Prove that every prime is irreducible in an ID.

Lat b be a prime and b = cd.  

Then b | cd so b|c or b|d (as b is prime).

If b | c  then cd | c. So d | 1 i.e. d is a unit.

If b | d  then cd | d. So c | 1 i.e. c is a unit.

In either case c or d is a unit.

So b cannot be factored into two nonunits.

Thus b is irreducible.

Let <p> be a nontrivial prime ideal ⊆ PID R.

We have ab ∈ <p> → a ∈ <p> or b∈ <p>

Thus p | ab → p | a or p | b. 

p is not a unit since <p> ≠ R.

p ≠ 0 since <p> is nontrivial

We've shown that p is is a nonunit, nonzero

element of R such that p | ab → p |a or p | b.

Thus p is a prime element in R.

Prove that the ideals in Z

are of the form <n> for integer n.

Case I = {0}. Then I = <0>

Case I ≠ {0}. 

Let n = smallest positive integer ∈ I.

(n exists, otherwise I = {0}.)

First note <n> ⊆ I since I is an ideal.

Now, for any m∈ I,  m = qn + r with 0 <= r < n.

Then r = m - qn ∈ I.

(I is an ideal so qn ∈ I. I is a group so m-qn ∈ I.)

Hence r=0 since r < n and n was smallest > 0 ∈ I.

That means m = qn. So I ⊆ <n>

Hence I = <n>.

An element of Z/4Z is a coset of 4Z in Z.

= 0+4Z, 1+4Z, 2+4Z, 3+4Z.

So the ideals in Z/nZ are...

<0+4Z> = 

{

(0+4Z)(0+4Z),

(0+4Z)(1+4Z),

(0+4Z)(2+4Z),

(0+4Z)(3+4Z) 

} 

= 

{0+4Z}

To be continued

Show that coset multiplication in a ring is well defined,

where we define (a+I)(b+I) to be ab+ I.

Given I, an ideal of R, and a,b,a'b' ∈ R,

We must show 

(a'+I) = (a+I) --> a'b + I =  ab+ I.

(b'+I) = (b+I) --> ab' + I =  ab + I.

Suppose (a'+I) = (a+I), i.e. a'- a ∈ I.

Then a'b - ab = (a' - a)b ∈ I since I is an left ideal.

Thus a'b+I = ab+I.

Suppose (b'+I) = (b+I), i.e. b'- b ∈ I.

Then ab' - ab = a(b' - b) ∈ I since I is a right ideal.

Thus ab'+I = ab+I. 

Herstein pg. 135 #2

Prove that the only ideals in a field are (0) and the field itself.

{0} is an ideal in F since ({0},+) is a subgroup of (F,+)

and 0y = y0 = 0 ∈ {0}, for all y ∈ F.

F is an ideal in F since (F,+) is a subgroup of (F,+)

and for any x in F, xy ∈ F and yx ∈ F, for all y ∈ F.

Now, let I≠ {0} be a nontrivial ideal in F and b≠0 ∈ I.

Since F is a field b^-1 ∈ F. 

Thus  1 = bb^-1 ∈ I.  

For any x in F,  x = 1x ∈ I.

Hence I = F.

We've shown there can be no other ideals in F besides the trivial ideal and F itself.

Herstein Page 135 #3 

Prove that any homomorphism of a field is either an isomophism or takes each element to 0.

Herstein Page 135 #3) 

Prove that any homomorphism of a field is either an isomophism or takes each element to 0.

Let φ be a homom. from field F to R.

From problem 2 we know

the only ideals of F are {0} and F itself.

Since ker(φ) is an ideal of F,

we must have ker(φ) = {0},

in which case F is a one-to-one (thus an isomorphism)

or else ker(φ) = F, in which case φ takes F to {0}.

Herstein Page 135 #5

If U,V are ideals of R, let U+V = {u+v | u in U and v in V}.

Prove that  U+V is also an ideal

Herstein Page 135 #5)

If U,V are ideals of R,

let U+V = {u+v | u in U and v in V}.

Prove that  U+V is also an ideal.

U+V is nonempty since 0 = 0+0 ∈ U+V

(Since U and V are subgroups of R, 0∈U and 0∈V.)

