Term
Prove that every prime is irreducible in an ID. 

Definition
Lat b be a prime and b = cd.
Then b  cd so bc or bd (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. 


Term
Prove that in a PID, every nontrivial prime ideal is generated by a prime element. 

Definition
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.



Term
Prove that every nontrivial prime ideal is maximal in a PID. 

Definition
See proof in Comp Solutions set. 


Term
Prove that the ideals in Z
are of the form <n> for integer n. 

Definition
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 mqn ∈ I.)
Hence r=0 since r < n and n was smallest > 0 ∈ I.
That means m = qn. So I ⊆ <n>
Hence I = <n>.



Term
What are the ideals in Z/4Z ? Which ideals are maximal? 

Definition
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



Term
Show that coset multiplication in a ring is well defined,
where we define (a+I)(b+I) to be ab+ I. 

Definition
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. 


Term
Herstein pg. 135 #2
Prove that the only ideals in a field are (0) and the field itself. 

Definition
{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. 


Term
Herstein Page 135 #3
Prove that any homomorphism of a field is either an isomophism or takes each element to 0. 

Definition
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 onetoone (thus an isomorphism)
or else ker(φ) = F, in which case φ takes F to {0}. 


Term
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 

Definition
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') = (uu') + (vv') ∈ U+V.
(Since U and V are subgroups, uu' ∈U and vv'∈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.



Term
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. 

Definition
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 u_{1} ... u_{m, } x_{1} ... x_{n} be arbitrary elements of U
and v_{1} ... v_{m, } y_{1} ... y_{n} be arbitrary elements of V.
Then u_{1}v_{1}+...+ u_{m}v_{m} and x_{1}y_{1}+...+ x_{n}y_{n}
represent two arbitrary elements of UV.
Now, 0 ∈ UV since 0 = (0)(0) with 0∈ U, 0∈V.
Also, (u_{1}v_{1}+...+ u_{m}v_{m})  (x_{1}y_{1}+... +x_{n}y_{n})
= u_{1}v_{1}+...+u_{m}vm + (x_{1})y_{1}+...+(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
(u_{1}v_{1}+... + u_{m}v_{m})r = u_{1}v_{1}r +... + u_{m}v_{m}r ∈ 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(u_{1}v_{1}+... + u_{j}v_{j})r = ru_{1}v_{1}+... + ru_{j}v_{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 u_{1}v_{1}+...+ u_{m}v_{m} , an arbitrary element of UV.
For each u_{i}v_{i} in the sum...
u_{i}v_{i} ∈ U since u_{i}∈U, v_{i}∈R and U is a (right) ideal of R.
u_{i}v_{i} ∈ V since v_{i}∈V, u_{i}∈R and V is a (left) ideal of R.
Therefore u_{i}v_{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
u_{1}v_{1}+...+ u_{m}v_{m} ∈ U ∩ V.
We've shown that UV ⊆ U ∩ V.



Term
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'. 

Definition
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 b^{2} = (φ(1))^{2} = φ(1^{2}) = φ(1) = b.
Thus b^{2}  b = 0.
Moreover for any r in R',
b(br  r) = b^{2}r  br = (b^{2}  b)r = 0r = 0.
Since R' in an integral domain,
either b = 0 or brr = 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 brr = 0 which gives br = r i.e. b is the unit in R'.



Term
Herstein Page 136 #14.
For a∈R, let Ra= {xa  x∈R}.
Prove that Ra is a leftideal of R. 

Definition
Herstein Page 136 #14.
For a∈R, let Ra= {xa  x∈R}.
Prove that Ra is a leftideal of R.
First note 0 = 0a ∈ Ra so Ra is nonempty.
Next consider arbitrary xa, ya ∈ Ra.
Then xaya = (xy)a ∈ Ra.
Hence xaya ∈ 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. 


Term
Prove that the intersection of
two left ideals of R is a left ideal of R. 

Definition
Let I_{1} and I_{2} be left ideals of R.
Since I_{2} and I_{2} are subgroups of R,
their intersection is a subgroup of R.
Now consider i in I_{1} ∩ I_{2}.
For any x in R,
x(i) ∈ I_{1} (since i ∈ I_{1} and I_{1} is an ideal of R),
x(i) ∈ I_{2} (since i ∈ I_{2} and I_{2} is an ideal of R).
Thus x(i) ∈ I_{1} ∩ I_{2}.
We've shown that I_{1} ∩ I_{2 } is a left ideal of R.



