Term
06F1a_J
Prove...
Given G, a finite group of order n with multiplicative identity e, if G is abelian and n is odd, then the product of all the elements of G is e. 

Definition
There is one element of order 1: the identity e.
There are no elements of order 2, since G has odd order and the order of an element must divide the order of the group.^{*}
Each element of order > 2 may be paired with its inverse. (Since G is abelian we can freely rearrange the factors.)
So the product is e.
* Thankyou Amarnath! 


Term
06F1b_J
Prove...
If n is even, then there is an a ∈ G such that a ≠ e and a^{2} = e.


Definition
Let A ⊆ G, B ⊆ G contain the order=2 and order>2 elements, respectively.
Then n = A + B + 1 (where 1 is contributed from the identity).
B is even since each element of B may be paired with its inverse.
So, to make n even, A must be odd and therefore nonzero.
That means there exists at least one a ∈ G with a ≠ e but a^{2} = e.



Term
07S1.1_J
List all nonisomorphic abelian groups of order 72 

Definition
By the FTFAG,
any finite abelian group of order n is isomorphic to Z_{m1} x ... x Z_{mk} ,
where {m_{1},...,m_{k}} is a set prime powers whose product is n.
Moreover, distinct {m_{1},...,m_{k}} yield nonisomorphic groups.
72 can be written as a product of prime powers in the following ways:
2×2×2×3×3
2×4×3×3
8×3×3
2×2×2×9
2×4×9
8×9.
This gives exactly six nonisomorphic groups:
Z_{2}Z_{2}Z_{2}Z_{3}Z_{3}
Z_{2}Z_{4}Z_{3}Z_{3}
Z_{8}Z_{3}Z_{3}
Z_{2}Z_{2}Z_{2}Z_{9}
Z_{2}Z_{4}Z_{9}
Z_{8}Z_{9.} 


Term
07S1.2_J
Determine the number of nonisomorphic abelian groups of order 2592, and explain your answer. 

Definition
2592 = 2^{5}× 3^{4}
There are p(5) = 7 ways to write 5 as a sum of positive integers:
5, 4+1, 3+2, 3+1+1, 2+2+1, 2+1+1+1, 1+1+1+1+1.
This corresponds to seven ways in which 2^{5} can be written using factors that are powers of 2:
2^{5}, 2^{4}2^{1}, 2^{3}2^{2}, 2^{2}2^{2}2^{1}, 2^{2}2^{1}2^{1}2^{1}, 2^{1} 2^{1}2^{1}2^{1}2^{1}.
There are p(4) = 5 ways to write 4 as a sum of positive integers:
4, 3+1, 2+2, 2+1+1, 1+1+1+1.
This corresponds to five ways in which 3^{4} can be written using factors that are powers of 3:
3^{4}, 3^{3}3^{1}, 3^{2}3^{2}, 3^{2}3^{1}3^{1}, 3^{1}3^{1}3^{1}3^{1}.
Combining the results above, we have (7)(5) = 35 ways in which
2592 can be written as a product of prime powers.
Each of these factorings corresponds to one of 35 nonisomorphic abelian groups of order 2592.
For example, by choosing the third partition of 5 (3+2) and the fourth partition of 4 (2+1+1) we get 2^{5} = 2^{3}2^{2} and 3^{4} = 3^{3}3^{1}3^{1} i.e.
2592 = (8)(4)(9)(3)(3), corresponding to the group Z_{8}Z_{4}Z_{9}Z_{3}Z_{3}.
(Note: p(n) is the partition function, which counts the number of ways n can be written as a sum of positive integers.) 


Term
07F1a_J
Prove...
If G is abelian then
φ: G → G  φ(x) = x^{1} is a group automorphism of G.


Definition
φ is a homomophism:
φ(xy) = (xy)^{1} =(yx)^{1} = x^{1}y^{1} = φ(x)φ(y).
φ is injective:
φ(x) = φ(y) ⇒ x^{1 }= y^{1} ⇒ yx^{1}x = yy^{1}x ⇒ ye = ex ⇒ y = x
φ is surjective:
b ∈ G ⇒ b = (b^{1})^{1} = φ(b^{1}) where b^{1} ∈ G
(Note: b ∈ G ⇒ b^{1}∈ G since is a group.)
Thus φ is a bijective homomorphism from G to itself, i.e. it's an automorphism on G.



Term
07F1b_J
Prove...
φ: S_{3} → S_{3}  φ(x) = x^{1} is not a group automorphism on S_{3}. 

