# Shared Flashcard Set

## Details

Ring Proofs JCT
Ring Theory
42
Mathematics
04/19/2016

Term
 Prove that every prime is irreducible in an ID.
Definition
 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.
Term
 Prove that in a PID, every nontrivial prime ideal is generated by a prime element.
Definition
 Let

be a nontrivial prime ideal ⊆ PID R. We have ab ∈

→ a ∈

or b∈

Thus p | ab → p | a or p | b.  p is not a unit since

≠ R. p ≠ 0 since

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 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 ⊆ 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 ⊆ Hence I = .
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 one-to-one (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') = (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.
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 u1 ... um,  x1 ... xn be arbitrary elements of U and v1 ... vm,  y1 ... yn be arbitrary elements of V.   Then  u1v1+...+ umvm  and  x1y1+...+ xnyn    represent two arbitrary elements of UV.    Now, 0 ∈ UV since 0 = (0)(0) with 0∈ U, 0∈V. Also,  (u1v1+...+ umvm) - (x1y1+... +xnyn)  = u1v1+...+umvm + (-x1)y1+...+(-xn)yn ∈ 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+... + umvm)r = u1v1r +... + umvmr ∈ 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+... + ujvj)r = ru1v1+... + rujvj ∈ 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+...+ umvm , an arbitrary element of UV. For each uivi in the sum... uivi ∈ U since ui∈U, vi∈R and U is a (right) ideal of R. uivi ∈ V since vi∈V, ui∈R and V is a (left) ideal of R.   Therefore uivi ∈ 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+...+ umvm ∈ 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 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'.
Term
 Herstein Page 136 #14.   For a∈R, let Ra= {xa | x∈R}. Prove that Ra is a left-ideal of R.
Definition
 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.
Term
 Prove that the intersection of two left ideals of R is a left ideal of R.
Definition
 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.
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, (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.
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 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.
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 ∈ we must have = R. Thus 1∈ ,  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 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 ∈ .   ≠ <0> so we must have = R.   Then 1 ∈ so xy = 1 for some y  ∈  R.   Moreoever y ≠ 0 (otherwise 0 = xy = 1).   So = 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

the ideal of Z consisting of all multiples of p. Prove   a) Z/

is isomorphic to Zp, the ring of integers mod p.     b)  Zp is a field.

Definition
 Herstein Page 140 #3 Let Z be the ring of integers, p a prime number, and

the ideal of Z consisting of all multiples of p. Prove   a) Z/

is isomorphic to Zp, the ring of integers mod p. b) Zp is a field.    a) Define φ: Z/

→ Zp  with φ(n+

) = n mod p (i.e. the remainder when we divide n by p).     First show that φ is well-defined: n +

= m +

→ n - m ∈

→ n - m = pk for some k in Z → n ≅ m (mod p)  → n mod p = m mod p → φ(n+

) = φ(m+

).   Next show that  φ is one-to-one: φ(n+

) = φ(m+

) → 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 +

= m +

.      Next show that  φ is onto Zp: For any k in {0,1,...,p-1} k = k mod p. So φ(k+

) = k.   Next show φ(m+

+ n+

) = φ(m+

) + φ(n+

):   φ(m+

+ n+

) = φ(m+n +

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

) + φ(n+

).   Next show  φ((m+

)( n+

)) = φ(m+

)φ(n+

):   φ((m+

)( n+

)) = φ(mn +

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

)φ(n +

)   We've shown that φ is an isomorphism from Z/

onto Zp. Therefore  Z/

and Zp are isomorphic.     b) Z is a commutative ring with unity. To show that Z/

is a field it suffices to show

maximal in Z.   Since Zp ≅ Z/

it will follow directly that Zp is a field.   First note that

is indeed an ideal of Z: 0 = p(0) ∈

so

is nonempty. pk ∈

→ -(pk) = p(-k) ∈

. So

is an additive subgroup of Z. Further: pk ∈

→ (pk)n = p(kn) ∈

, ∀n ∈ Z. So

is an ideal of Z.   Now we will show that

is maximal: All ideals in Z are of the form , so if there is an ideal between

and Z, we have

⊆ Z for some n ∈ Z. WLOG assume n > 0 since = <-n>. Now p ∈  so p = kn for some k > 0. Then, since p is prime, n = 1 or n = p. If n = 1 then = Z. If n = p then =

. So there is no ideal strictly between

and Z. Hence

is maximal.    We've established that

is maximal in Z.  Therefore Z/

is field, as is Zp.