Let u,u' ∈U and v,v∈V

so u+v and u'+v' are elements of U+V.

Then (u+v) - (u'+v') = (u-u') + (v-v') ∈ U+V.

(Since U and V are subgroups, u-u' ∈U and v-v'∈V.)

We've shown that U+V is a subgroup of R.

Now for any r in R, consider that

ur∈U (since u∈U and U is an ideal of R) 

and vr∈V (since v∈V and V is an ideal of R) .

Thus

(u+v)r = ur+vr ∈ U+V.

Similarly, since ru∈U and rv∈V.

r(u+v) = ru +rv ∈ U+V.

We've shown that U+V is an ideal of R.

Herstein pg. 135 6, 7)

If U, V are ideals of R

let UV be the set of all elements

that can be written as finite sums of elements

of the form uv

where u in U and v in V.

6) Prove that UV is an ideal of R.  

7) Prove that UV ⊆ U ∩ V.

6, 7) If U, V are ideals of R, let UV be the set of all elements that can be written as finite sums of elements of the form uv where u in U and v in V.

6) Prove that UV is an ideal of R.  

For m,n > 0... 

Let u₁ ... u_m,  x₁ ... x_n be arbitrary elements of U

and v₁ ... v_m,  y₁ ... y_n be arbitrary elements of V.

Then  u₁v₁+...+ u_mv_m  and  x₁y₁+...+ x_ny_n 

represent two arbitrary elements of UV. 

Now, 0 ∈ UV since 0 = (0)(0) with 0∈ U, 0∈V.

Also,  (u₁v₁+...+ u_mv_m) - (x₁y₁+... +x_ny_n) 

= u₁v₁+...+u_mvm + (-x₁)y₁+...+(-x_n)y_n ∈ UV

since each term is of the form uv, with u∈U, v∈V.

Thus UV is a subgroup of R.

Now, let r be an element of R. Then

(u₁v₁+... + u_mv_m)r = u₁v₁r +... + u_mv_mr ∈ UV

To justify membership in UV, consider that

each term in the sum

is of the form u(vr)

where u∈U

and vr∈V.

(Recall v∈V, r∈V and V is an ideal of R.)  

Similarly

r(u₁v₁+... + u_jv_j)r = ru₁v₁+... + ru_jv_j ∈ UV .

This time membership in UV follows because

each term is of the form (ru)v

with ru∈U and v∈V.

We've established that UV is an ideal in R.

7) Prove that UV ⊆ U ∩ V.

As in queston 6, consider u₁v₁+...+ u_mv_m , an arbitrary element of UV.

For each u_iv_i in the sum...

u_iv_i ∈ U since u_i∈U, v_i∈R and U is a (right) ideal of R.

u_iv_i ∈ V since v_i∈V, u_i∈R and V is a (left) ideal of R.

Therefore u_iv_i ∈ U ∩ V.

Now, since U and V are subgroups of R, so is their intersection U ∩ V.

That means U ∩ V is closed under + and 

u₁v₁+...+ u_mv_m ∈ U ∩ V.

We've shown that UV ⊆ U ∩ V.

Herstein Page 135 #21.

If R is a ring with unit element 1

and φ is a homomorphism of R

into an integral domain R'

with I(φ)≠R,

prove that  φ(1) is the unit element of R'.

Herstein Page 135 #21.

If R is a ring with unit element 1 and φ is a homomorphism of R into an integral domain R' with ker(φ)≠R, prove that  φ(1) is the unit element of R'.

Let b = φ(1).  

Then b² = (φ(1))² = φ(1²) = φ(1) = b.

Thus b² - b = 0.

Moreover for any r in R',  

b(br - r) = b²r - br = (b² - b)r = 0r = 0.

Since R' in an integral domain,

either b = 0 or br-r = 0.

If b=0  then for any  in R,

φ(x) = φ(1x) = φ(1)φ(x) = bφ(x) = 0φ(x) = 0,

But ker(φ)≠R so this is not possible.

Thus br-r = 0 which gives br = r i.e. b is the unit in R'.

Herstein Page 136 #14.

For a∈R, let Ra= {xa | x∈R}.