Definition
φ((1 2)(1 3)) = φ((1 3 2)) = (1 2 3)
but
φ((1 2)) φ((1 3)) = (1 2)(1 3) = (1 3 2)
The homomophism property fails so φ cannot be an automorphism.



Term
08F1a_J
Prove...
Let N be a normal subgroup of a group G and let G/N the quotient group of left cosets of N in G.
If G is cyclic, then G/N is cyclic, and the converse is false. 

Definition
Let g be a generator of G.
Consider the cosets gN, g^{2}N, g^{3}N etc.
Since N is normal (in fact G is abelian, since cyclic),
we have g^{i}N = g^{i}(N^{i}) = (gN)^{i}.
This shows that the set of all cosets is generated by coset gN.
Hence G/N is cyclic if G is cyclic.
The converse is false: G/N cyclic doesn't force G to be cyclic.
For example, S_{3 }is not cyclic
but S_{3}/S_{3} and S_{3}/<(1 2 3)> are both cyclic.
(Note: <(1 2 3)> is normal in S_{3} so we may form quotient group.)



Term
08F1b_J
Prove...
G/N is abelian, where N◁ G,
if and only if
a^{1}b^{1}ab ∈ N, ∀a,b ∈ G, 

Definition
(aN)(bN) = (bN)(aN)
iff
abN =baN
iff
a^{1}b^{1}abN =a^{1}b^{1}baN
iff
a^{1}b^{1}abN = N
iff
a^{1}b^{1}ab ∈ N
Notes:
The first equivalence holds because N is normal in G, thus Nb = bN. So aNbN = abNN = abN.
The last equivalence doesn't require N to be normal.



Term
FALL 2008 PROBLEM 1 PART c
Prove:
If φ: G → G definined by a ∈ G ⇒ φ(a) = a^{2} is a homomorphism, then G is abelian. 

Definition
Applying the definition of φ and the homomorphism property:
(ab)^{2} = φ(ab) = φ(a)φ(b) = a^{2}b^{2}
Now
ab = eabe
= a^{1}aabbb^{1}
= a^{1} a^{2}b^{2} b^{1}
= a^{1} (ab)^{2} b^{1}
= a^{1}ababb^{1}
= ebae = ba.
Hence G is abelian.



Term
SPRING 2009, PROBLEM 1 PART a
Prove...
If G is abelian then ∀ a,b ∈ G and n ∈ ℤ, (ab)^{n} = a^{n}b^{n}.


Definition
First show (ab)^{n} = a^{n}b^{n} for n ≥ 0 by induction.
Basis step: (ab)^{0} = e = ee = a^{0}b^{0}.
Inductive step: (ab)^{n+1} = (ab)^{n}(ab) = a^{n}b^{n}ab = a^{n}ab^{n}b = a^{n+1}b^{n+1}.
Now, for n<0 ...
(ab)^{n}
= ((ab)^{1})^{n}
= (b^{1}a^{1})^{n}
= (b^{1})^{n}(a^{1})^{n}
= b^{n}a^{n}
= a^{n}b^{n}.
Notice we could use the first result since n > 0. 


Term
SPRING 2009, PROBLEM 1 PART b
Prove...
If G is a finite abelian group of order G and n is an integer relatively prime to G,
then the map φ : G → G defined by x ∈ G ⇒ φ(x) = x^{n}
is an automorphism.
Moreover, for each g ∈ G there exists x ∈ G such that g = x^{n} .


Definition
φ is a homomorphism on G:
φ(xy) = (xy)^{n} = x^{n}y^{n} (since G abelian)
φ is onetoone:
φ(x) = φ(y)
→ x^{n} = y^{n}
→ x^{n}y^{n} = e
→ (xy^{1})^{n} = e
→ o(xy^{1})  n
→ o(xy^{1})  gcd(G,n) = 1 ^{*}
→ o(xy^{1}) = 1
→ xy^{1} = e
→ x = y.
* Since xy^{1} ∈ G, o(xy^{1}) divides G.
φ is onto G:
Since φ is onetoone φ(G) = G.
But G is finite and φ(G) ⊆ G, so φ(G) must be G itself.
We've shown that φ is a bijective homomorphism from G to G, thus φ is an automorphism of G.
Since φ is onto G, by defintion, for each g ∈ G,
there exists x ∈ G such that g = φ(x), i.e. g = x^{n}.



Term
FALL 2009 PROBLEM 1 (→ direction of "if and only if")
Prove
If H,K are subgroups of G and HK is a subgroup of G, then HK = KH. 

