# Shared Flashcard Set

## Details

AA COMP SOLUTIONS JCT
Past Algebra Comps JCT
41
Mathematics
01/23/2016

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 a2 = 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 a2 = 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 Zm1 x ... x Zmk , where {m1,...,mk} is a set prime powers whose product is n. Moreover, distinct {m1,...,mk} 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: Z2Z2Z2Z3Z3  Z2Z4Z3Z3   Z8Z3Z3 Z2Z2Z2Z9 Z2Z4Z9 Z8Z9.
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 07S1.2_J  Determine the number of nonisomorphic abelian groups of order 2592, and explain your answer.
Definition
 2592 = 25× 34   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 25 can be written using factors that are powers of 2: 25, 2421, 2322, 222221, 22212121, 21 21212121. 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 34 can be written using factors that are powers of 3: 34, 3331, 3232, 323131, 31313131.   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 25 = 2322 and 34 = 333131 i.e. 2592 = (8)(4)(9)(3)(3), corresponding to the group Z8Z4Z9Z3Z3. (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-1y-1 =  φ(x)φ(y). φ is injective:  φ(x) = φ(y) ⇒ x-1 = y-1 ⇒ yx-1x = yy-1x ⇒ 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: S3 → S3  | φ(x) = x-1 is not a group automorphism on S3.
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, g2N, g3N etc.  Since N is normal (in fact G is abelian, since cyclic), we have giN = gi(Ni) = (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, S3 is not cyclic but S3/S3 and S3/<(1 2 3)> are both cyclic.  (Note: <(1 2 3)> is normal in S3 so we may form quotient group.)
Term
 08F1b_J Prove... G/N is abelian, where N◁ G, if and only if a-1b-1ab ∈ N,  ∀a,b ∈ G,
Definition
 (aN)(bN) = (bN)(aN) iff  abN =baN iff a-1b-1abN =a-1b-1baN iff a-1b-1abN = N iff a-1b-1ab ∈ 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) = a2 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) = a2b2 Now ab = eabe = a-1aabbb-1 = a-1 a2b2 b-1  = a-1 (ab)2 b-1  = a-1ababb-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 = anbn.
Definition
 First show (ab)n = anbn for n ≥ 0 by induction. Basis step: (ab)0  = e = ee = a0b0. Inductive step: (ab)n+1 = (ab)n(ab) = anbnab = anabnb = an+1bn+1. Now, for n<0 ... (ab)n = ((ab)-1)-n  = (b-1a-1)-n = (b-1)-n(a-1)-n = bnan = anbn. 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) = xn  is an automorphism.   Moreover, for each g ∈ G there exists x ∈ G such that g = xn .
Definition
 φ is a homomorphism on G: φ(xy) = (xy)n = xnyn (since G abelian) φ is one-to-one: φ(x) = φ(y)  →  xn = yn   →  xny-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 one-to-one |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 = xn.
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 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'-1h'-1 ∈ KH   (since k'-1∈ K, h'-1∈ H). We've shown HK ⊆ KH. Since KH ⊆ HK and HK ⊆ KH we have HK = KH.
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 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=h1k1 ∧ b=h2k2 → ab= h1k1h2k2 = h1h3k3k2 ∈ HK. (Note: k1h1 ∈ KH and KH = HK → k1h1 = h3k3 , for some h3∈H, k3∈K.) HK is closed for taking inverses: a ∈ HK → a=hk → a-1=k-1h-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  q1, q2 distinct ∈ Q∩[0,1).  Since q1-q2 ∉ Z we must have q1+Z ≠ q2+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 + n1) +(j/k + n2) + ... + (j/k+nk) which is just j + (n1 + n2 + ... + nk)  ∈ Z  Thus k·(q+Z) ⊆ Z. Also Z ⊆ k·(q+Z) since the ni may be chosen freely (e.g. to obtain 3, use n1 = -j, n2=3, ni>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 well-defined and an automorphism of Q/Z. φ is well-defined: Suppose q1+Z = q2+Z. Then  q1-q2 = n  ∈ Z and... φ(q1+Z) = -q1+Z = -q1+(n+Z) = (-q1+n)+Z = -q2+Z = φ(q2+Z).  φ is a homomorphism: φ((q1+Z) + (q2 +Z) )  = φ( (q1+q2) +Z) )  = -(q1+q2)+Z = (-q1+-q2)+Z = (-q1+Z) + (-q2+Z) = φ(q1+Z) + φ(q2+Z).   φ is one-to-one: φ(q1+Z) = φ(q2+Z)  ⇒ -q1+Z  =  -q2+Z  ⇒ -q1 - -q2 = q2-q1 ∈ Z ⇒ q1+Z = q2+Z φ is onto Q/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 one-to-one correspondence from Q/Z to Q/Z. Therefore φ is an automophism of Q/Z.