Prove that Ra is a left-ideal of R.

Herstein Page 136 #14.

For a∈R, let Ra= {xa | x∈R}.

Prove that Ra is a left-ideal of R.

First note 0 = 0a ∈ Ra  so Ra is nonempty.

Next consider arbitrary xa, ya  ∈ Ra.

Then xa-ya =  (x-y)a ∈ Ra. 

Hence xa-ya ∈ Ra.

We've shown that Ra is a subgroup of R.

Now consider arbitrary xa ∈ Ra.

For any y in R, y(xa) = (yx)a ∈ Ra.

So Ra is a left ideal in R.

Prove that the intersection of

two left ideals of R is a left ideal of R.

Let I₁ and I₂ be left ideals of R.

Since I₂ and I₂ are subgroups of R,

their intersection is a subgroup of R.

Now consider i in I₁ ∩ I₂.

For any x in R,  

x(i) ∈ I₁ (since i ∈ I₁ and I₁ is an ideal of R),

x(i) ∈ I₂ (since i ∈ I₂ and I₂ is an ideal of R).

Thus x(i) ∈ I₁ ∩ I₂.

We've shown that I₁ ∩ I₂  is a left ideal of R.

Herstein Page 136 #18.

If R is a ring and L is a left ideal of R,

let λ(L) = {x∈R | ax = 0 for all a in L}.  

Prove that λ(L) is a two sided ideal of F.

Herstein Page 136 #18.

If R is a ring and L is a left ideal of R,

let λ(L) = {x∈R | xa = 0 for all a in L}.  

Prove that λ(L) is a two sided ideal of F.

0a = 0 for all a in L so 0∈λ(L).

Suppose x, y ∈ λ(L), so xa = ya = 0 for any a in L.

Then for any a in L,

(x-y)a = xa - ya = 0 - 0 = 0. So x-y ∈λ(L).

We've shown that λ(L) is a subgroup of R.

Let x ∈ λ(L) and y ∈ R.

Now given any a in L,

(yx)a = y(xa) = y0 = 0. 

That means yx∈ λ(L).

So λ(L) is  left ideal.

Also (xy)a = x(ya) = 0

since ya ∈ L

(Recall L is an ideal.)

and x∈ λ(L). 

So λ(L) is a right ideal.

Thus λ(L) is a two-sided ideal.

Let R be a ring with unit element. Using its elements we define a ring R' by

defining a ⊕ b = a + b + 1

and

a • b = ab + a + b.

a) Prove that R' is a ring.

b) What is the zero in R' ?

c) What is the unit of R' ?

d) Prove that R' is isomorphic to R.

First note ⊕ and • are binary operations

over R'

since they are defined in terms of operations

(+ and times)  

over the same set of elements.

 ⊕  is commutative  since

a⊕b = a + b + 1 = b + a + 1 = b⊕a .

-1 ⊕ a  = a ⊕ -1 =  a + (-1) + 1 = a.

So -1 is an identity for ⊕.

 ⊕ is associative since

(a ⊕ b) ⊕ c

=  (a + b + 1) + c + 1

= a + (b + c + 1) + 1

= a ⊕ (b ⊕ c).

Now a ⊕ (-a - 2) = a + -a + -2 + 1 = -1.

So -a - 2 acts as an additive inverse of a.

We've shown that (R' , ⊕) is a group.

Now we will show that the multipication (•) for R'  is associative and admits an identity.

(a•b) • c

= (ab + a + b)• c

= (ab + a + b)c + (ab + a + b) + c

= abc + ac + bc + ab + a + b + c

a•(b•c)

= a•(bc + b + c)

= a(bc + b + c)  + a + bc + b + c

= abc + ab + ac + a + bc + b + c

= abc + ac +bc + ab +a + b + c

So • is associative.

Notice that •  is also commutative

since ab + a + b = ba + b + a.

  a • 0 = a0 + a + 0 = 0 = 0a + 0 + a = 0 • a

so 0 acts as a multiplicative identity in R'.

We still need to show that • distributes over ⊕.