Definition
Consider an arbitrary kh ∈ KH (where k ∈ K, h ∈ H).
Note that k ∈ HK since K = {e}K ⊆ HK.
Similarly, h ∈ HK since H = H{e} ⊆ HK.
Therefore kh ∈ HK, since HK is a group.
We've shown KH ⊆ HK.
Conversely, consider arbitrary hk ∈ HK.
(hk)^{1} ∈ HK, since HK is a group.
Therefore (hk)^{1} = h'k' (where h' ∈ H, k' ∈ K).
Thus hk = (h'k')^{1} = k'^{1}h'^{1} ∈ KH (since k'^{1}∈ K, h'^{1}∈ H).
We've shown HK ⊆ KH.
Since KH ⊆ HK and HK ⊆ KH we have HK = KH. 


Term
FALL 2009 PROBLEM 1 (← direction of "if and only if")
Prove
If H,K are subgroups of G and HK = KH then HK is a subgroup of G.


Definition
 HK is nonempty: e ∈ H ∧ e ∈ K → ee = e ∈ HK.
 HK is closed for the group operation:
a,b ∈ HK → a=h_{1}k_{1 }∧ b=h_{2}k_{2} → ab= h_{1}k_{1}h_{2}k_{2} = h_{1}h_{3}k_{3}k_{2} ∈ HK. (Note: k_{1}h_{1} ∈ KH and KH = HK → k_{1}h_{1} = h_{3}k_{3} , for some h_{3}∈H, k_{3}∈K.)
 HK is closed for taking inverses:
a ∈ HK → a=hk → a^{1}=k^{1}h^{1 }∈ KH=HK.
Hence HK is a subgroup of G.



Term
FALL 2010 PROBLEM 1 PART a
Prove...
The quotient group Q/Z is an infinite group
in which each element has finite order. 

Definition
Consider q_{1}, q_{2 }distinct ∈ Q∩[0,1).
Since q_{1}q_{2 }∉ Z we must have q_{1}+Z ≠ q_{2}+Z,
meaning each element of Q∩[0,1)
represents a distinct element of Q/Z.
Since Q∩[0,1) is infinite, Q/Z must also be infinite.
Now consider an arbitrary q+Z ∈ Q/Z.
q may be written as j/k where j∈Z, k∈Z^{*},
so q+Z may be expressed as {j/k+n  n∈Z}.
Let k·(q+Z) denote (q+Z) + (q+Z) + ... + (q+Z) (Using k terms.)
Elements of k·(q+Z) are obtained by adding k elements of q+Z.
We may write such an element as
(j/k + n_{1}) +(j/k + n_{2}) + ... + (j/k+n_{k})
which is just
j + (n_{1} + n_{2} + ... + n_{k}) ∈ Z
Thus k·(q+Z) ⊆ Z.
Also Z ⊆ k·(q+Z) since the n_{i} may be chosen freely
(e.g. to obtain 3, use n_{1} = j, n_{2}=3, n_{i>2} = 0).
So k·(q+Z) = Z, meaning q+Z has finite order.



Term
Fall 2010 Problem 1 PART b
Find an automorphism of Q/Z other than the identity. 

Definition
Define φ(q+Z) = q+Z.
We will show that φ is welldefined and an automorphism of Q/Z.
Suppose q_{1}+Z = q_{2}+Z. Then q_{1}q_{2} = n_{ }∈ Z and...
φ(q_{1}+Z) = q_{1}+Z = q_{1}+(n+Z) = (q_{1}+n)+Z = q_{2}+Z = φ(q_{2}+Z).
φ((q_{1}+Z) + (q_{2} +Z) )
= φ( (q_{1}+q_{2}) +Z) )
= (q_{1}+q_{2})+Z = (q_{1}+q_{2})+Z
= (q_{1}+Z) + (q_{2}+Z)
= φ(q_{1}+Z) + φ(q_{2}+Z).
φ(q_{1}+Z) = φ(q_{2}+Z)
⇒ q_{1}+Z = q_{2}+Z
⇒ q_{1}  q_{2 }= q_{2}q_{1} ∈ Z
⇒ q_{1}+Z = q_{2}+Z
For any q+Z ∈ Q/Z, q+Z = φ(q+Z) and q+Z ∈ Q/Z
We've shown that φ is well defined
and a onetoone correspondence from Q/Z to Q/Z.
Therefore φ is an automophism of Q/Z. 


Term
FALL 2010 PROBLEM 1 PART c
Find a nontrivial homomorphism
φ: Q/Z → Q/Z which is not injective.


