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.

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!

06F1b_J

Prove... 

If n is even, then there is an a ∈ G such that a ≠ e and a² = e.

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.

07S1.1_J

List all nonisomorphic abelian groups of order 72

By the FTFAG,

any finite abelian group of order n is isomorphic to Z_m1 x ... x Z_mk ,

where {m₁,...,m_k} is a set prime powers whose product is n.

Moreover, distinct {m₁,...,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:

Z₂Z₂Z₂Z₃Z₃ 

Z₂Z₄Z₃Z₃  

Z₈Z₃Z₃

Z₂Z₂Z₂Z₉

Z₂Z₄Z₉

Z₈Z_9.

07S1.2_J 

Determine the number of nonisomorphic abelian groups of order 2592, and explain your answer.

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.)

07F1a_J

Prove...

If G is abelian then

φ: G → G  | φ(x) = x^-1 is a group automorphism of G.

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.

07F1b_J

Prove... 

φ: S₃ → S₃  | φ(x) = x^-1 is not a group automorphism on S₃.

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.

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. 

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.)

08F1b_J

Prove...

G/N is abelian, where N◁ G,

if and only if

a^-1b^-1ab ∈ N,  ∀a,b ∈ G,

(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.

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.

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.

SPRING 2009, PROBLEM 1 PART a

Prove...

If G is abelian then ∀ a,b ∈ G and n ∈ ℤ, (ab)ⁿ = aⁿbⁿ.

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.

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 .

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.

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.

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.

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.

FALL 2010 PROBLEM 1 PART a

Prove... 

 The quotient group Q/Z is an infinite group

in which each element has finite order.

 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+n_k)

which is just

j + (n₁ + n₂ + ... + 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 n₁ = -j, n₂=3, n_i>2 = 0).

So k·(q+Z) = Z, meaning q+Z has finite order.

Fall 2010 Problem 1 PART b

Find an automorphism of Q/Z other than the identity.

Define φ(q+Z) = -q+Z.

We will show that φ is well-defined and an automorphism of Q/Z.

φ(q₁+Z) = φ(q₂+Z) 

⇒ -q₁+Z  =  -q₂+Z 

⇒ -q₁ - -q₂ = q₂-q₁ ∈ Z

⇒ q₁+Z = q₂+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.

FALL 2010 PROBLEM 1 PART c

Find a nontrivial homomorphism  

φ: Q/Z → Q/Z which is not injective.

Define φ: Q/Z → Q/Z as follows: φ(q+Z) = 2q+Z

φ(q₁ + q₂)

= 2(q₁+q₂)+Z

= (2q₁+2q₂)+Z

= (2q₁+Z) + (2q₂+Z)

= φ(q₁) + φ(q₂)

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.

 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 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 = <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) 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.

11F1_J

Prove...

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

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.

___________________________________________

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.

The statement is true, as shown below.

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.

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.

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.

11F1_R

Prove...

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

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'}.

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.

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. 

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.  ■

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}.

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}.   ■

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.

a) Show that N is an ideal in R.

•  0∈N (0¹ = 0).

•  Let x ∈ N, y∈ N with x^m = 0, yⁿ = 0. Then...

(x - y)^m+n = Σ[ a_i x^m+n-iyⁱ ],

where a_i = 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 x^m = 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.  ■

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 : Z₆ → Z₆ such that φ(1) ≠ 1. 

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).

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. 

Let φ be ring a homom. from R to S and b = φ(1_R).  Then...

   b²  = φ(1_R²) = φ(1_R) = b.

Thus b² - b  = 0. 

Factoring: 

0 = b² - b = (b)(b) - (b)(1_S) =  b(b-1_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_Rx) = φ(1_R)φ(x)=0φ(x) = 0 .

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.

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. 

10F2_J

Let R be a principal ideal domain.

Prove that each nontrivial prime ideal in R is a maximal ideal. 

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.

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.

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?

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.

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.

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. 

11F3_J 

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

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

A_i Aⁱ ≠ ≤ ≥ ≅ ≡ → ⇒ ↔ ⇔ ∪ ∩ ¬ ∧ ∨ φ Φ α β Σ ∈ ⊂ ⊆ ⊃ ⊊  ⊋ ⋅ ⊲ ∣ ∋ ℝ ℚ ℤ ∅

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.

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+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 <f(x)> is not maximal

since <f(x)> is stricly included in <r(x)>

and <r(x)> != F[x].

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.

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)).

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.

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.

12S3_J

3. Consider the polynomial p(x) = x² + 1

a. If Z₂ is the field of integers mod 2, then the ideal I = <p(x)> 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 = <p(x)> in Q[x], then Q[x]/I is a field.

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 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 <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.