Since •  is commutative we only need to show left distributivity.

a•(b⊕c) = a•(b + c + 1)

= a(b+c+1) + a + b+c + 1 =

ab + ac + 2a + b + c + 1

(a•b)⊕(a•c) =( ab + a + b)⊕(ac + a + c)

= ab + a + b + ac + a + c + 1

= ab + ac + 2a + b + c + 1.

So • distributes over ⊕.

We've shown that (R' ,⊕,  •) is a commutative ring with additive identity -1 and multiplicative identity 0.

Now we want to find an isomorphism from R to R'.

Recall that as sets R and R' are identical. 

Let φ(b) = b - 1 where subtraction is as in R.

φ is onto since for any c in R, c = (c+1) -1 .

φ is one-to-one since for any b,c in R,

φ(b) = φ(c) --> b-1 = c-1 --> b = c.

So φ is a bijection from the set of elements to itself, i.e from R to R'.

Now we want to show that φ is a ring homomorphism from R to R'.

φ(a+b) = (a+b)-1.

φ(a) ⊕ φ(b)

= (a-1)⊕(b-1)

= a - 1 + b - 1 + 1

= (a+b)-1.

So φ(a+b) = φ(a) ⊕ φ(b)

φ(ab) = ab - 1 

φ(a) • φ(b) = (a-1)•(b-1)

= (a-1)(b-1) + (a-1) + (b-1) = ab -a -b + 1 +a +b -2 

= ab - 1.

So φ(ab) = φ(a) • φ(b) .

We've shown that  φ is bijective ring homomorphism from R to R'. Hence R and R' are isomorphic.

First note that R cannot be the trivial ring, otherwise there is only one ideal, the ring itself.

Since R != {0}

Then <1> = R is the one and only nontrivial ideal.

Now, consider x ≠ 0 ∈ R.

Since x = 1x ∈ <x> we must have <x> = R.

Thus 1∈ <x>,  so xy = 1 for some y. 

Thus x has an inverse.

Yes if there is no mult. identity. 

For example 4Z/2Z has trivial multiplication. So the only ideal is the trivial one. 

To  get trivial multiplication in nontrivial ring, you can't have an identity. For  1 != 0 --> 1 = (1)(1) != 0. 

Math 6121 Assignment 5.                                               Traugott

Herstein Page 139 Problem 1.

Notation: In what follows <b> will always denote the right ideal generated by b, an element of R.

Let R be a ring with unit element, R is not necessarily commutative,  such that the only right ideals of R are <0> and R itself.  Prove that R is a division ring.

In order for the property to hold, we must assume that R is not the trivial ring. Otherwise 1 = 0 and all right ideals are indeed <0> or R itself (since <0> is the only ideal). Yet R is not a division ring since it has no group of nonzero elements.

So assume R is a nontrivial ring with unit 1 whose only ideals are <0> and R itself.

First note that 1 ≠ 0. Otherwise R = <1> = <0>, in which case R would be the trivial ring.

Now since R is not the trivial ring, there exists x ≠ 0  ∈ R.  Moreoever x = x(1) so x ∈ <x>.

<x> ≠ <0> so we must have <x> = R.

Then 1 ∈ <x> so xy = 1 for some y  ∈  R.

Moreoever y ≠ 0 (otherwise 0 = xy = 1).

So <y>= R and  yz = 1 for some z  y  ∈  R.

Then z = 1z = xyz = x1 = x.

So xy = yx = 1.

We've shown that every nonzero element in R is a unit.

So R is a ring with multiplicative identity in which every nonzero element is a unit. Thus R is a division ring.

Herstein Page 140 #3

Let Z be the ring of integers, p a prime number, and <p> the ideal of Z consisting of all multiples of p. Prove

a) Z/<p> is isomorphic to Zp, the ring of integers mod p.

b)  Z_p is a field.

Herstein Page 140 #3

Let Z be the ring of integers, p a prime number, and <p> the ideal of Z consisting of all multiples of p.

Prove

a) Z/<p> is isomorphic to Z_p, the ring of integers mod p.

b) Z_p is a field. 

a) Define φ: Z/<p> → Z_p  with φ(n+<p>) = n mod p

(i.e. the remainder when we divide n by p).