Definition
Define φ: Q/Z → Q/Z as follows: φ(q+Z) = 2q+Z
q_{1}+Z = q_{2}+Z
⇒ q_{1}q_{2} ∈ Z
⇒ 2q_{1}  2q_{2} ∈ Z
⇒ 2q_{1} + Z = 2q_{2} + Z
⇒ φ(q_{1}+Z) = φ(q_{2}+Z)
φ(q_{1} + q_{2})
= 2(q_{1}+q_{2})+Z
= (2q_{1}+2q_{2})+Z
= (2q_{1}+Z) + (2q_{2}+Z)
= φ(q_{1}) + φ(q_{2})
φ(0.6 + Z)= 1.2 + Z = 0.2 + Z = φ(0.1 + Z)
but 0.6+Z ≠ 0.1+Z since 0.6  0.1 = 0.5 ∉ Z 


Term
11S1_J
Prove...
If abelian group G has subgroups H and K of orders three and five, respectively, then G has an element of order fifteen. 

Definition
Proof 1
First recall: HK is a subgroup of G (since G is abelian) and o(HK) = o(H)o(K) / o(H∩K) = (3)(5)/o({e}) = 15/1 = 15.
(H∩K = {e} since only e has an order dividing 3 and 5.) ^{*}
Next, note that an abelian group of order pq (where p and q are distinct primes) is isomorphic to Z_{pq} (e.g. see here).
As HK is abelian (inherited from G) and of order 15 = (3)(5), it must be isomorphic to Z_{15} , thus generated by an element d of order 15.
Since d ∈ HK ⊆ G, the theorem is proved.
*Thank you Laura!
Proof 2
H and K are of prime orders, hence cyclic with H = <b>, K = <c>, where o(b) = 3 and o(c) = 5.
Now consider bc ∈ HK ⊆ G. By the Lemma below, o(bc) = 15.
Thus the theorem is proved.
Lemma: O(bc) = 15
Since (bc)^{15}= (by commutativity) b^{15}c^{15} = (b^{3})^{5}(c^{5})^{3} = e^{5}e^{3 }= e, we must have o(bc)  15, meaning o(bc) ∈ {1,3,5,15}.
o(bc)≠1, otherwise e=b^{3}c^{5}= (bc)^{3}c^{2} = c^{2}, contradicting o(c)=5.
o(bc)≠3, otherwise e=(bc)^{3} = b^{3}c^{3} = c^{3} contradicting o(c)=5.
o(bc)≠5, otherwise e=(bc)^{5 }= b^{5}c^{5} = b^{5}= b^{2} contradicting o(b)= 3.
Thus o(bc) = 15, as desired.



Term
11F1_J
Prove...
If G has a unique element x of order 2 then xy = yx for every y in G. 

Definition
Theorem: If G has x of order 2 and no other element of ord. 2, then ∀y ∈ G, xy=yx.
Proof
The statement above is of form p ^ ¬q → r which is equivalent to p ^ ¬r → q.
Thus we may rewrite the statement as follows:
If G has x of order 2 and ∃y∈G with xy≠yx then G has another element of ord. 2.
So assume x ∈ G with o(x)=2 and for some y ∈ G xy≠yx.
From xy≠yx we conclude y^{1}xy≠ x. (Since y^{1}xy = x → xy = yx.)
From o(x)=2 we conclude o(y^{1}xy) = 2. (See Lemma below.)
This completes the proof.
Lemma: For any x,y ∈ G, o(x) = 2 → o(y^{1}xy) = 2.
Proof:
(y^{1}xy)( y^{1}xy) = y^{1}x(y y^{1})xy = y^{1}xxy = y^{1}y = e. Therefore o (y^{1}xy) ≤ 2.
Also, o(y^{1}xy) ≠1. Otherwise, y^{1}xy = e → xy = y → x = e, contradicting o(x)=2.
Thus o(y^{1}xy) = 2.
___________________________________________ 


Term
14S1_J
Prove or disprove...
If f and g are homomophisms from G to H
then the set S ={a  f(a) = g(a) } is a subgroup of H. 

Definition
The statement is true, as shown below.
 S is nonempty since it contains e' = f(e) = g(e).
 Suppose a∈S and b∈S. Then...
f(ab^{1})
=f(a)f(b^{1})
= f(a)(f(b))^{1}
= g(a)(g(b))^{1}
= g(a)g(b^{1})
= g(ab^{1}).
So ab^{1 }∈ S.
We've shown that S is a subgroup of H.