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 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   φ is well defined:  q1+Z = q2+Z ⇒ q1-q2 ∈ Z ⇒ 2q1 - 2q2 ∈ Z ⇒ 2q1 + Z = 2q2 + Z ⇒ φ(q1+Z) = φ(q2+Z)   φ is a homomorphism φ(q1 + q2) = 2(q1+q2)+Z = (2q1+2q2)+Z = (2q1+Z) + (2q2+Z) = φ(q1) + φ(q2)     φ is not injective 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 Zpq  (e.g. see here).   As HK is abelian (inherited from G) and of order 15 = (3)(5), it must be isomorphic to Z15 , 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 = , K = , 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) b15c15 = (b3)5(c5)3 = e5e3 = e,  we must have o(bc) | 15, meaning o(bc) ∈ {1,3,5,15}. o(bc)≠1, otherwise e=b3c5= (bc)3c2 = c2, contradicting o(c)=5.  o(bc)≠3, otherwise e=(bc)3 = b3c3 = c3 contradicting o(c)=5. o(bc)≠5, otherwise e=(bc)5 = b5c5 = b5= b2 contradicting o(b)= 3.    Thus o(bc) = 15, as desired.
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 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-1xy≠ x.  (Since y-1xy = x →  xy = yx.) From o(x)=2 we conclude  o(y-1xy) = 2.     (See Lemma below.)      This completes the proof.   Lemma:  For any x,y ∈ G,  o(x) = 2 → o(y-1xy) = 2. Proof: (y-1xy)( y-1xy) = y-1x(y y-1)xy =  y-1xxy = y-1y = e.  Therefore  o (y-1xy) ≤ 2. Also, o(y-1xy) ≠1.  Otherwise, y-1xy = e → xy = y → x = e,  contradicting o(x)=2.  Thus o(y-1xy) = 2. ___________________________________________
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 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.
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 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
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 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
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 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
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 11F1_R Prove...   If G has a unique element x of order 2 then xy = yx for every y in G.
Definition
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 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
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 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 skew-symmetric matrix   then each element on the main diagonal of A is zero.
Definition
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 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 ∈ R-I.)   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 + I2 = (a+I)(b+I) = I.  Hence ab ∈ I. But ab ∈ I → a∈I or b∈I which contradicts a,b ∈ R-I. 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.  ■
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 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}.   ■
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 13F2a_J   2. Let R be a commutative ring. Prove  the following. a. N = {x ∈ R | xn = 0 for some positive integer n}, is an ideal of R.
Definition
 a) Show that N is an ideal in R.    •  0∈N (01 = 0). •  Let x ∈ N, y∈ N with xm = 0, yn = 0. Then... (x - y)m+n = Σ[ ai xm+n-iyi ], where ai = C(m+n-i, i) and i ∈ {0,...,m+n}. If m+n-i < m then n-i < 0 i.e. i > n.   So for any term in the expansion, either x is raised to a power ≥ m, giving 0 (since xm = 0)  or y is raised to a power ≥ n, giving 0 (since yn = 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 x-y ∈ N.    Thus N is a subgroup of R.   • Let x ∈ N with xn = 0. Then ∀r ∈ R ...  (xr)n = xnrn =0rn = 0   and  (rx)n = rnxn =rn0 = 0.   We've shown that N is an ideal in R.  ■
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 12F2b_J 2. Let R be a ring with multiplicative identity 1R. Let S be a ring with multiplicative identity 1S and without zero divisors.   b. Find a nontrivial ring homom. φ : Z6 → Z6 such that φ(1) ≠ 1.
Definition
 Let k = φ(1).     Then k2 = φ(1)2 = φ(12) = φ(1) = k.   Thus φ maps 1 to an element equal to its own square.    Now... 02 =0, 12 =1, 22 = 4, 33 = 3, 42 = 4, 52 = 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) = 32ab = 3a3b = φ(a)φ(b).  ■ Interesting fact: In the ring Z2 x Z3  ≅ Z6   φ(1) =3 corresponds to φ(1,1) = (1,0), φ(1) =4 corresponds to φ(1,1) = (0,1).
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 12F2a_J   2. Let R be a ring with multiplicative identity 1R. Let S be a ring with multiplicative identity 1S and without zero divisors.   a. Prove that if φ : R→S is a nontrivial ring homom., then φ(1R)=1S.
Definition
 Let φ be ring a homom. from R to S and b = φ(1R).  Then...      b2  = φ(1R2) = φ(1R) = b. Thus b2 - b  = 0.    Factoring:  0 = b2 - b = (b)(b) - (b)(1S) =  b(b-1S).   S has no zero divisors, so we must have b=0 or b=1S .   Taking b=0 gives the trivial homomorphism.* Taking b=1S gives φ(1R) = 1S .  Thus if φ is nontrivial we must have φ(1R) = 1S.  ■ * ∀x∈R, φ(x) = φ(1Rx) = φ(1R)φ(x)=0φ(x) = 0 .
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 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.
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 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 = for some r∈R. Let