First show that φ is well-defined:

n +<p> = m +<p>

→ n - m ∈ <p>

→ n - m = pk for some k in Z

→ n ≅ m (mod p)

 → n mod p = m mod p

→ φ(n+<p>) = φ(m+<p>).

Next show that  φ is one-to-one:

φ(n+<p>) = φ(m+<p>)

→ n mod p = m mod p

→ For some r, q₁, q₂, r = n-pq₁ = m-pq₂

→ n - m = p(q₁ - q₂)

→ n-m in p

→ n +<p> = m +<p>.

 Next show that  φ is onto Z_p:

For any k in {0,1,...,p-1}

k = k mod p. So φ(k+<p>) = k.

Next show

φ(m+<p> + n+<p>) = φ(m+<p>) + φ(n+<p>):

φ(m+<p> + n+<p>) = φ(m+n + <p>)

= (m+n) mod p

= ((m mod p) + (n mod p)) mod p

= φ(m+<p>) + φ(n+<p>).

Next show 

φ((m+<p>)( n+<p>)) = φ(m+<p>)φ(n+<p>):

φ((m+<p>)( n+<p>))

= φ(mn + <p>)

= mn mod p

= ((m mod p)(n mod p)) (mod p)

= φ(m +<p>)φ(n +<p>)

We've shown that φ is an isomorphism from Z/<p> onto Z_p.

Therefore  Z/<p> and Z_p are isomorphic.

b) Z is a commutative ring with unity. To show that Z/<p> is a field it suffices to show <p> maximal in Z.  

Since Z_p ≅ Z/<p> it will follow directly that Z_p is a field.

First note that <p> is indeed an ideal of Z:

0 = p(0) ∈ <p> so <p> is nonempty.

pk ∈ <p> → -(pk) = p(-k) ∈ <p>.

So <p> is an additive subgroup of Z.

Further: pk ∈ <p>→ (pk)n = p(kn) ∈ <p>, ∀n ∈ Z.

So <p> is an ideal of Z.

Now we will show that <p> is maximal:

All ideals in Z are of the form <n>, so if there is an ideal between <p> and Z, we have

<p> ⊆ <n> ⊆ Z for some n ∈ Z.

WLOG assume n > 0 since <n> = <-n>.

Now p ∈ <n> so p = kn for some k > 0.

Then, since p is prime, n = 1 or n = p.

If n = 1 then <n> = Z.

If n = p then <n> = <p>.

So there is no ideal strictly between <p> and Z.

Hence <p> is maximal. 

We've established that <p> is maximal in Z. 

Therefore Z/<p> is field, as is Z_p.

Herstein

Show that Zm x Zn has an element of order >= lcm(m,n). Thus if gcd(m,n) = 1, Zm xZn has element of order mn -> ZmxZn is cyclic

In an ID

if  ax = x for some a, and x!=0,

then the ID has a multiplicative identity.

Assume ax = x, x!=0/

Let y be arbitrary.

ax = x

0 = y0 = y(ax-x) = axy - xy = x(ay - y) 

Since ID and x != 0 --> ay = y.  

So a acts as identity on y.

x^2 + 1 is irreducible over Z₁₁.

Note that x^2+1 can only be factored into two polynomials of lesser degree if those factors are linear, i.e. if x^2 + 1 has roots in Z₁₁. That's not the case since no perfect square is congruent to -1 (mod 11).  (The squares of 0, through 10 are congruent to 0, 1, 4, 9, 

there would ne since no perfect square is congruent to -1 (mod 11):

(0^2 == 1

Let p be an irreducible element and <p> in <n>.  

Then p =  kn. But then k or n is a unit.

If k is a unit then n = p/k so <n> in <p>

hence <p> = < n>.

If n is a unit then <n> is R.

So <p> is maximal.

(p) is maximal

= (p) has no proper container (a)

= p  has no proper divisor a

= p  is irreducible

= (p) is prime, by PID⇒UFD, so ireducible=prime

Herstein 3.10.4

Show that fp(x) = x^p + x^p-1 +...+ x + 1

is irreducible over Q, where p is a prime.

Problem 3  For p prime, f_p-1(x) = x^p-1 +...+ x¹+x⁰  

is irreducible over Q. 

