isomorphic to the ring of n n matrices over F . urthermor moree the ring ring of F . Further n n matrices contains an isomorphic copy of every extension of F of degree less than or equal to n.
×
×
Sec 13.3 4. Prove that Q( 2) and Q( 3) are not isomorphic.
√
√
√ → Q(√3).
Proof. Suppose by contradiction that there exists an isomorphism ϕ : Q( 2)
Then observe that ϕ(1) = 1 and
√
√ √ √ √ √ So ϕ( 2) is either 2 or − 2. Since every element α ∈ Q( 3) is of the form √ 2 = 2ϕ(1) = ϕ(2) = ϕ(( 2)2 ) = (ϕ( 2))2 .
α = a + b 3,
Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 13.2 and 13.3.
Homework #9
Masaya Sato
there are some a and b in Q such that
√
√
2 = a + b 3.
Then
√
√
√ ⇒ 4 = a + 3b 3b + 2ab 2ab 3 √ ⇒ 3 = 4 − a2ab− 3b . √ irrational. Therefore √ √ However this contradicts that 3 is irrational. Therefore Q( 2) and Q( 3) are not isomor2
2= a+b 3
2
2
2
phic.
√
√
4
2)/Q( 2) explicitly. explicitly. 5. Determine the automorphisms of the extension Q( 2)/
√ √
− √2 ∈ Q(√2)[x 2)[x].
Solution: Observe first that 2 is a root of an irreducible polynomial x2 So the automorphisms of Q( 2) are the identity map id and ϕ that maps 4
4
√ 4
√ 4
So ϕ(a + b 2) = a
− b√2.
2
→
√ 4
2 or
√ 4
2
→ −
√ 4
2.
4
Abstract Abstract Algebra Algebra by Dummit and Foote 4
