Problem 1. Let R be a unital ring.
a) Suppose that I, J, K are ideals of R such that both I, K and J, K are coprime.
Prove that IJ and K are coprime. Conclude that I ∩ J and K are coprime.
b) Suppose that I1 , I2 , ..., Im are ideals of R which are pairwise coprime (i.e. Ii +Ij =
R for i 6= j). Prove that the map
f : R/(I1 ∩I2 ∩...∩Im ) −→ (R/I1 )⊕(R/I2 )⊕...⊕(R/Im ), f (r+(I1 ∩I2 ∩...∩Im )) = (r+I1 , ..., r+Im )
is an isomorphism (use the second problem from problem set 16 and induction).
This result is often called Chinease Remainder Theorem for rings (can you see the
Chinese remainder theorem as a special case of this result?)
c) Suppose that R is commutative (and unital). Let I1 , I2 , ..., Im be pairwise coprime
ideals of R. Prove that I1 ∩ I2 ∩ ... ∩ Im = I1 I2 ...Im .
Solution: a) Since R is unital and I + K = R we can write 1 = i + k for some
i ∈ I and k ∈ K. Similarly, since J + K = R we have 1 = j + k1 for some j ∈ J
and k1 ∈ K. Thus
1 = 1 · 1 = (i + k)(j + k1 ) = ij + ik1 + kj + kk1 ∈ IJ + K
since ij ∈ IJ and ik1 , kj, kk1 are all in K. Since the ideal IJ + K contains 1, it
coincides with R, i.e. IJ + K = R. In other words, IJ and K are coprime. Recall
now that IJ ⊆ I ∩ J, so R = IJ + K ⊆ (I ∩ J) + K. It follows that (I ∩ J) + K = R,
i.e. I ∩ J and K are coprime.
b) We first note that if each of the ideals I1 , I2 , ..., Im is coprime to an ideal K then
I1 ∩ I2 ∩ ... ∩ Im and K are coprime. This follows from a) by induction. In fact,
a) is exactly the case of m = 2. Assuming that the claim holds for some m ≥ 2
consider m + 1 ideals I1 , I2 , ..., Im , Im+1 , each of which is coprime to K. By the
inductive assumption, I1 ∩ I2 ∩ ... ∩ Im and K are coprime. Also Im+1 and K are
coprime. Applying now a) to I = I1 ∩ I2 ∩ ... ∩ Im and J = Im+1 we see that
I ∩ J = I1 ∩ I2 ∩ ... ∩ Im ∩ Im+1 and K are coprime.
Now we are ready to prove the claim b). Note that for m = 2 this was proved in
part c) of Problem 2 of problem set 16. Suppose now that the result is true for some
1
m ≥ 2. Cosider m + 1 pairwise coprime ideals I1 , I2 , ..., Im , Im+1 of R. By inductive
assuption, the map
f : R/(I1 ∩ I2 ∩ ... ∩ Im ) −→ (R/I1 ) ⊕ (R/I2 ) ⊕ ... ⊕ (R/Im )
given by
f (r + (I1 ∩ I2 ∩ ... ∩ Im )) = (r + I1 , ..., r + Im )
is an isomorphism. It follows that the map
g : R/(I1 ∩ I2 ∩ ... ∩ Im ) ⊕ (R/Im+1 ) −→ (R/I1 ) ⊕ (R/I2 ) ⊕ ... ⊕ (R/Im ) ⊕ (R/Im+1 )
given by
g(r + (I1 ∩ I2 ∩ ... ∩ Im ), t + Im+1 ) = (r + I1 , ..., r + Im , t + Im+1 )
is also an isomorphism. Since I1 ∩ I2 ∩ ... ∩ Im and Im+1 are coprime (this is what
we proved at the beginning of our solution to b)) we see (by the case m = 2) that
the map
h : R/(I1 ∩ I2 ∩ ... ∩ Im ) ∩ Im+1 −→ R/(I1 ∩ I2 ∩ ... ∩ Im ) ⊕ (R/Im+1 )
given by
h(r + (I1 ∩ I2 ∩ ... ∩ Im ∩ Im+1 )) = (r + (I1 ∩ I2 ∩ ... ∩ Im ), r + Im+1 )
is an isomorphism. Thus the composition
hg : R/(I1 ∩ I2 ∩ ... ∩ Im ∩ Im+1 ) −→ (R/I1 ) ⊕ (R/I2 ) ⊕ ... ⊕ (R/Im ) ⊕ (R/Im+1 )
is also an isomorphism and it is given by
hg(r + (I1 ∩ I2 ∩ ... ∩ Im ∩ Im+1 )) = (r + I1 , ..., r + Im , r + Im+1 ).
