Homework 2 Solutions
Problem 4.2.3. Find the greatest common divisor of f (x) and f 0 (x), over Q.
(d) f (x) = x4 + 2x3 + 3x2 + 2x + 1
Solution. Using the Euclidean algorithm for polynomials (c.f. Example 4.2.4), gcd(f (x), f 0 (x)) = x2 + x + 1.
Problem 4.2.4. Find the greatest common divisor of the given polynomials, over Q.
(b) 4x3 − 2x2 − 3x + 1 and 2x2 − x − 2.
Solution. Again using the Euclidean algorithm for polynomials, gcd(4x3 − 2x2 − 3x + 1, 2x2 − x − 2) = 1
Problem 4.2.5. Find the greatest common divisor of the given polynomials, over the given field.
(b) x3 − 2x2 + 3x + 1 and x3 + 2x + 1 over Z5
Solution. The Euclidean algorithm works over any field, thus it can be verified that gcd(x3 − 2x2 + 3x + 1, x3 +
2x + 1) = 1.
Problem 4.2.15. Show that for any real number a 6= 0 the polynomial xn − a has no multiple roots in R.
Proof. Let f (x) = xn − a then f 0 (x) = nxn−1 and hence gcd(f (x), f 0 (x)) = 1. Therefore f (x) has no repeated
factor and thus no repeated roots in R.
Problem 4.3.3. Let E be a field, and let F be a subfield of E. Prove that the multiplicative identity of F must
be the same as that of E.
Proof. Let 0F , 1F ∈ F and 0E , 1E ∈ E be the additive and multiplicative identities of F and E respectively.
First note that 0F + 0E = 0F = 0F + 0F and thus 0F = 0E .
Since 1F · 1F = 1F = 1F · 1E it follows that 1F (1F − 1E ) = 0, but 1F 6= 0F = 0E so 1F − 1E = 0E ⇒ 1F =
1E .
Problem 4.3.5. Let φ : F1 → F2 be an isomorphism of fields. Prove that φ(1) = 1.
Proof. Since φ is an isomorphism of fields it follows that φ(1) = φ(1 · 1) = φ(1) · φ(1). If φ(1) = 0 then φ would
not be one to one since φ is a group homomorphism with respect to addition and hence φ(0) = 0. Thus φ(1) 6= 0
so φ(1) = φ(1) · φ(1) ⇒ φ(1) = 1.
Problem 4.3.6. Let F be a field, let p(x) be an irreducible polynomial in F [x], and let
E = {[a] ∈ F [x]/hp(x)i | a ∈ F }.
Show that E is a subfield of F [x]/hp(x)i. Then show that φ : F → E defined by φ(a) = [a], for all a ∈ F , is an
isomorphism of fields.
Proof. Since F [x]/hp(x)i is a field it suffices to show that E is nonempty, closed under subtraction, closed under
multiplication and multiplicative inverses of nonzero elements.
Since [0] ∈ E, E is nonempty. Given α, β ∈ E, α = [a] and β = [b] for some a, b ∈ F . Therefore
α − β = [a] − [b] = [a − b] which is in E since a − b ∈ F . Similarly, αβ = [a][b] = [ab] ∈ E since ab ∈ F .
If α 6= [0] then a 6= 0 and thus a−1 ∈ F so [a−1 ] ∈ E and is the inverse of α.
If α ∈ E then α = [a] = φ(a) for some a ∈ F , hence φ is onto. Suppose [a] = φ(a) = φ(b) = [b], then
[a − b] = [0] so a − b = 0, i.e. a = b and φ is injective. Lastly φ(a + b) = [a + b] = [a] + [b] = φ(a) + φ(b) and
φ(ab) = [ab] = [a][b] = φ(a)φ(b). Therefore φ is an isomorphism.
Problem 4.3.8. Prove that R[x]/hx2 + 2i is isomorphic to C
√
Proof. Define a map φ : C → R[x]/hx2 + 2i by φ(a + ib) = [a + (b/ 2)x]. Verifying that φ is an isomorphism is
left to the reader.
Problem 4.3.10. Is Q[x]/hx2 + 2i isomorphic to Q[x]/hx2 + 1i?
1
Proof. The fields are not isomorphic, proceed by contradiction. Suppose that there is an isomorphism φ :
Q[x]/hx2 + 2i → Q[x]/hx2 + 1i then φ([x]) = [a + bx] for some a, b ∈ Q. On one hand φ([x2 ]) = [a + bx]2 =
[a2 + 2abx + b2 x2 ] = [a2 − b2 + 2abx]. On the other hand since φ is an isomorphism, φ([x2 ]) = φ([−2]) = [−2].
Thus [a2 − b2 + 2abx] = [−2] which implies that a2 − b2 = −2 and 2ab = 0 by uniqueness of representatives with
degree
less than 2. However, 2ab = 0 ⇒ a = 0 or b = 0. The former case implies that b2 = 2 which contradicts
√
2 being irrational while the latter case implies that a2 = −2 which contradicts a2 ≥ 0 for a ∈ Q. Therefore no
such isomorphism can exist.
√
√
2
Problem 4.3.14.
√ Show that the polynomial
√ x − 3 has a root in Q( 3) but not in Q( 2). Explain why this
implies that Q( 3) is not isomorphic to Q( 2).
√
√
√
for a contradiction that x2 − 3 has a root a + b 2 in Q( 2).
Proof. Clearly x2 − 3 has a root in Q( √3). Suppose
√
This would imply that a2 + 2b2 + 2ab 2 = (a + b 2)2 = 3 which implies a2 p
+ 2b2 = 3 and 2ab = 0. Therefore
2
a = 0 or b = 0. The former case implies that√b = 3/2 which contradicts that 3/2 is irrational.√The latter case
implies that a2 = 3 which contradicts that 3 is irrational. Therefore x2 − 3 has no root in Q( 2).
√
√
√ 2
√ 2
Note that if there
√isomorphism φ : Q( 3) → Q( 2) then2 φ(3) = 3 and φ( 3) − 3 = φ( 3 − 3) =
√ were an
φ(0) = 0. Hence φ( 3) ∈ Q( 2) would be a root of the polynomial x − 3 which contradicts the previous result.
Thus the two fields are not isomorphic.
2