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
 [image]
Term
 Prove Wilson's Theorem (p-1)! == -1 (mod p)
Definition
Term
 Show p = 4k + 1 → exists x | x^2 = -1 (mod p)
Definition
 use wilson's. See smith.
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(ax-x) = axy - xy = x(ay - y)  Since ID and x != 0 --> ay = y.   So a acts as identity on y.
Term
 Prove that is a maximal ideal in Z11[x] and that Z11[x] /  is a finite field with 121 elements.
Definition
 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
Term
 Prove that in a PID an irreducible element generates a maximal ideal.
Definition
 Let p be an irreducible element and

in .   Then p =  kn. But then k or n is a unit. If k is a unit then n = p/k so in

hence

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

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) = xp + xp-1 +...+ x + 1 is irreducible over Q, where p is a prime.
Definition
 Problem 3  For p prime, fp-1(x) = xp-1 +...+ x1+x0   is irreducible over Q.  By Lemma: fp-1(x+1) = C(p,p)xp-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, fp-1(x+1) is irreducible over Q.  That means fp-1(x)  is also irreducible over Q, since fp-1(x+1)=g(x)h(x)  →  fp(x)=g(x-1)h(x-1), with degrees unchanged when we substitute x by x-1.   Lemma: If fn(x) = xn + xn-1+ xn-2 +...+ x + 1 then fn(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 fk(x+1) = C(k+1,k+1)xk +...+ C(k+1,1)x0. Then fk+1(x+1) = (x+1)k+1 + C(k+1,k+1)xk +...+ C(k+1,1)x0 = C(k+1,k+1)xk+1 + [(C(k+1,k)+C(k+1,k+1))xk +...+ (C(k+1,1)+C(k+1,0))x0]  =  xk+1 + C(k+2,k+1)xk + ... + C(k+2,1)x0 = C(k+2,k+2)xk+1 + C(k+2,k+1)xk + ... + C(k+2,1)x0.
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) = amxm +...+a0x0 g(x) = bnxn +...+b0x0 where am,..., a0, bn ,..., b0 ∈ R, am≠0≠bn  and m=deg(f),  n=deg(g).  The leading term of f(x)g(x) will be ambnxm+n with ambn 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.
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 f0(x)/a wth f0(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 f0(x)/a wth f0(x) in R[x] and a in R.   f(x) may be written (an/bn)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...bn)f(x) =  = (b0..bnan/bn)xn +..+(b0..bi..bnai/bi)xi +..+ (b0..bna0/b0)x0  = (b0...bn-1an)xn +..+ (b0...bi-1bi+1abnai)xi ...+ (b1...bna0)x0 . So f0(x) is in R[x] and f(x) = (1/a)f0(x) as required, taking a = b0...bn.
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 ={xf(x,y)+yg(x,y) s.t. f(x,y), g(x,y) ∈ F[x,y]}.   I'll first show that is an ideal of F[x,y].   is nonempty since 0 = x0 + y0 ∈ 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) ∈ . To see that 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))∈.   Now I'll show that is not principal.   Suppose, by way of contradiction, that =, f(x,y) ∈ F[x,y]. That means x and y are in . (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   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 cannot be equal to any principal ideal . Thus F[x,y] is not a PID.
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 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.
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 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.
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 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
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 Prove In a (possibly) non-unital ID of finite characteristic, the characteristic is prime.
Definition
 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, 00, 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 1p =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 Zp . Here nothing maps to x since powers of x are only increased by φ. However if F(x) is perfected = F(x1/p^∞) then we get surjective e.g. φ(x1/p) = x.   If R is an ID we still get injective since ap = bp  → ap - bp = 0  →  (a-b)p = 0 →  a-b = 0 →  a = b.   Suppose R is not an ID e.g. Fp[x]/ The  φ(x) = xp + = = 0p + = 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
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 Find all the automorphisms of F8
Definition
 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 |
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 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
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 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
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 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/an)nx   Then p(y/rn) = 0 So
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 Prove Every ED is a PID
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∖{0R}→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.
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 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
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 Prime Ideal is Maximal in PID
Definition


is prime ideal in R. Suppose  contains

. Then p = jk. Thus j in

or k in

If j is in

then =

If k is in

then k = px. So p = jpx i.e. 1 = jx.   That means = R. stack exchange

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