Term
14S1_R
Given two group homomorphisms
f : G > H and g : G > H
and given
K = { a in G  f(a) = g(a) }
prove or disprove:
K is a subgroup of G. 

Definition


Term
11F2a_C
Let G be a group, and let Z be the center of G.
Prove
Z is a subgroup of G, and Z is normal in G.


Definition


Term
10S1_R
Prove each of the following:
a) Each subgroup of a cyclic group is cyclic.
b) If N is a cyclic normal subgroup of a group G
then each subgroup of N is normal in G. 

Definition


Term
11F1_R
Prove...
If G has a unique element x of order 2 then xy = yx for every y in G. 

Definition


Term
12S1_R
1. Consider groups G and G' and a group homomorphism φ : G → G' with kernel K.
Prove if N' is normal in G' then
(1) K ⊆ N
(2) N ≤ G
(3) N is normal in G
where N = φ^{−1}(N')
= {x ∈ G  φ(x) ∈ N'}. 

Definition


Term
12F4_R
4. Let A and B be n × n matrices with real entries.
a. Prove that if A and B are symmetric matrices
then AB is symmetric if and only if A and B commute.
b. Prove that if A is a skewsymmetric matrix
then each element on the main diagonal of A is zero. 

Definition


Term
14S2_J
2. Let I be an ideal in a commutative ring R with multiplicative identity 1 such that I ≠ R
and
∀a, b ∈ R, ab ∈ I ⇒ a ∈ I or b ∈ I.
Prove that R/I is an integral domain.


Definition
Let aI,bI be arbitrary nonzero elements of R/I.
(So a,b ∈ RI.)
Suppose a+I or b+I is a divisor of zero in R/I,
i.e. (a+I)(b+I) = I.
Then ab+I = ab + aI + Ib + I^{2} = (a+I)(b+I) = I.
Hence ab ∈ I.
But ab ∈ I → a∈I or b∈I
which contradicts a,b ∈ RI.
Thus neither aI nor bI can be a divisor of zero.
We've shown that R/I has no divisors of zero,
i.e. that R/I is an integral domain. ■



Term
13F2b_J
2. Let R be a commutative ring.
Prove the following.
b. If R is an integral domain and I, J are nonzero ideals of R,
then I ∩ J ≠ {0}. 

Definition
Consider a ≠ 0 ∈ I and j ≠ 0 ∈ J.
ij ∈ I (since I is an ideal and j∈R).
ij ∈ J (since J is an ideal and i∈R).
So ij ∈ I ∩ J.
But ij ≠ 0, since R has no zero divisors.
Thus I ∩ J ≠ {0}. ■



Term
13F2a_J
2. Let R be a commutative ring.
Prove the following.
a. N = {x ∈ R  x^{n} = 0 for some positive integer n},
is an ideal of R.


Definition
a) Show that N is an ideal in R.
• 0∈N (0^{1} = 0).
• Let x ∈ N, y∈ N with x^{m} = 0, y^{n} = 0. Then...
(x  y)^{m+n} = Σ[ a_{i} x^{m+ni}y^{i} ],
where a_{i} = C(m+ni, i) and i ∈ {0,...,m+n}.
If m+ni < m then ni < 0 i.e. i > n.
So for any term in the expansion,
either x is raised to a power ≥ m, giving 0 (since x^{m} = 0) or y is raised to a power ≥ n, giving 0 (since y^{n} = 0).
In either case the term evaluates to 0.
So all terms of the expansion are 0 i.e.
(x  y)^{m+n }= 0 and xy ∈ N.
Thus N is a subgroup of R.
• Let x ∈ N with x^{n} = 0. Then ∀r ∈ R ...
(xr)^{n} = x^{n}r^{n} =0r^{n} = 0
and
(rx)^{n} = r^{n}x^{n} =r^{n}0 = 0.
We've shown that N is an ideal in R. ■



Term
12F2b_J
2. Let R be a ring with multiplicative identity 1_{R}. Let S be a ring with multiplicative identity 1_{S} and without zero divisors.
b. Find a nontrivial ring homom. φ : Z_{6} → Z_{6} such that φ(1) ≠ 1. 

