Download Midterm Exam 1. Prove that for any positive integer n, ⋂ Ri = { 0 } for

Midterm Exam
1. Prove that for any positive integer n,
Ri of Q, i = 1, 2, . . . , n.
Tn
i=1 Ri
6= { 0 } for every choice of nonzero subrings
Solution: Suppose that R is a nonzero subring of Q. Then there exist positive integers r, s with r/s ∈ R,
and so r = s(r/s) ∈ R since R is closed under addition. But then the additive subgroup rZ of Z (which is
an ideal of Z) is contained in R. This proves that every nonzero subring of Q contains a nonzero ideal of
Z. For any nonzero subrings R1 , R2 , . . . , Rn of Q, let m1 , m2 , . . . , mn ∈ Z be such that (mi ) is a nonzero
ideal of Z contained in Ri for each i = 1, 2, . . . , n. Then (0) 6= (m1 m2 · · · mn ) = (m1 )(m2 ) · · · (mn ) ⊆
(m1 ) ∩ (m2 ) ∩ · · · ∩ (mn ) ⊆ R1 ∩ R2 ∩ · · · ∩ Rn .
2. Let f : R → S be a ring homomorphism, and let I be an ideal of R. Prove that f −1 ◦f (I) = I
if and only if ker(f ) ⊆ I.
Solution: First, suppose that f −1 ◦f (I) = I. Since f (0) = 0, we have 0 ∈ f (I) and thus ker(f ) = f −1 (0) ⊆
f −1 ◦f (I) = I.
Conversely, suppose that ker(f ) ⊆ I. Since I ⊆ f −1 ◦f (I), it suffices to prove that f −1 ◦f (I) ⊆ I. Let
x ∈ f −1 ◦f (I), so there exists i ∈ I with f (x) = f (i). Thus j = x − i ∈ ker(f ) ⊆ I, and so x = j + i ∈ I,
as required.
3. Let R be a commutative ring with identity 1, and for any positive integer n, let Mn (R)
denote the ring of n × n matrices with entries from R, with the usual definition of addition
and multiplication. For any A ∈ Mn (R), and any i, j with 1 ≤ i, j ≤ n, let A(i, j) denote
the entry in row i, column j of A. For each i and j with 1 ≤ i, j ≤ n, let Ei,j denote the
n × n matrix with all entries equal to 0 except for the entry in row i, column j, which is
equal to 1 (that is, Ei,j (r, s) = 0 for any (r, s) with (r, s) 6= (i, j), while Ei,j (i, j) = 1).
(a) Prove that if I is an ideal of R, then Mn (I), the subset of Mn (R) consisting of all n × n
matrices with entries from I, is an ideal of Mn (R).
Solution: Since 0 ∈ I, then the n × n matrix with all entries equal to 0 is an element of Mn (I); that
is, the zero of Mn (R) belongs to Mn (I). Suppose now that A, B ∈ Mn (I). Then for each i, j with
1 ≤ i, j ≤ n, (A − B)(i, j) = A(i, j) − B(i, j) ∈ I since A(i, j), B(i, j) ∈ I. This establishes that
A − B ∈ Mn (I). Finally, suppose that A ∈ Mn (I) and B ∈ Mn (R). Let i, j be such that 1 ≤ i, j ≤ n.
Then for each i with
Pn1 ≤ k ≤ n, A(ik ), A(k, j) ∈ I and so A(i,
Pn k)B(k, j) ∈ I, B(i, k)A(k, j) ∈ I.
Thus (AB)(i, j) = k=1 A(i, k)B(k, j) ∈ I, and (BA)(i, j) = k=1 B(i, k)A(k, j) ∈ I, so AB, BA ∈
Mn (I).
(b) Let J be an ideal of Mn (R), and let
JR = { a ∈ R | there exists A ∈ J and there exist i, j with a = A(i, j) }.
Prove that JR is an ideal of R and that J = Mn (JR ).
Solution: Since the zero matrix belongs to J, 0 ∈ JR . Suppose that a, b ∈ Jr . Then there exist
A, B ∈ J with a = A(i, j) and (since B ∈ J implies that −B ∈ J) −b = −B(r, s) for some indices
i, j, r, s. But then E1,i A ∈ J, and E1,i A has as its first row the ith row of A, while all subsequent rows
are 0, and thus E1,i AEj,1 ∈ J is the n × n matrix whose first row, first column entry is A(i, j) = a,
while all other entries are 0. Similarly, E1,r BEs,1 has −b = B(r, s) as its first row, first column
entry, while all other entries are 0. But then a − b is the entry in the first row, first column of
E1,i AEj,1 + E1,r BEs,1 , and so a − b ∈ JR . Finally, let c ∈ R, and let C be the n × n matrix with first
row, first column entry c, and all other entries 0, so C ∈ Mn (R). Then ac is the entry in the first
row, first column of E1,i AEj,1 C ∈ J, so ac ∈ JR . Since R is commutative, ca ∈ JR as well, and so JR
is an ideal of R.