be a nontrivial prime ideal in R.  Then for any a,b in R we know ab∈

⇔ a∈

∨ b∈

. Also note that p is nonzero since

nontrivial.  Now consider an ideal with

⊆R.  We have p∈ so p=qk for some k∈R.  Thus qk∈

and, since

is prime, q∈

∨ k∈

. If q∈

then

so =

. If k∈

then k=pj so p=pjq. Recall that every PID is an ID (with unity). So p=pjq --> 1=jq → 1∈ →  = R.  (Recall p nonzero so we can cancel it.)    We've shown that there is no strictly between

and R. Hence

is maximal in R.

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 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?
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 11S2 Let I = and J = 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) ∈  R-I. 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) ∈ K-J. 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.
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 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 Zp for some prime integer p. b. F contains exactly pk 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, gk = 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,+) ≅ Zn1 x Zn2 x ... x Znk  for some n1,...,nk. Moreover n1 = n2 = ... = nk = p. (Otherwise (F,+) would have a subgroup of order nj≠p, namely 0 x...x 0 x Znj x 0 x...x 0.) Therefore (F,+) ≅ (Zp)k  for some k. Thus F has exactly pk elements for some finite k.
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 11F3_J Let I = {p(x) ∈ Z[x] | p(0) = 0}. Prove that I is a prime ideal in Z[x].
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 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].
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 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]/⟨4x2+4x−3⟩ is a field.
Definition
 a) We will show that must be maximal. Consider an ideal of F[x] containing .   Then p(x) is in 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 hence = . If r(x) = c, then = F[x].  We've shown there is no principle ideal strictly between and F.  Then, since F is a PID,  there is no ideal in general between and F. Hence is maximal and F[x]/ is a field.   b)  The discriminant of f(x) =  4x^2+4x-3 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 is not maximal since is stricly included in and != F[x].
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 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/  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)).
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 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 = 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 = 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 but   but (x+1) is not in hence is not prime.   Since Z_2[x] is an commutative ring with unity and 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 is maximial in Q[x] . Since Q[x] is a commutative ring and is maximal, Q[x]/ is a field.
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 12S3_J   3. Consider the polynomial p(x) = x2 + 1   a. If Z2 is the field of integers mod 2, then the ideal I = in Z2[x] is not a prime ideal of Z2[x]; hence Z2[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 = 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 but   but (x+1) is not in hence is not prime.   Since Z_2[x] is an commutative ring with unity and is not a prime ideal in Z2[x], the quotient Z2[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 is maximial in Q[x] . Since Q[x] is a commutative ring with unity and is maximal, Q[x]/ is a field.
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