Definition
Let k = φ(1).
Then k^{2} = φ(1)^{2} = φ(1^{2}) = φ(1) = k.
Thus φ maps 1 to an element equal to its own square.
Now...
0^{2} =0, 1^{2 }=1, 2^{2} = 4, 3^{3} = 3, 4^{2} = 4, 5^{2} = 1.
So φ(1) must be 0, 1, 3 or 4.
φ(1) = 0 gives the trivial homom.
φ(1) = 1 gives the map φ(n) = n.
φ(1) = 3 gives the map φ(n) = 3n.
φ(1) = 4 gives the map φ(n) = 4n.
The last two cases satisfy our requirements that φ be nontrivial with φ(1) ≠ 1.
We can verify that each map is a homom.
For example, if φ(n) = 3n then
φ(ab) = 3(ab) = 3^{2}ab = 3a3b = φ(a)φ(b). ■
Interesting fact:
In the ring Z_{2} x Z_{3} ≅ Z_{6}
φ(1) =3 corresponds to φ(1,1) = (1,0),
φ(1) =4 corresponds to φ(1,1) = (0,1).



Term
12F2a_J
2. Let R be a ring with multiplicative identity 1_{R}. Let S be a ring with multiplicative identity 1_{S} and without zero divisors.
a. Prove that if φ : R→S is a nontrivial ring homom., then φ(1_{R})=1_{S}. 

Definition
Let φ be ring a homom. from R to S and b = φ(1_{R}). Then...
b^{2} = φ(1_{R}^{2}) = φ(1_{R}) = b.
Thus b^{2}  b = 0.
Factoring:
0 = b^{2}  b = (b)(b)  (b)(1_{S}) = b(b1_{S}).
S has no zero divisors, so we must have
b=0 or b=1_{S }.
Taking b=0 gives the trivial homomorphism.^{*}
Taking b=1_{S} gives φ(1_{R}) = 1_{S} .
Thus if φ is nontrivial we must have φ(1_{R}) = 1_{S}. ■
* ∀x∈R, φ(x) = φ(1_{R}x) = φ(1_{R})φ(x)=0φ(x) = 0 .



Term
10S2_J
2. Let R be a commutative ring with multiplicative identity 1,
and let I and J be ideals in R.
Prove: I + J = R ⇒ I ∩ J = IJ. 

Definition
Assume I + J = R.
a) Show I ∩ J ⊆ IJ
Choose i∈I and j∈J such that i+j = 1.
(We know i and j exist since I+J = R.)
Now consider k ∈ I ∩ J.
ki = ik ∈ IJ (since i∈I, k∈J).
and
kj ∈ IJ (since k∈I, j∈J ).
Therefore,
k = k1= k(i + j) = ki + kj ∈ IJ. ■
b) Show IJ ⊆ I ∩ J
Consider k ∈ IJ
i.e. k = ij = ji for some i∈I and j∈J.
Then
k ∈ I since i∈I,
k ∈ J since j∈J.
So k ∈ I ∩ J. ■
Note: This property is analogous to
gcd(i,j) = 1 ⇒ lcm(i,j) = ij. 


Term
10F2_J
Let R be a principal ideal domain.
Prove that each nontrivial prime ideal in R is a maximal ideal. 

Definition
Since R is a PID, each ideal I is generated by a single element,
i.e. I = <r> for some r∈R.
Let <p> be a nontrivial prime ideal in R.
Then for any a,b in R we know ab∈<p> ⇔ a∈<p> ∨ b∈<p>.
Also note that p is nonzero since <p> nontrivial.
Now consider an ideal <q> with <p>⊆<q>⊆R.
We have p∈<q> so p=qk for some k∈R.
Thus qk∈<p> and, since <p> is prime, q∈<p> ∨ k∈<p>.
If q∈<p> then <q>⊆<p> so <q>=<p>.
If k∈<p> then k=pj so p=pjq.
Recall that every PID is an ID (with unity).
So p=pjq > 1=jq → 1∈<q> → <q> = R.
(Recall p nonzero so we can cancel it.)
We've shown that there is no <q> strictly between <p> and R.
Hence <p> is maximal in R.



Term
09F2_J
Let R be a commutative ring with multiplicative identity 1.
Prove that the set
N = {a ∈ R  an = 0 for some positive integer n}
is an ideal of R,
and that N is a subset of each prime ideal of R. 