By Lemma: f_p-1(x+1) = C(p,p)x^p-1 +...+ C(p,2)x¹ + C(p,1)x⁰. 

Now since p is prime, p | C(p,k) for 1 ≤ k ≤ p-1.

That's true because the numerator of p! /(k!(p-k)!)

has p as a factor and the denominator doesn't.

Moreover p ∤ C(p,p) = 1 and p² ∤ C(p,1) = p.

So, by Eisenstein's criterion, f_p-1(x+1) is irreducible over Q. 

That means f_p-1(x)  is also irreducible over Q, since

f_p-1(x+1)=g(x)h(x)  →  f_p(x)=g(x-1)h(x-1),

with degrees unchanged when we substitute x by x-1.

Lemma: If f_n(x) = xⁿ + x^n-1+ x^n-2 +...+ x + 1 then

f_n(x+1) = C(n+1,n+1)xⁿ +...+ C(n+1,2)x¹ + C(n+1,1)x⁰ 

Basis step n = 0:  f₀(x+1) = 1 =  C(0+1,0+1)x⁰. 

Inductive step: Suppose f_k(x+1) = C(k+1,k+1)x^k +...+ C(k+1,1)x⁰.

Then f_k+1(x+1) = (x+1)^k+1 + C(k+1,k+1)x^k +...+ C(k+1,1)x⁰

= C(k+1,k+1)x^k+1 + [(C(k+1,k)+C(k+1,k+1))x^k +...+ (C(k+1,1)+C(k+1,0))x⁰]

 =  x^k+1 + C(k+2,k+1)x^k + ... + C(k+2,1)x⁰

= C(k+2,k+2)x^k+1 + C(k+2,k+1)x^k + ... + C(k+2,1)x⁰.

Herstein 3.11.4

If R is a ring with unity,

prove that any unit in R[x] must already be a unit in R.

Math 6121 Assignment 6B    Traugott

Problem 4

[image]If R is a ring with unity,

prove that any unit in R[x] must already be a unit in R.

Case 1 = 0:  

Then R = {0} and R[x] = R.

The property is trivial true. 

Case 1 != 0:  

Let f and g be units in R[x], with fg = 1.

Since neither f nor g is 0, we can write

f(x) = a_mx^_m +...+a₀x^₀

g(x) = b_nx^_n +...+b₀x^₀

where a_m,..., a₀, b_n ,..., b₀ ∈ R, a_m≠0≠b_n 

and m=deg(f),  n=deg(g). 

The leading term of f(x)g(x) will be a_mb_nx^m+n

with a_mb_n nonzero since R is an integral domain.

Thus  m+n= deg(fg) = deg(1) = 0.  

So m = n = 0.

Then f(x)=a₀∈R, g(x)=b₀∈R with a₀b₀ =1.

So f and g are units in R.

Herstein  3.11.11

If R is an integral domain and F is the field of quotients of R, prove that any f(x) in F[x] can be written f₀(x)/a wth f₀(x) in R[x] and a in R.

Problem 11

If R is an integral domain and F is the field of quotients of R, prove that any f(x) in F[x] can be written f₀(x)/a wth f₀(x) in R[x] and a in R.

f(x) may be written (a_n/b_n)xⁿ +...+ (a₀ /b₀)x⁰ 

Where we use fraction notation to denote elements of F.

Recall, the usual properties of fractions apply in the field of quotients.

In particular (as)/(bs) = a/b since (as)b = (bs)a.

That means we can clear denominators of f(x):

f₀(x) = (b₀...b_n)f(x) = 

= (b₀..b_na_n/b_n)xⁿ +..+(b₀..b_i..b_na_i/b_i)xⁱ +..+ (b₀..b_na₀/b₀)x⁰ 

= (b₀...b_n-1a_n)xⁿ +..+ (b₀...b_i-1b_i+1ab_na_i)xⁱ ...+ (b₁...b_na₀)x⁰ .

So f₀(x) is in R[x] and f(x) = (1/a)f₀(x) as required,

taking a = b₀...b_n.

Herstein 3.11.8

Prove that when F is a field F[x, y] is not a principal ideal ring  (i.e.not a principal ideal domain).

