Download Math. 5363, exam 2, solutions 1. Give an example of a ring which is

Math. 5363, exam 2, solutions
1. Give an example of a ring which is an integral domain but not a unique factorization domain.
√
One such example is the ring Z[ −5], as explained in the text on page 385.
2. Give an example of a ring which is a unique factorization domain (UFD) but
not a principal ideal domain (PID).
Since R[x] is a UFD, R[x, y] = R[x][y] is a UFD. As we have seen in the solution to
problem 2 of homework 8, the ideal I ⊆ R[x, y] of all the polynomials vanishing at
the origin is not principal. Hence, R[x, y] is not a PID.
3. Give an example of a ring R and a prime ideal I ⊆ R which is not a maximal
ideal.
Let R = R[x, y] and let I ⊆ R be the ideal of all the polynomials vanishing on the
y-axis. Thus I = (x) = xR[x, y]. This is a prime ideal because if f (x, y)g(x, y) is
divisible by x then f (x, y) is divisible by x or g(x, y) is divisible by x. On the other
hand, I is not a maximal ideal because it is properly contained in the ideal of all
the polynomials vanishing at the origin.
4. Let φ : R → R0 be a homomorphism of commutative rings with identity such
that φ(1) = 1. Prove that if P 0 ⊆ R0 is a prime ideal then P = φ−1 (P 0 ) ⊆ R is a
prime ideal.
Suppose a, b ∈ R are such that ab ∈ P . Then φ(a)φ(b) = φ(ab) ∈ P 0 . Since P 0 is
prime, we have φ(a) ∈ P 0 or φ(b) ∈ P 0 . In the first case a ∈ φ−1 (P 0 ) = P and in
the second case b ∈ P .
5. Determine the maximal ideals of the following rings:
(a) R[x]/(x2 ),
1
2
Let φ : R[x] → R[x]/(x2 ) be the quotient homomorphism, which maps g(x) to
g(x) + (x2 ). Let I ⊆ R[x]/(x2 ) be a non-zero proper ideal (different than zero or
the whole ring). Since φ is surjective, φ−1 (I) ⊆ R[x] is a proper ideal. Since R[x] is
a PID, there is a polynomial f (x) ∈ R[x] such that φ−1 (I) = (f (x)). Since (x2 ) is
the kernel of φ, we have (x2 ) ⊆ (f (x)). Thus f (x) divides x2 . Therefore, up to constant multiples, f (x) = 1 or f (x) = x or f (x) = x2 . In the first case φ−1 (I) = R[x]
- a contradiction. In the last case I = 0 - a contradiction. Therefore (f (x)) = (x).
Thus I is the ideal generated by x + (x2 ). The result is that our ring has only one
proper ideal. In particular only one maximal ideal, generated by x + (x2 ).
(b) R[x]/(x2 − 3x + 2),
Notice that (x2 − 3x + 2) = (x − 1)(x − 2). As before we have the quotient homomorphism
φ : R[x] 3 g(x) → g(x) + ((x − 1)(x − 2)) ∈ R[x]/((x − 1)(x − 2)).
Let I ⊆ R[x]/((x − 1)(x − 2)) be a proper non-zero ideal. Then φ−1 (I) = (f (x)) ⊆
Rx is a proper ideal. Then f (x) divides (x − 1)(x − 2). We exclude the cases
f (x) = 1 and f (x) = (x − 1)(x − 2) as before. Hence f (x) = x − 1 or f (x) = x − 2.
Thus I is generated either by x − 1 + ((x − 1)(x − 2)) or by x − 2 + ((x − 1)(x − 2)).
There is no inclusion between these two ideals. Hence both are maximal. The zero
ideal is not maximal because it is contained in any of the two.
(c) R[x]/(x2 + x + 1).
The polynomial x2 + x + 1 is prime in R[x]. Therefore the ring R[x]/(x2 + x + 1)
is a field. In particular the only maximal ideal is the zero ideal.
6. Let a + bi be prime in Z[i]. Prove that
(a) a − bi is a prime.
If a − bi = (c + di)(e + f i) is a non-trivial decomposition in Z[i], then a + bi =
3
(c − di)(e − f i) is a non-trivial decomposition. Hence, a + bi is not prime - a
contradiction.
(b) N (a + bi) is a power of some positive prime in Z.
Suppose, there are two positive integers m and n such N (a + bi) = mn. Then
(a + bi)(a − bi) = mn
in the ring Z[i]. Since a + bi is prime, it has to divide m or n. Suppose it divides
m. Then
m = (a + bi)(c + di)
in the ring Z[i]. Therefore
m2 = (a2 + b2 )(c2 + d2 ) = mn(c2 + d2 ).