Definition
Let 0 and 1 denote the additive and multiplicative identities of R, respectively.
(Note: integer 1 will be written in italics.)
First show that (N,+)≤(R,+).
• N is nonempty since 10 = 0 so 0∈R.
• Let a,b∈R with am=0, bn=0.
(a+b)(mn) = a(mn) + b(m n) = (am)n + (bn)m = 0n + 0m = 0.
So a+b ∈R.
• Finally (a)n = a+...+ a = 1a+...+ 1a
= 1(an) = 1(0) = 0.
Hence (N,+) ≤ (R,+).
Now, if a∈N with an = 0 and r∈R, then
(ar)n = (ar)+...+(ar) = (a+...+a)r = (an)r = 0r = 0.
Hence ar∈R.
We've shown that N is an ideal in R.
Next we must show that N⊆I, for any prime ideal I in R.
First note that 0∈I since (I,+)≤(R,+).
Now let a∈N with an = 0.
Then a(1n)=a(1+...+1)= a+...+a = an = 0.
So a(1n)∈I
But I is prime, so either a∈I or 1n∈I.
Unfortunately we can't conclude directly that
a∈I which is needed to show N⊆I.
Where to go from here?



Term
11S2
Let I = <x> and J = <x,y> be ideals
in the polynomial ring R = Q[x,y],
where Q is the field of rational numbers.
Prove that I is a prime ideal, but is not maximal,
and that J is a maximal ideal in R. 

Definition
Let f(x,y), g(x,y) ∈ RI.
Then at least one term of each polynomial does not contain x as a factor.
The product of those terms will contribute to a term of fg without x as a factor. Hence fg is not in I.
We've shown that I is prime.
I is not maximal since I⊂J⊂R where the inclusions are strict
(e.g. y is in J but not I, 1 is in R but not J).
Now, suppose J is properly contained in an ideal K of R
and let k(x,y) ∈ KJ.
The only elements of R not in J are polynomials with a nonzero constant term.
So k(x,y) = j(x,y) + c with j(x,y) in J and c != 0.
Since both k and j are in K, their difference c is also in K.
That means K contains a unit, hence K = R.
We've shown that there is no ideal strictly between J and R
so J is maximal. 


Term
11S3b_J
Let F be a finite field and φ : Z → F be the map defined by
φ(n) = n · 1 for each n∈Z, where 1 is the multiplicative identity in F.
Prove the following.
a. φ is a ring homomorphism such that φ(Z) is a field isomorphic to Z_{p} for some prime integer p.
b. F contains exactly p^{k} elements for some positive integer k.


Definition
Proof of part b.
In part a proof we showed that ker(φ) =pZ for some prime p.
Thus for any b in F, p · b = p · (1b) = (p · 1)b = φ(p)b = 0b = 0.
That means the order of b in (F,+) divides p. (In a group, g^{k} = e iff ord(g) divides k.)
Moreover, since p is prime, ord(b)= 1 (i.e. b = 0) or ord(b) = p.
Thus every nonzero element of (F,+) has order p, as do all nontrivial cyclic subgroups.
Now, (F,+) is finite abelian, so (F,+) ≅ Z_{n1} x Z_{n2} x ... x Z_{nk} for some n_{1},...,n_{k}.
Moreover n_{1} = n_{2} = ... = n_{k} = p.
(Otherwise (F,+) would have a subgroup of order n_{j}≠p, namely 0 x...x 0 x Z_{nj} x 0 x...x 0.)
Therefore (F,+) ≅ (Z_{p})^{k} for some k.
Thus F has exactly p^{k} elements for some finite k.



Term
11F3_J
Let I = {p(x) ∈ Z[x]  p(0) = 0}. Prove that I is a prime ideal in Z[x]. 

Definition
Let I = {p(x) ∈ Z[x]  p(0) = 0}. Prove that I is a prime ideal in Z[x].
Let φ: Z[x]→Z be such that φ(p(x)) = p(0).
φ(p(x)+q(x)) = p(0)+q(0) = φ(p(x))+φ(q(x))
φ(p(x)q(x)) = p(0)q(0) = φ(p(x))φ(q(x)).
So φ is a homomorphism from Z[x] to Z.
ker(φ) = {p(x) s.t. φ(p(x)) = 0}
= {{p(x) s.t. φ(0) = 0} = I.
So I is an ideal of Z[x].
I is prime since pq(0) = 0 > p(0)q(0) = 0 > p(0) = 0 or q(0) = 0 since Z is an ID.
Thus I is a prime ideal of Z[x].



Term
A_{i} A^{i} ≠ ≤ ≥ ≅ ≡ → ⇒ ↔ ⇔ ∪ ∩ ¬ ∧ ∨ φ Φ α β Σ ∈ ⊂ ⊆ ⊃ ⊊ ⊋ ⋅ ⊲ ∣ ∋ ℝ ℚ ℤ ∅ 

Definition


Term
12F3_J
Let F[x] be the ring of polynomials with coefficients in field F.
a. Prove that if p(x) is irreducible in F[x], then F[x]/⟨p(x)⟩ is a field.
b. Prove or disprove: If Q is the field of all rational numbers, then Q[x]/⟨4x^{2}+4x−3⟩ is a field. 