Problem 8: Prove that when F is a field,

F[x, y] is not a principal ideal domain.  

Consider the set

<x,y>={xf(x,y)+yg(x,y) s.t. f(x,y), g(x,y) ∈ F[x,y]}.

I'll first show that <x,y> is an ideal of F[x,y].  

<x,y> is nonempty since 0 = x0 + y0 ∈ <x,y>

<x,y> is closed under subtraction since

(xf₁(x,y)+yg₁(x,y)) - (xf₂(x,y)+yg₂(x,y))

= x(f₁-f₂)(x,y) + y(g₁-g₂)(x,y) ∈ <x,y>.

To see that <x,y> is closed under supermultiplication

let h(x,y) be an abitrary element of F[x,y]. Then

(xf(x,y)+yg(x,y))h(x,y) = x(fh(x,y))+y(gh(x,y))∈<x,y>.

Now I'll show that <x,y> is not principal.  

Suppose, by way of contradiction, that

<x,y>=<f(x,y)>, f(x,y) ∈ F[x,y].

That means x and y are in <f(x,y)>.

(1 ∈ F[x,y] so the ideals contain their generators.)

Hence f(x,y) | x  and f(x,y) | y.  

That's only possible if f(x,y) is a nonzero constant in F.

But f(x,y) is a member of <x,y> 

so it must be of the form xf(x,y)+yg(x,y),

meaning either 0 or of degree ≥ 1. 

We've arrived at a contradiction

so <x,y> cannot be equal to any principal ideal <f(x,y)>.

Thus F[x,y] is not a PID.

Prove

In an integral domain, every prime element is irreducible.

Let p be a prime element and pose p = ab.

Then p | ab  and, since p is a prime, either p | a or p | b.

WLOG say p | a.  Then a = pc for some c.

Hence p = pcb. 

Since we are in an integral domain (with unity) we may cancel p to get 1 = cb.

That means b is a unit.

Since p cannot be factored into nonunits, 

p is irreducible.

Prove

Maximal Ideal is Prime in Commutative Ring with Unity.

Here’s a proof that doesn’t involve the quotient R/J.

Suppose that J is not prime;

then there are a,b∈R∖J such that ab∈J .

Let A be the ideal generated by

J∪{a}; A={j+ar: j∈J and r∈R}.

Clearly J⊊A,  so A=R, 1∈A,

and hence 1=j+ar for some j∈J and r∈R.

Then b = b1 = b(j+ar) = bj+bar.

But bj ∈ bJ ⊆ RJ = J,

and bar ∈ Jr ⊆ JR = J, so b∈J.

This contradiction shows that J is prime.

Prove

In an ID without unity every nonzero element has the same additive order. Therefore, we may talk about the characteristic as in a unital ID.

n·a = 0 for a ≠ 0 

→ a(n·b) = n·ab =(n·a)b = 0b = 0

→ n·b = 0

Prove

In a (possibly) non-unital ID of finite characteristic,

the characteristic is prime.

The characteristic of a non-unital ring is simply 

the exponent of the additive group, 

i.e. the smallest n such that na=0 for every a∈A.

Then the argument that the characteristic is prime for 

integral domains follows the usual way:

So assume that n=kl, 0<k,l<n, is a composite number. 

Choose a∈A such that ra≠0 for every r=1,2,…,n−1 

(if there is no such element, the characteristic is necessarily 

smaller than n).

Thus, ka≠0, la≠0 but (ka)⋅(la) = kla² = na² = (na)a =0a = 0,

contradicting the fact that A was an integral domain. 

Thus, if the characteristic >0, it is necessarily prime.

Prove:

Frobenius Endomorphism φ: R→R is injective

when R is a field but not necessarily surjective

(unless R is finite of course).

If R is a ring only, it is not necessarily injective.

Frobenius is not trivial homom, since 1^p =1 ≠ 0.

(Note char(R) = p ≠ 1 so R not trivial ring.)

Therefore, if R s a field, ker(φ) = {0} so φ injective.

(Recall a field only has two ideals.)

Example where φ is not surjective, even with R a field:

R = F(x) = rational functions over Z_p .