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November 11, 2011
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It remains to prove that J = Mn (JR ). Evidently, J ⊆ Mn (JR ), so it suffices to prove that
Mn (JR ) ⊆ J. Let A ∈ Mn (JR ). For each i, j with 1 ≤ i, j ≤ n, there exists Ai,j ∈ J and indices ri , si
with A(i, j) = Ai,j (ri , si ). ButP
then Ei,ri Ai,j Esi ,j ∈ J and has (i, j) entry equal to A(i, j), while all
other entries are 0. Thus A = i,j Ei,ri Ai,j Esi ,j ∈ J, as required.
4. Determine whether or not Z/(4)[x] is a principal ideal ring (of course, Z/(4)[x] is not an
integral domain since Z/(4) is not an integral domain). Hint: consider the ideal generated
by 2 and x. Be careful; remember that for example, 2x + 1 is a unit in Z/(4)[x], since
(2x + 1)2 = 4x2 + 4x + 1 = 1.
Solution: Suppose that the ideal generated by 2 and x is principal, say generated by p. Then p divides 2
and p divides x, so 2 = pf and x = pg for some f, g ∈ Z/(4)[x]. Note that the product of the constant
terms of p and f is 2, so neither p nor f has 0 constant term. But pg has 0 constant term, so either g
has 0 constant term, or else both p and g have constant term 2. Suppose first that g has constant term
′
0, which means that g = xg ′ for some gP
∈ Z/(4)[x].
then x = xpg ′ , or x(1 − pg ′ ) = 0, and x is not a
P But
i
i+1
zero divisor in Z/(4)[x] (that is, 0 = x ai x = ai x
means that ai = 0 for every i), so pg ′ = 1. But
then p is a unit of Z/(4)[x], and so (p) = Z/(4)[x]. Consequently, 1 = 2u + xv for some u, v ∈ Z/(4)[x].
But then 2 = 2xv, which is not possible since 2xv has 0 constant term.
Now suppose that both p and g have constant term 2, so p = 2 + xq and g = 2 + xh for some
q, h ∈ Z/(4)[x]. Then x = pg = 2x(q + h) + x2 qh, and so 2(q(0) + h(0)) = 1, which is not possible since
2 is not a unit of Z/(p)[x]. Thus the assumption that (2, x) is a principal ideal in Z/(4)[x] has led to a
contradiction, and so (2, x) is not a principal ideal in Z/(4)[x].
5. Prove that if k is a field, then k[x] has infinitely many irreducible elements.
Q
Solution: Suppose not, and let p1 , p2 , . . . , pn denote the irreducible elements of k[x]. Let f = 1 + ni=1 pi .
Since k is a field, k[x] is a Euclidean domain and thus f has a unique representation as a product of
irreducible elements of k[x] (since f is not zero nor a unit of k[x]). But then for some i, pi is a divisor
of f , and so pi divides 1. But then pi is a unit, which is not the case. This contradiction stems from
the assumption that k[x] had only finitely many irreducible elements, and so we conclude that k[x] has
infinitely many irreducible elements.
6. Prove that if R is a commutative simple ring with R2 6= { 0 }, then R is a field.
Solution: Let a, b ∈ R with ab 6= 0 and consider the ideal aR. Since aR 6= { 0 }, aR = R. Let u ∈ R be
such that au = a. Let x ∈ R. Then there exists y ∈ R with x = ay, and so ux = xu = ayu = yau = ya =
ay = x. It follows that u is the identity for R. But now for any nonzero x ∈ R, x = xu 6= 0, and xu ∈ xR,
so xR is a nonzero ideal of R, hence xR = R. Thus there exists y ∈ R with xy = u = yx, and so every
nonzero element of R is a unit.
7. Let R be a commutative ring with identity, and let S be a multiplicatively closed subset
of R that contains 1 but does not contain 0 or any zero divisiors. Let S −1 R denote the
ring of fractions; that is, S −1 R = { r/s | r ∈ R, s ∈ S }. You have already proven in
Assignment 3 that for any ideal I of S −1 R, I ∩ R = { r ∈ R | r/1 ∈ I } is an ideal of R
and I is the ideal of S −1 R that is generated by I ∩ R. These facts may be used in this
problem if you wish.
(a) Let p be a prime ideal of R for which p∩S = ∅. Prove that S −1 p = { r/s | r ∈ p, s ∈ S }
is a proper prime ideal of S −1 R (proper means not the whole ring).
Solution: Since 0 ∈ p, and 0 = 0/s for any s ∈ S, we have 0 ∈ S −1 p. Let r1 , r2 ∈ p and s1 , s2 ∈ S.
Then r1 /s1 − r2 /s2 = (r1 s2 − r2 s1 )/s1 s2 and r1 , r2 ∈ p means that r1 s2 − r2 s1 ∈ p, while s1 s2 ∈ S.