Definition
a) We will show that <p(x)> must be maximal.
Consider an ideal <r(x)> of F[x]
containing <p(x)>.
Then p(x) is in <r(x)>
so p(x) = q(x)r(x).
But p(x) is irreducible
so either q(x) or r(x) is a nonzero constant c.
If q(x) = c then r(x) = p(x)/c
so r(x) is in <p(x)> hence <r(x)> = <p(x)>.
If r(x) = c, then <r(x)> = F[x].
We've shown there is no principle ideal
strictly between <p(x)> and F.
Then, since F is a PID,
there is no ideal in general between <p(x)>and F.
Hence <p(x)> is maximal and F[x]/<p(x)> is a field.
b) The discriminant of f(x) = 4x^2+4x3 is 8
so f(x) has roots in Q.
That means f(x) = r(x)s(x)
where r and s are linear polynomials in Q[x].
But then <f(x)> is not maximal
since <f(x)> is stricly included in <r(x)>
and <r(x)> != F[x]. 


Term
13S3_J
Let Q be the field of rational numbers and Q(sqrt(5)) = {a + b(sqrt(5)), a,b in Q}
a. Prove that Q(sqrt(5)) is a subfield of R
b. Find all possible nontrivial ring homomorphisms
φ : Q(sqrt(5)) > Q(sqrt(5))
such that φ(a) = a for each a ∈ Q.


Definition
Q(sqrt(5)) is a field since it is isomorphic to Q/<x^2  5> where x^2  5 is irreducible over Q.
Prove directly by showing it's a subring and exhibiting inverses = (a  b(sqrt(5))/(a^2 + 5b^2)
b) Ring homoms: 5 = phi(5) = phi(sqrt(5))^2. So phi takes sqrt(5) to +sqrt(5) or to sqrt(5).
Thus
phi1(a + bsqrt(5)) = a + b(sqrt(5)) a
phi2(a + bsqrt(5)) = a  b(sqrt(5)).



Term
12S3_J
3. Consider the polynomial p(x) = x^2 + 1
a. If Z_2 is the field of integers mod 2, then the ideal I = <p(x)> in Z_2[x] is not a prime ideal of Z_2[x]; hence Z_2[x]/I is not an integral domain.
b. If Q is the field of all rational numbers, then the polynomial p(x) is irreducible in Q[x] and hence if I is the ideal I = <p(x)> in Q[x], then Q[x]/I is a field.


Definition
a) p(x) = (x+1)^2 so (x+1)^2 = p(x)1 is in <p(x)> but but (x+1) is not in <p(x)> hence <p(x)> is not prime.
Since Z_2[x] is an commutative ring with unity and <p(x)> is not a prime ideal in Z_2[x], the quotient Z_2[x]//I is not an integral domain.
b) x^2 + 1 has only roots +i and i in C. Roots in Q would also be roots in C so there are no roots in Q. Hence x^2 + 1 is irreducible over Q, meaning <x^2 +1> is maximial in Q[x] .
Since Q[x] is a commutative ring and <x^2+1> is maximal, Q[x]/<x^2+1> is a field.



Term
12S3_J
3. Consider the polynomial p(x) = x^{2} + 1
a. If Z_{2} is the field of integers mod 2, then the ideal I = <p(x)> in Z_{2}[x] is not a prime ideal of Z_{2}[x]; hence Z_{2}[x]/I is not an integral domain.
b. If Q is the field of all rational numbers, then the polynomial p(x) is irreducible in Q[x] and hence if I is the ideal I = <p(x)> in Q[x], then Q[x]/I is a field.


Definition
a) p(x) = (x+1)^2 so (x+1)^2 = p(x)1 is in <p(x)> but but (x+1) is not in <p(x)> hence <p(x)> is not prime.
Since Z_2[x] is an commutative ring with unity and <p(x)> is not a prime ideal in Z_{2}[x], the quotient Z_{2}[x]//I is not an integral domain.
b) x^2 + 1 has only roots +i and i in C. Roots in Q would also be roots in C so there are no roots in Q. Hence x^2 + 1 is irreducible over Q, meaning <x^2 +1> is maximial in Q[x] .
Since Q[x] is a commutative ring with unity and <x^2+1> is maximal, Q[x]/<x^2+1> is a field.