Term
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. 

Definition
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,
(xy)a = xa  ya = 0  0 = 0. So xy ∈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 twosided ideal.



Term
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. 

Definition
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 onetoone since for any b,c in R,
φ(b) = φ(c) > b1 = c1 > 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)
= (a1)⊕(b1)
= a  1 + b  1 + 1
= (a+b)1.
So φ(a+b) = φ(a) ⊕ φ(b)
φ(ab) = ab  1
φ(a) • φ(b) = (a1)•(b1)
= (a1)(b1) + (a1) + (b1) = 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.



Term
If R is a commutative ring with unitiy and R has exactly two ideals then R is a field. 

Definition
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.



Term
Can a nontrivial ring have only 1 ideal? 

Definition
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. 


Term
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. 

Definition
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. 


Term
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. 

Definition
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 welldefined:
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 onetoone:
φ(n+<p>) = φ(m+<p>)
→ n mod p = m mod p
→ For some r, q_{1}, q_{2}, r = npq_{1} = mpq_{2}
→ n  m = p(q_{1}  q_{2})
→ nm in p
→ n +<p> = m +<p>.
Next show that φ is onto Z_{p}:
For any k in {0,1,...,p1}
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}.



Term
Prove that polynomials over a field form a PID. 

Definition


Term
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 

Definition


Term
Prove Wilson's Theorem (p1)! == 1 (mod p) 

Definition


Term
Show p = 4k + 1 → exists x  x^2 = 1 (mod p) 

Definition


Term
In an ID
if ax = x for some a, and x!=0,
then the ID has a multiplicative identity. 

Definition
Assume ax = x, x!=0/
Let y be arbitrary.
ax = x
0 = y0 = y(axx) = axy  xy = x(ay  y)
Since ID and x != 0 > ay = y.
So a acts as identity on y. 


Term
Prove that <x^2 + 1> is a maximal ideal in Z_{11}[x] and that Z_{11}[x] / <x^2 + 1> is a finite field with 121 elements. 

Definition
x^2 + 1 is irreducible over Z_{11}.
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 Z_{11}. 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 


Term
Prove that in a PID an irreducible element generates a maximal ideal. 

Definition
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



Term
Herstein 3.10.4
Show that fp(x) = x^{p} + x^{p1} +...+ x + 1
is irreducible over Q, where p is a prime.


Definition
Problem 3 For p prime, f_{p1}(x) = x^{p1} +...+ x^{1}+x^{0}
is irreducible over Q.
By Lemma: f_{p1}(x+1) = C(p,p)x^{p1}^{ }+...+ C(p,2)x^{1} + C(p,1)x^{0}.
Now since p is prime, p  C(p,k) for 1 ≤ k ≤ p1.
That's true because the numerator of p! /(k!(pk)!)
has p as a factor and the denominator doesn't.
Moreover p ∤ C(p,p) = 1 and p^{2 }∤ C(p,1) = p.
So, by Eisenstein's criterion, f_{p1}(x+1) is irreducible over Q.
That means f_{p1}(x) is also irreducible over Q, since
f_{p1}(x+1)=g(x)h(x) → f_{p}(x)=g(x1)h(x1),
with degrees unchanged when we substitute x by x1.
Lemma: If f_{n}(x) = x^{n} + x^{n1}+ x^{n2} +...+ x + 1 then
f_{n}(x+1) = C(n+1,n+1)x^{n}^{ }+...+ C(n+1,2)x^{1} + C(n+1,1)x^{0}
Basis step n = 0: f_{0}(x+1) = 1 = C(0+1,0+1)x^{0}.
Inductive step: Suppose f_{k}(x+1) = C(k+1,k+1)x^{k}^{ }+...+ C(k+1,1)x^{0}.
Then f_{k+1}(x+1) = (x+1)^{k+1} + C(k+1,k+1)x^{k}^{ }+...+ C(k+1,1)x^{0}
= 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))x^{0}]
= x^{k+1} + C(k+2,k+1)x^{k} + ... + C(k+2,1)x^{0}
= C(k+2,k+2)x^{k+1} + C(k+2,k+1)x^{k} + ... + C(k+2,1)x^{0}.