This proves the result for m + 1, so it is true for all m by induction.
c) We use induction on m. For m = 2 this result was established in part d) of
Problem 2 of problem set 16. Suppos the result holds for some m ≥ 2 and consider
m + 1 pairwise coprime ideals I1 , I2 , ..., Im , Im+1 of R. By the inductive assuption,
2
I1 ∩ I2 ∩ ... ∩ Im = I1 I2 ...Im . As we observed in our solutuion to b), the ideals
I1 ∩ I2 ∩ ... ∩ Im and Im+1 are coprime. Thus
(I1 ∩ I2 ∩ ... ∩ Im ) ∩ Im+1 = (I1 ∩ I2 ∩ ... ∩ Im )Im+1 .
Threfore
I1 I2 ...Im Im+1 = (I1 I2 ...Im )Im+1 = (I1 ∩ I2 ∩ ... ∩ Im )Im+1 = (I1 ∩ I2 ∩ ... ∩ Im ) ∩ Im+1
which proves the result for m + 1. By induction, the result holds for all m.
Solution to problem 2.16. a) Let Z[i] = {a + bi : a, b ∈ Z} (this is a standard
notation for the ring of Gaussian integers). This is clearly a subset of the field of
complex numbers and we need to prove that it is a ring. Let a + bi, c + di ∈ Z[i].
Then
(a + bi) − (c + di) = (a − c) + (b − d)i ∈ Z[i]
and
(a + bi) · (c + di) = (ac − bd) + (ad + bc)i ∈ Z[i]
(since a − c, b − d, (ac − bd), (ad + bc) are integers). Thus Z[i] is a subring of C.
√
√
Remark: In a similar way one can prove that the set Z[ m] = {a+b m : a, b ∈ Z}
is a subring of C for any integer m (for positive m it is even a subring of R).
√
√
√
b) Let ω = (−1 + −3)/2 (here −3 = 3i). Note that
ω2 + ω + 1 = 0
Define Z[ω] = {a + bω : a, b ∈ Z}. We claim that Z[ω] is exactly the set of Eisenstain
b
b√
integers. In fact, a + bω = (a − ) +
−3. If b is even then both (a − 2b ) and 2b
2
2
are integers, and if b is odd then both (a − 2b ) − 12 and 2b − 12 are integers. In both
cases a + bω is an Eisenstain integer as defined in the problem. Thus every element
√
of Z[ω] is an Eisenstain integer. Conversely, note that x + y −3 = (x + y) + 2yω.
If either x, y are integrs or x − 21 , y − 21 are inetegrs then both (x + y) and 2y are
√
integers and therefore x + y −3 ∈ Z[ω]. This proves that every Eisenstain integer
belongs to Z[ω]. We see then that the set of Eisenstain integers is equal to Z[ω]. We
need to prove that this set is a subring of C. Let a + bω, c + dω ∈ Z[ω]. Then
(a + bω) − (c + dω) = (a − c) + (b − d)ω ∈ Z[ω]
3
and
(a+bω)·(c+dω) = ac+(ad+bc)ω+bdω 2 = ac+(ad+bc)ω+bd(−1−ω) = (ac−bd)+(ad+bc−bd)ω ∈ Z[ω]
(since a − c, b − d, (ac − bd), (ad + bc − bd) are integers and ω 2 = −1 − ω). Thus Z[ω]
is a subring of C.
Remark. In a similar way one can show that if m is an integer such that m ≡
√
1 (mod 4) and w = (−1 + m)/2 then the set Z[w] = {a + bw : a, b ∈ Z} is a
subring of C. In fact this is a special case of the following general result.
Suppose that u is a complex number such that
uk + a1 uk−1 + a2 uk−2 + ... + ak−1 u + ak = 0
for some k > 0 and some integers a1 , ..., ak . Then the set
Z[u] = {c0 + c1 u + c2 u2 + ... + ck−1 uk−1 : c0 , c1 , ..., ck−1 ∈ Z}
is a subring of C.
4