Here nothing maps to x

since powers of x are only increased by φ.

However if F(x) is perfected = F(x^{1/p^∞}) then we get surjective

e.g. φ(x^1/p) = x.

If R is an ID we still get injective since

a^p = b^p  → a^p - b^p = 0 

→  (a-b)^p = 0 →  a-b = 0 →  a = b.

Suppose R is not an ID e.g. F_p[x]/<x^p>

The  φ(x) = x^p + <x^p> = <x^p> = 0^p + <x^p> = φ(0) . 

If R is a finite field then  φ is a bijection, so an automorphism (and a permutation of R).

Example F₄  0, 1, x, 1+x   

x² = 1+x.

char(F₄) = 2.  

φ(x) = 1+x,  

φ(1+x) = (1+x)² = 1+ x² = 1+1+x = x.

We have φ surjective for Field of rational functions

F₈ = Z₂[b] where b³ = b + 1.

φ(1) = 1 so φ fixes Z₂.

Now φ(b) = b --> identity

φ(b) = b² is Frobenius

φ(b) = b+1 -->

φ(b+1) = b+2 = b

φ(b²) = (b+1)² = b²+1

φ(b²+1) = b²+1 + 1 = b²

φ(b²+b) = b²+1 + b+ 1 = b²+b

φ(b²+b+1) = b²+b + 1 

So it fixes four elements and flips two pairs. This is not a homomorphism

Can tell because b is generator to phi(b) = bⁱ means i | 

Prove in a ring

0x = x0 = 0

-(ab) = (-a)b = a(-b)

(-a)(-b) = ab

If identity exists it's unique

0 = 0x - 0x = (0 + 0)x - 0x = 0x + 0x - 0x + 0 = 0x.

Similarly for x0 = 0.

a(-b) + ab = a(-b + b) = a0 = 0

This shows a(-b) = -(ab).

Now (-a)b + ab = (-a + a)b = 0b = 0

This shows (-a)b = -(ab).

(-a)(-b) = -(a(-b)) = --(ab) = ab

Prove that if I is a prime ideal of CR1 R,

then R/I is an integral domain.

Suppose that I is a prime ideal of R.

Let 

If α is an algebraic number,

prove that there is an n such that

nα is an algegraic integer.

If α is an algebraic number,

prove that there is an n such that

nα is an algegraic integer.

Since α is an algebraic number, it is a root of some polynomial  p(x) over Q of degree n>=1.

Let r be the product of all the denominators  and numerators 

of coefficients in p(x).

Let y = (r/a_n)ⁿx

Then p(y/rⁿ) = 0

So 

Prove

Every ED is a PID

Proof Let (D,+,×) be a Euclidean domain whose zero is 0

and whose Euclidean valuation is ν.

We need to show that every ideal of (D,+,×) is a principal ideal.

Let U be an ideal of (D,+,×) such that U≠{0}.

Let d∈U such that d≠0 and ν(d) is as small as possible for elements of U.

(By definition, ν is defined as ν:D∖{0_R}→N,

so the codomain of ν is a subset of the natural numbers.

By the Well-Ordering Principle, d exists as an element of the

preimage of the least member of the image of U.)

Let a∈U.

Let us write a=dq+r where either r=0 or ν(r)<ν(d).

Then r=a−dq and so r∈U.

Suppose r≠0. That would mean ν(r) < ν(d)

contradicting d as the element of U with the smallest ν.

So r=0, which means a=qd.

That is, every element of U is a multiple of d.

So U is the principal ideal generated by d.

This deduction holds for all ideals of D. Hence the result.

Irreducible generate maximal ideal in a PID

Proof 1

i is irreducible so i = ab --> a or b is a unit.

Now consider (j) containing i.  So i = jk. 

Hence j or k is a unit.

Than (j) is R or j = i(k^-1) so (j) = (i).

We've shown (i) maximal

Prime Ideal is Maximal in PID

<p> is prime ideal in R.

Suppose <j>  contains <p>.

Then p = jk.

Thus j in <p>  or k in <p>

If j is in <p> then <j> = <p>

If k is in <p> then k = px.

So p = jpx i.e. 1 = jx.

That means <j> = R.

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