Term
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. 

Definition
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_{m}x^{m} +...+a_{0}x^{0}
g(x) = b_{n}x^{n} +...+b_{0}x^{0}
where a_{m},..., a_{0}, b_{n} ,..., b_{0} ∈ R, a_{m}≠0≠b_{n}
and m=deg(f), n=deg(g).
The leading term of f(x)g(x) will be a_{m}b_{n}x^{m+n}
with a_{m}b_{n} nonzero since R is an integral domain.
Thus m+n= deg(fg) = deg(1) = 0.
So m = n = 0.
Then f(x)=a_{0}∈R, g(x)=b_{0}∈R with a_{0}b_{0} =1.
So f and g are units in R.



Term
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 f_{0}(x)/a wth f_{0}(x) in R[x] and a in R.


Definition
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 f_{0}(x)/a wth f_{0}(x) in R[x] and a in R.
f(x) may be written (a_{n}/b_{n})x^{n }+...+ (a_{0} /b_{0})x^{0 }
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):
f_{0}(x) = (b_{0}...b_{n})f(x) =
= (b_{0}..b_{n}a_{n}/b_{n})x^{n }+..+(b_{0}..b_{i}..b_{n}a_{i}/b_{i})x^{i} +..+ (b_{0}..b_{n}a_{0}/b_{0})x^{0 }
= (b_{0}...b_{n1}a_{n})x^{n }+..+ (b_{0}...b_{i1}b_{i+1}ab_{n}a_{i})x^{i} ...+ (b_{1}...b_{n}a_{0})x^{0 }.
So f_{0}(x) is in R[x] and f(x) = (1/a)f_{0}(x) as required,
taking a = b_{0}...b_{n}.



Term
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). 

Definition
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
(xf_{1}(x,y)+yg_{1}(x,y))  (xf_{2}(x,y)+yg_{2}(x,y))
= x(f_{1}f_{2})(x,y) + y(g_{1}g_{2})(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.



Term
Prove
In an integral domain, every prime element is irreducible. 

Definition
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. 


Term
Prove
Maximal Ideal is Prime in Commutative Ring with Unity. 

Definition
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. 


Term
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. 

Definition
n·a = 0 for a ≠ 0
→ a(n·b) = n·ab =(n·a)b = 0b = 0
→ n·b = 0



Term
Prove
In a (possibly) nonunital ID of finite characteristic,
the characteristic is prime. 

Definition
The characteristic of a nonunital 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) = kla^{2} = na^{2} = (na)a =0a = 0,
contradicting the fact that A was an integral domain.
Thus, if the characteristic >0, it is necessarily prime. 


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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. 

Definition
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
→ (ab)^{p} = 0 → ab = 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 F_{4} 0, 1, x, 1+x
x^{2} = 1+x.
char(F_{4}) = 2.
φ(x) = 1+x,
φ(1+x) = (1+x)^{2} = 1+ x^{2} = 1+1+x = x.
We have φ surjective for Field of rational functions



Term
Find all the automorphisms of F_{8} 

Definition
F_{8} = Z_{2}[b] where b^{3} = b + 1.
φ(1) = 1 so φ fixes Z_{2}.
Now φ(b) = b > identity
φ(b) = b^{2} is Frobenius
φ(b) = b+1 >
φ(b+1) = b+2 = b
φ(b^{2}) = (b+1)^{2} = b^{2}+1
φ(b^{2}+1) = b^{2}+1 + 1 = b^{2}
φ(b^{2}+b) = b^{2}+1 + b+ 1 = b^{2}+b
φ(b^{2}+b+1) = b^{2}+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) = b^{i} means i 



Term
Prove in a ring
0x = x0 = 0
(ab) = (a)b = a(b)
(a)(b) = ab
If identity exists it's unique 

Definition
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



Term
Prove that if I is a prime ideal of CR1 R,
then R/I is an integral domain. 

Definition
Suppose that I is a prime ideal of R.
Let 


Term
If α is an algebraic number,
prove that there is an n such that
nα is an algegraic integer. 

Definition
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})^{n}x
Then p(y/r^{n}) = 0
So



Term

Definition
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 WellOrdering 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.



Term
Irreducible generate maximal ideal in a PID 

Definition
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



Term
Prime Ideal is Maximal in PID 

Definition
<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.
stack exchange 