Thus
m = n(c2 + d2 ).
In particular every prime integer that divides n has to divide m. This shows that
N (a + bi) is a power of a single prime integer.
(c) N (a + bi) equals p or p2 , where p is as in (b).
We know from (b) that there is a prime p ∈ Z and a positive integer k such that
N (a + bi) = pk . Let us consider p as an element of Z[i]. Suppose p is the product
of m primes in Z[i]. Then pk is the product of km primes in Z[i]. But according
to (a), N (a + bi) = (a + bi)(a − bi) is the product of two primes in Z[i]. Since
N (a + bi) = pk , we see that km = 1 or km = 2. Therefore either k = 1 or k = 2.
(d) N (a + bi) = p2 in (c) forces a + bi = p, apart from a unit factor.
The assumption is that
(a + bi)(a − bi) = p2 .
4
Since a + bi is a prime in Z[i], it divides p. Thus p = (a + bi)(c + di). Hence,
p = p = (a − bi)(c − di). Therefore,
p2 = (a + bi)(c + di)(a − bi)(c − di) = N (a + bi)(c2 + b2 ),
which shows that c + di = ±1.
7. Prove that no prime a + bi in Z[i] has N (a + bi) = p, with p of the form 4n + 3.
Conclude that every positive prime in Z of the form 4n + 3 is a prime in Z[i].
Suppose N (a + bi) = p, with p a prime integer of the form 4n + 3. Then
(7.1)
a2 + b2 = p = 4n + 3.
Let us map both sides of the equation (7.1) to the finite ring Z/4Z = {0, 1, 2, 3} by
the quotient map (i.e. modulo 4). The result is
2
a2 + b = 3.
(7.2)
Since in Z/4Z,
12 = 1, 22 = 0, 32 = 1,
the equation (7.2) has no solutions. Therefore (7.1) has no solutions. Thus N (a +
bi) = p is impossible. Therefore, by Problem 6(c), N (a + bi) = p2 and Problem
6(d) shows that p is prime in Z[i].
8. Suppose R is an integral domain and F ⊆ R is a subring that is a field, so that
R may be considered as a vector space over F . Prove that if the dimension of R
over F is finite then R is a field.
Let a ∈ R be a non-zero element. Then the map
(8.1)
R 3 r → ra ∈ R
is an injective ring homomorphism. In particular (8.1) is an injective linear map.
Since the dimension is finite, this map is surjective. Hence 1 is in the range of (8.1).
In other words there is b ∈ R such that
ab = 1.
5
This shows that a is invertible. Thus R is a field.
9. What is a necessary and sufficient condition on an integer N > 0 for the positive
square root of N to be in Q[21/3 ] ⊆ R?
The polynomial x3 − 2 ∈ Q[x] is prime. Hence, Q[21/3 ] is a field extension of Q of
dimension 3 ([Q[21/3 ] : Q] = 3). Suppose N is not the square of an integer. Then
the polynomial x2 − N ∈ Q[x] is prime. Hence, Q[N 1/2 ] is a field extension of Q of
√
dimension 2 ([Q[21/2 ] : Q] = 2). If N ∈ Q[21/3 ] then Q[N 1/2 ] ⊆ Q[21/3 ]. In this
case
3 = [Q[21/3 ] : Q] = [Q[21/3 ] : Q[N 1/2 ]][Q[N 1/2 ] : Q] = [Q[21/3 ] : Q[N 1/2 ]]2,
which is impossible. Thus
√
N∈
/ Q[21/3 ]. We see that
√
N ∈ Q[21/3 ] if and only if
N is the square of an integer.
10. Suppose K is a finite field extension of k of the form k(r) with [k : K] odd.
Prove that K = k(r2 ).
Let f (x) = xn + cn−1 xn−1 + ... + c1 x + c0 ∈ k[x] be the minimal polynomial of r.
Then n = [k : K] is odd and
rn + cn−1 rn−1 + ... + c1 r + c0 = 0.
Equivalently,
(10.1) rn +cn−2 rn−2 +cn−4 rn−4 +...+c1 r = −cn−1 rn−1 −cn−3 rn−3 +...−c2 r2 −c0 .
Notice that
rn−1 + cn−2 rn−3 + cn−5 rn−4 + ... + c1 6= 0
because the degree is smaller than n. Hence we may compute from (10.1) that
(10.2)
r = (−cn−1 rn−1 −cn−3 rn−3 +...−c2 r2 −c0 )(rn−1 +cn−2 rn−3 +cn−5 rn−4 +...+c1 )−1 .
The right hand side of (10.2) contains only even powers of r. Thus r ∈ k(r2 ).
Therefore k(r) ⊆ k(r2 ), which implies the desired equality.