Thus r1 /s1 − r2 /s2 ∈ S −1 p. As well, for r ∈ R, s ∈ S, we have (r1 /s1 )(r/s) = (rr1 )/(ss1 ) and
rr1 ∈ p, ss1 ∈ S, so (r1 /s1 )(r/s) ∈ S −1 p. Thus S −1 p is an ideal of S −1 R. Suppose now that
r1 , r2 ∈ R and s1 , s2 ∈ S are such that (r1 /s1 )(r2 /s2 ) ∈ S −1 p. Then (r1 r2 )/(s1 s2 ) ∈ S −1 p, which
means that (r1 r2 )/(s1 s2 ) = r/s for some r ∈ p, s ∈ S. But then sr1 r2 = rs1 s2 ∈ p, and since
Midterm Exam
November 11, 2011
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p ∩ S = ∅, s ∈
/ p. Thus r1 r2 ∈ p, and so either r1 ∈ p, in which case r1 /s1 ∈ S −1 p, or else r2 ∈ p, in
which case r2 /s2 ∈ S 1 p. Thus S −1 p is a prime ideal of S −1 R. If S −1 p is not proper, then 1 ∈ S −1 p,
so 1 = r/s for some r ∈ p, s ∈ S. But then s = r ∈ p ∩ S = ∅, not possible. Thus S −1 p is a proper
ideal of S −1 R.
(b) Prove that if I is a prime ideal of S −1 R, then I ∩ R is a prime ideal of R.
Solution: Suppose that I is a prime ideal of S −1 R, and let x, y ∈ R be such that xy/1 ∈ I ∩ R. Then
(x/1)(y/1) ∈ I, and I is prime, so either x/1 ∈ I or y/1 ∈ I. Thus either x ∈ I ∩ R or else y ∈ I ∩ R,
which proves that I ∩ R is prime.
(c) Prove that p 7→ S −1 p is a bijective mapping from the set of all prime ideals of R with
empty intersection with S onto the set of proper prime ideals of S −1 R with inverse
given by I 7→ I ∩ R.
Solution: We have proven above that if p is a prime ideal of R with p ∩ S = ∅, then S −1 p is a proper
prime ideal of S −1 R, and we have proven above that if I is a prime ideal of S −1 R, then I ∩ R is an
ideal of R. Thus it suffices to prove that if I is a proper prime ideal of S −1 R, then I ∩ R ∩ S = ∅
and S −1 (I ∩ R) = I; and that if p is a prime ideal of R with p ∩ S = ∅, then p = S −1 p ∩ R.
Suppose then that I is a proper prime ideal of S −1 R, and let x ∈ I ∩ R ∩ S = I ∩ S. Then
1 = (x)(1/x) ∈ I, which means that I = S −1 R, which is not the case. Thus I ∩ R ∩ S = ∅. Now to
prove that S −1 (I ∩ R) = I. Let r/s ∈ S −1 (I ∩ R), so r ∈ I ∩ R and s ∈ S. Then r/s = (r)(1/2) ∈ I,
so S −1 (I ∩ R) ⊆ I. Next, let a ∈ I, say a = r/s for some r ∈ R, s ∈ S. Then r = as ∈ I, so r ∈ I ∩ R.
Thus a = r/s ∈ S −1 (I ∩ R), and so I ⊆ S −1 (I ∩ R).
Now let p be a prime ideal of R with p ∩ S = ∅. We are to prove that S −1 p ∩ R = p. Since
p ⊆ S −1 p, it suffices to prove that S −1 p ∩ R ⊆ p. Let r/s ∈ S −1 p ∩ R. Then r/s = t/1 for some
t ∈ R. We have ts = r = (r/s)s ∈ p, and so either t ∈ p or else s ∈ p. Since p ∩ S = ∅, we have t ∈ p
and so r/s = t ∈ p.
8. Let R be a UFD. Prove that a nonzero prime ideal I of R is minimal in the set of all
nonzero prime ideals of R if and only if I is principal.
Solution: Suppose first of all that I is principal, say I = (a). Let J be a prime ideal of R with J ⊆ I.
Suppose that a ∈
/ J. Let j ∈ J, so j = ar for some r ∈ R. Since a ∈
/ J, we have r ∈ J since J is prime.
Thus j ∈ aJ, and so J ⊆ aJ ⊆ J, which means that J = aJ. But then J = an J for any n ≥ 1, and so for
any j ∈ J, an divides j for each n ≥ 1. This is not possible, as R is a UFD, so the assumption that a ∈
/J
results in a contradiction. Thus a ∈ J and so J = I.
Conversely, suppose that I is minimal in the set of all nonzero prime ideals of R. Let a ∈ I, a 6= 0. If a
is a unit, then I = R, which means that the only prime ideals of R are { 0 } and R itself. But then R has
no irreducible elements, so every nonzero element is a unit and R is a field.QOtherwise, a is not a unit.
Let the decomposition of a as a product of irreducible elements of R be a = i pi . Then since I is prime,
pi ∈ I for some i, and so (pi ) ⊆ I. Since pi is prime in R, (pi ) is a nonzero prime ideal of R, and so by
the minimality of I, we have I = (pi ).