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of R as is the inverse of ±. The identity function id = idR : R ’’ R is an automorphism. Hence
Aut(R) forms a group under composition. The argument for AutR0 (R) is similar.
Proposition 1.53. Let R be one of the core rings Z or Z/n with n > 1. Then
(i) The only automorphism of R is the identity, i.e., Aut(R) = {id}.
(ii) If S is a ring containing a core ring R and ± ∈ Aut(S), then ± restricts to the identity
on R, i.e., ±(r) = r for all r ∈ R. Hence, Aut(S) = AutR (S).
Proof. (i) For such a core ring R, every element has the form k1 for some k ∈ Z. For an
automorphism ± of R,
±
 ±(1) + · · · + ±(1) if k > 0,




 k
±(k1) = ’(±(1) + · · · + ±(1)) if k < 0,



 ’k


±(0) if k = 0
±
 1 + · · · + 1 if k > 0,




 k
= ’(1 + · · · + 1) if k < 0,



 ’k

 0 if k = 0

=k1.
1.5. AUTOMORPHISMS OF RINGS AND FIELDS 17

Thus ± = id.
(ii) For ± ∈ Aut(S), ±(1) = 1 and a similar argument to that for (i) shows that ±(r) = r for all
r ∈ R.
Proposition 1.54. Let D be an integral domain and ± : D ’’ D be an automorphism.
Then the induced homomorphism gives an automorphism ±— : Fr(D) ’’ Fr(D).
Proof. Given ±, the induced homomorphism ±— : Fr(D) ’’ Fr(D) exists and we need
to show it has an inverse. The inverse automorphism ±’1 : D ’’ D also gives rise to an
induced homomorphism (±’1 )— : Fr(D) ’’ Fr(D). Since ±’1 —¦ ± = id = ± —¦ ±’1 , we can apply
Corollary 1.19 to show that
(±’1 )— —¦ (±)— = id = (±)— —¦ (±’1 )— .
Hence (±)— is invertible with inverse (±’1 )— .
Corollary 1.55. There is a monomorphism of groups
( )— : Aut(D) ’’ Aut(Fr(D)); ± ’’ ±— .
Note that this monomorphism need not be an epimorphism since it is possible that an
automorphism of Fr(D) could map some elements of D ⊆ Fr(D) out of D.
Example 1.56. The ¬eld of fractions of the ring of integers Z is the ¬eld of rationals Q.
The homomorphism
( )— : Aut(Z) ’’ Aut(Q); ± ’’ ±—
is an isomorphism and hence Aut(Q) = {id}.
Combining this example with Proposition 1.53(ii) we obtain a result which will prove useful
later.
Proposition 1.57. Let k be one of the prime ¬elds Q or Fp with p > 0 prime. If R is a
ring containing k as a subring, then every automorphism of R restricts to the identity on k, i.e.,
Aut(R) = Aut| (R).
Let k be a ¬eld. The group of invertible 2 — 2 matrices over k is the 2 — 2 general linear
group over k,
a11 a12
GL2 (k) = : aij ∈ k, a11 a22 ’ a12 a21 = 0
a21 a22
The scalar matrices form a normal subgroup
Scal2 (k) = {diag(t, t) : t ∈ k, t = 0} GL2 (k).
The quotient group is called the 2 — 2 projective general linear group over k,
PGL2 (k) = GL2 (k)/ Scal2 (k).
Notice that GL2 (k) has another interesting subgroup called the a¬ne subgroup,
ab
A¬ 1 (k) = : a, b ∈ k, a = 0 GL2 (k).
01
Example 1.58. Let k be a ¬eld and X an indeterminate. Then Aut| (k[X]) and hence
Aut| (k(X)), contains a subgroup isomorphic to A¬ 1 (k). In fact, Aut| (k[X]) ∼ A¬ 1 (k).
=
Proof. We begin by showing that to each a¬ne matrix
ab
A= ∈ A¬ 1 (k)
01
there is an associated automorphism ±A : k[X] ’’ k[X].
For this we use the element aX + b ∈ k[X] together with the extension result of Theo-
rem 1.21(i) to obtain a homomorphism ±A : k[X] ’’ k[X] with ±A (X) = aX + b. Using the
inverse matrix
a’1 ’a’1 b
’1
A=
0 1
18 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

we similarly obtain a homomorphism ±A’1 : k[X] ’’ k[X] for which ±A’1 (X) = a’1 X ’ a’1 b.
Using the same line of argument as in the proof of Proposition 1.54 (or doing a direct calculation)
we see that ±A’1 is the inverse of ±A an so ±A ∈ Aut| (k[X]). It is straightforward to check
that for A1 , A2 ∈ A¬ 1 (k),
±A2 A1 = ±A1 —¦ ±A2 ,
(note the order!) hence there is a homomorphism of groups
A¬ 1 (k) ’’ Aut| (k[X]); A ’’ ±A’1 ,
which is easily seen to be a monomorphism. Composing with ( )— we see that there is a
monomorphism A¬ 1 (k) ’’ Aut| (k(X)). In fact, this is also an epimorphism and we leave the
proof of this as an exercise.
Example 1.59. Let k be a ¬eld and X an indeterminate. Then
(i) Aut| (k(X)) contains a subgroup isomorphic to PGL2 (k).
(ii) In fact, Aut| (k(X)) ∼ PGL2 (k).
=

