Name:
Math 417 Midterm Exam
Answer each question in the space provided. To receive full credit, you must clearly present
your solution and any necessary computations. Scratch work will not be counted towards
your score.
1. Find a generator for the ideal (11+2i, 6-3i) in Z[i].
Z[i] is a Euclidean Domain, so we can compute a greatest common divisor using the
Euclidean Algorithm. The ideal will be generated by any greatest common divisor of
the two ring elements.
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First, if we compute 11+2i
6−3i in Q(i) we get 3 + i. Rounding the real and imaginary
parts, we get 1 + i ∈ Z[i]. Then we can write
11 + 2i = (1 + i)(6 − 3i) + (2 − i).
Then, if we repeat the process, we see
6−3i
2−i
= 3 ∈ Z[i]. So
6 − 3i = 3(2 − i).
The algorithm ends, because the remainder is zero. The last nonzero remainder was
2 − i. Hence (11 + 2i, 6 − 3i) = (2 − i).
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2. Give an example of each of the following. No proof or explanation is required.
(a) A field extenstion which is finitely generated but not algebraic.
A possible example is Q(π).
(b) A nonprincipal ideal in a UFD.
One example is (2, x) in Z[x].
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3. Determine whether the following polynomials are irreducible in Z[x]. Briefly explain
your answer.
(a) x4 − 12x3 + 14x2 − 21x − 3.
If we reduce the polynomial in the quotient (Z/2Z)[x], we have x4 + x + 1.
This has no roots in the field Z[x], so it has no linear factors. Hence, the
only possible factors are irreducible quadratic polynomials. The only irreducible
quadratic polynomial in (Z/2Z)[x] is x2 + x + 1, whose square is not x4 + x + 1.
Hence x4 + x + 1 is irreducible in the quotient. This proves it is irreducible in
Z[x].
(b) x4 + 14x2 + 24.
This polynomial factors as x4 + 14x2 + 24 = (x2 + 12)(x2 + 2) in Z[x] so it is
reducible.
(c) x4 − 12x3 + 15x2 − 21x − 3.
By Eisenstein’s Criterion with = 3, this polynomial is irreducible in Z[x].
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4. Compute the minimal polynomial of
√
3+
√
−2 over Q.
Solution: We can determine the minimal polynomial by taking powers of
until we find a nonzero linear combination that gives us 0.
√
√
√
√
( 3 + −2)1 = 3 + −2.
√
√
√
( 3 + −2)2 = 2 −6 + 1.
√
√
√
√
( 3 + −2)3 = −3 3 + 7 −2.
√
√
√
( 3 + −2)4 = 4 −6 − 23
√
√
. Then 3 + −2 is a root of x4 − 2x2 + 25.
√
3+
√
−2
This is the minimal polynomial if it is irreducible. Since the polynomial is positive
for all x, it has no roots in Q. If it factors, it must factor as a product of irreducible
quadratics. Write
(x2 + ax + b)(x2 + cx + d) = x4 − 2x2 + 25.
Then
bd = 25,
ad + bc = 0,
ac + b + d = −2,
a + c = 0.
If a = c = 0, then b + d = −2 and bd = 25, which has no solutions. So a = −c and
b = d. But then there are no rational solutions to ac + b + d = −2. So x4 − 2x2 + 25
is irreducible.
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5. Prove that Z[2i] is not a Euclidean Domain.
Consider the ideal (2, 2i) in R = Z[2i]. All Euclidean Domains are Principal Ideal
Domains, hence if R is a Euclidean Domain, we can write (2, 2i) = (a+2bi) for some
a, b ∈ Z. Then there must exist α, β ∈ R so that
2 = (a + 2bi)α,
2i = (a + 2bi)β.
There is a multiplicative norm on R given by N (a + 2bi) = a2 + 4b2 . The above
equations show that N (a + 2bi) divides 4. Hence a2 + 4b2 = 1, 2, or 4. There are no
integer solutions to a2 + 4b2 = 2.
If a2 + 4b2 = 4, then a + 2bi = 2 or 2i. But since 2 6∈ (2i) and 2i 6∈ (2), a2 + 4b2 6= 4.
If a2 + 4b2 = 1, then it is a unit, and (2, 2i) = R. However, 1 6∈ (2, 2i). If it were,
there would exist x, y ∈ R so that 1 = 2x + 2iy. We can see all coefficients on the left
are divisible by 2, but 1 is not, so no such x, y exist. Hence (2, 2i) is not principal
and R is not a Euclidean Domain.
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6. Suppose a and b are algebraic over a field F , the degree of a over F is m, and the
degree of b over F is n. Show that if m and n are relatively prime, then [F (a, b) :
F ] = mn.
Solution: Suppose [F (a) : F ] = m, [F (b) : F ] = n, and (m, n) = 1. First, we know
[F (a, b) : F ] = [F (a, b) : F (b)][F (b) : F ], and so [F (a, b) : F ] is divisible by n. The
minimal polynomial of a over F must be divisible by the minimal polynomial of a
over F (b). Hence [F (a, b) : F (a)] ≤ mn. On the other hand, [F (a, b) : F ] = [F (a, b) :
F (a)][F (a) : F ] is divisible by m. Since (m, n) = 1. It must be divisible by the least
common multiple of m and n which is mn. Hence [F (a, b) : F ] = mn.
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