Proof. (i) We begin by showing that to each invertible matrix
™™¦


a11 a12
A= ∈ GL2 (k)
a21 a22
there is an associated automorphism ±A : k(X) ’’ k(X).
We begin by choosing the element (a11 X + a12 )/(a21 X + a22 ) ∈ k(X) and then using Theo-
rem 1.21(i) to obtain a homomorphism k[X] ’’ k(X) that sends X to (a11 X+a12 )/(a21 X+a22 ).
By applying ( )— to this we obtain a homomorphism (known as a fractional linear transforma-
tion) ±A : k(X) ’’ k(X) for which
a11 X + a12
±A (X) = .
a21 X + a22
Again we ¬nd that
±A2 A1 = ±A1 —¦ ±A2 .
’1
There is an associated homomorphism of groups GL2 (k) ’’ Aut| (k(X)) sending A to ±A .
However, this is not an injection in general since for each scalar matrix diag(t, t),
tX
±diag(t,t) (X) = = X,
t
showing that ±diag(t,t) is the identity function.
In fact it is easy to see that Scal2 (k) GL2 (k) is the kernel of this homomorphism. Therefore
passing to the quotient PGL2 (k) = GL2 (k)/ Scal2 (k) we obtain a monomorphism PGL2 (k) ’’
Aut| (k(X)). There is one case where Scal2 (k) is the trivial group, namely k = F2 .
(ii) To show that every automorphism of k(X) is a fractional linear transformation is less
elementary. We give a sketch proof for the case of k = C; actually this argument can be modi¬ed
to work for any algebraically closed ¬eld, but an easy argument then shows the general case.
Let ± ∈ AutC (C(X)). There is an associated rational (hence meromorphic) function f given
by z ’’ f (z), where ±(X) = f (X), de¬ned on C with the poles of f deleted. If we write
p(X)
f (X) =
q(X)
where p(X), q(X) ∈ C[X] have no common factors of positive degree, then the order of f (X) is
ord f = max{deg p(X), deg q(X)}.
Now let c ∈ C. Then the number of solutions counted with algebraic multiplicity of the equation
f (z) = c turns out to be ord f . Also, if deg p(X) deg q(X) then the number of poles of f
1.5. AUTOMORPHISMS OF RINGS AND FIELDS 19

counted with algebraic multiplicity is also ord f . Finally, if deg p(X) > deg q(X) then we can
write
p0 (X)
f (X) = p1 (X) + ,
q(X)
where p0 (X), p1 (X) ∈ C[X] and deg p0 (X) < deg q(X). Then the number of poles of f counted
with algebraic multiplicity is
p0
deg p1 (X) + ord .
q
Now it is easy to see that since ± is invertible so is the function f . But this can only happen
if the function f is injective which means that all of these numbers must be 1, hence ord f = 1.
Thus
aX + b
f (X) = = constant
cX + d
ab
and the matrix must be invertible.
cd
Clearly not every fractional linear transformation ±A : k(X) ’’ k(X) maps polynomials to
polynomials so ( )— : Aut| (k[X]) ’’ Aut| (k(X)) is not an epimorphism.
Now we turn to a more familiar ¬eld R, the real numbers.
Proposition 1.60. The only automorphism of the ¬eld R is the identity function, hence
Aut(R) = {id}.

Proof. First we note that Q ⊆ R is a subring and if ± ∈ Aut(R) then ±(q) = q for
™™¦

q ∈ Q by Example 1.56.
We recall from Analysis that the rational numbers are dense in the real numbers in the
sense that each r ∈ R can be expressed as a limit r = limn’∞ qn , where qn ∈ Q. Then for a
continuous function f : R ’’ R, its value at r depends on its values on Q since
f (r) = f ( lim qn ) = lim f (qn ).
n’∞ n’∞
We will show that an automorphism ± ∈ Aut(R) is continuous.
First recall that for x, y ∈ R,
y ’ x = t2 for some non-zero t ∈ R.
x<y ⇐’ 0<y’x ⇐’
Now for ± ∈ Aut(R) and s ∈ R, we have ±(s2 ) = ±(s)2 . Hence,
x < y =’ ±(y) ’ ±(x) = ±(t)2 for some non-zero t ∈ R =’ ±(x) < ±(y).
So ± preserves order and ¬xes rational numbers.
Now let x ∈ R and µ > 0. Then we can choose a rational number q such that 0 < q µ.
Taking δ = q we ¬nd that for y ∈ R with |y ’ x| < δ (i.e., ’δ < y ’ x < δ) we have
’δ = ±(’δ) < ±(y) ’ ±(x) < ±(δ) = δ,
hence
|±(y) ’ ±(x)| < δ µ.
This shows that ± is continuous at x.
Thus every automorphism of R is continuous function which ¬xes all the rational numbers,
hence it must be the identity function.
Remark 1.61. If we try to determine Aut(C) the answer turns out to be much more com-
plicated. It is easy to see that complex conjugation ( ) : C ’’ C is an automorphism of C
and ¬xes every real number, i.e., ( ) ∈ AutR (C); in fact, AutR (C) = {id, ( )}. However, it is
not true that every ± ∈ Aut(C) ¬xes every real number! The automorphism group Aut(C) is
actually enormous but it is hard to ¬nd an explicit element other than id and ( ). Note that
given an automorphism ± ∈ Aut(C), the composition ± —¦ ( ) —¦ ±’1 is also self inverse, so there
are many elements of order 2 in the group Aut(C).
20 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Exercises on Chapter 1

1.1. Let R be a ring. Show that
{n ∈ Z : n > 0 and n1 = 0} = {n ∈ Z : n > 0 and nr = 0 for all r ∈ R}.
Deduce that if char R > 0 then these sets are non-empty and
char R = min{n ∈ Z : n > 0 and nr = 0 for all r ∈ R}.

1.2. Let R be an integral domain.
(a) Show that every subring S ⊆ R is also an integral domain. What is the relationship
between char S and char R ?
(b) If R is a ¬eld, give an example to show that a subring of R need not be a ¬eld.
1.3. For each of the following rings R, ¬nd the characteristic char R and the characteristic
subring of R. Determine which of these rings is an integral domain. In (b) and (c), A is a
commutative ring.
(a) Any subring R ⊆ C.
(b) The polynomial ring R = A[X].
(c) The ring of n — n matrices over A,
±® 

 a11 . . . a1n 
 
. . :a ∈A .
..
R = Matn (A) = ° . . » ij
.
. . 
 
an1 . . . ann

1.4. If R is a commutative ring with unit containing the prime ¬eld Fp for some prime p > 0,
show that the function • : R ’’ R given by •(t) = tp , de¬nes a ring homomorphism. Give
examples to show that • need not be surjective or injective.
1.5. Let R and S be rings with unity and Q S a prime ideal.
(a) If • : R ’’ S is a ring homomorphism, show that
•’1 Q = {r ∈ R : •(r) ∈ Q} ⊆ R
is a prime ideal of R.
(b) If R ⊆ S is a subring, show that Q © R is a prime ideal of R.
(c) If the word ˜prime™ is replaced by ˜maximal™ throughout, are the results in parts (a)
and (b) still true? [Hint: look for a counterexample.]
(d) If R ⊆ S is a subring and P R is a maximal ideal, suppose that Q S is a prime ideal
for which P ⊆ Q. Show that Q © R = P .
1.6. Let k be a ¬eld, R be a ring with unit and • : k ’’ R a ring homomorphism. Show that
• is a monomorphism.
1.7. Consider the sets
Z(i) = {u + vi : u, v ∈ Z} ⊆ C, Q(i) = {u + vi : u, v ∈ Q} ⊆ C.
(a) Show that Z(i) and Q(i) are subrings of C. Also show that Z(i) is an integral domain,
Q(i) is a ¬eld and Z(i) is a subring of Q(i).
(b) Show that the inclusion homomorphism inc : Z(i) ’’ Q(i) extends to a monomorphism
inc— : Fr(Z(i)) ’’ Q(i).
(c) Show that inc— is an isomorphism, so Fr(Z(i)) = Q(i).
1.8. Let R be a commutative ring.
(a) If a, b ∈ R, show that there is a unique ring homomorphism ψa,b : R[X] ’’ R[X] for
which ψa,b (r) = r if r ∈ R and ψa,b (X) = aX + b. If c, d ∈ R, determine ψa,b —¦ ψc,d . If
a is a unit, show that ψa,b is an isomorphism and ¬nd its inverse.
(b) Now suppose that R = k is a ¬eld and a, b ∈ k with a = 0. Prove the following.
(i) If f (X) ∈ k[X], the deg ψa,b (f (X)) = deg f (X).
21

(ii) If p(X) ∈ k[X] is a prime then so is ψa,b (p(X)).
(iii) If p(X) ∈ k[X] is an irreducible then so is ψa,b (p(X)).
1.9. Let k be a ¬eld and k[[X]] be the set consisting of all power series

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