33
MATH 4281 Introduction to Modern Algebra 3/3
MATH 4281Introduction to Modern Algebra
3/3
13. Rings ,
modules.
and algebras ( Ch.
19 )
* RingsDef . Let A be a set with two operations
• and t . Then ( A,
t,
. ) is called a ring if- ( A
, t ) is an abelian group ,
- • is associative,
- for tab, c EA ,
a Cbtc ) = abtac and
( btc )a= bat ca .C distribution law )
Def . Let CA,
t,
. ? bea ring . If there exists an
identity element of CA,
. ) ,then it is called
the multiplicative identity of CA ,t
.
. ).
denoted I.
In this case ( A. t.
. ) is called
a ring with unity or a ring with I.
Convention,
From now on , a ring always have 1.
e . g .
21,
IR.
Q is a ring .
But 221 is not a ring .
Def . Let CA , -1 .. ) be a ring -
Then,
• I A.
t.
. ) is called commutative if . is commutative.
• The identity element of ( A. t ) is called
the additive identity of CA,
t.
. ) .
denoted O.
Lemma. Let A he a ring .
I ) ta E A, a . o = o . a = o .
2) V-a.be A ,C - a ) . b= a - C- b ) = - ab .
3) V-a.BE A , C- a ) . C - b ) = ab.
pf . I ) A1 = A . a -
- ACA -10 ) = data . o = Atta . o.
-
'
. A . O = o . Similarly O . a = O .
2) abt C- a ) . b = ( a - a ) . b= o . b=o.
-
'
. ( - a ) . b= ab. Similarly a - C- b) = - ab
.
37 C- a) C- b) = - ( at - b ) ) = -C - Cab ) ) = ab . xD.
Lemma. Let A be a ring
. Then A = 903 if
and only if 0=1 in A .
pf . If A = lol ,then o is also the multiplicative identity .
Conversely , if 1=0 then ta EA ,
A= A . I = a - o = o.
a
'
. A = fo ).
Def . Let ( A. t .o ) be a ring .
- If a. BEA satisfy Ato , bto,
and ab=o,
then
a. b are called zero - divisors of A.
. If A is commutative and does not have a zero - divisor,
then it is called an I integral ) domain.
- If AEA satisfy a"
-_ o for some n E Zso,
then a is called nilpotent .
Lem.
A is an integral domain if and only if
for #a. b. c E A at . ato
,ab =ac ⇐ be ⇐ ba -
- Ca.
Pf . If A is an integral domain,
then
ab= ac ⇐ acts - a ) =o # b - co ⇐ b -- c .
Similarly b=c €7 ba-
- Ca . Conversely ,
if the condition above holds then
ab -_ o 2=7 ab -
- do -
- ob ←→ a=o or 6=0 .
Thus A is an integral domain.
IDK
e. g . 1) ( 21,
t,
. ).
C IQ ,t ,
. ),
C IR,
-1,
- 7,
C Cl,
-1 .
. ) .
2) ( In,
t,
. ).
3) Mat nxn ( ? ) where ? is a ring .
4) Ht : { atbitcjtdk I a. b. c. DER },
i. j . k behave like elements in ⑦ 8
5) For a set A . PCA ) w/ operationsXt Y = I X -
' C) UH - X ) ,X . Y -
- XNY.
6) For an abeliangp
G.
let End CG ) be
the set of all endomorphisms of G .Then
for fig C- End CG ), Fey is well - defined
.
( ft g) ca ) :-c feast geol . Multiplication is
given by f. g : = fog C composition )
9) . Let A be a ring and X be a set .
Then the set of functions from X to A,
denoted Map L X. A ) is a ringItg) C a) = fca ) + goal . Cfg )Ca3= fed gca )
.
→ Map I IR .1127
, Maple . R ). Map ( R
.Z )
,. - -
8) functions w/ special properties . e. g.
{ f : Cl → fl f is anti,
or
f is holomorphic , or
f is constant , etc. }
Ck ( IR ) : -- I f : IR - Rl f -
is differentiable k times
and f' "
is continuous }.
. r. /
Def .
Let A be a ring .
Then
UEA is called a unit if Fu ' EA S.t.
un'
= u 've =L.
We denote the set of
units in A by AX.
Exercise.
show that At is a group .
e. g.DZ/x.-IIl3.QX--Q-9o3.lRX-- R - 103.
. .
.
2) Zn× ?
37 Matron CAI ×= : Glu CA )
.
"
general lineargroup
"
47 End CGI ×= Aut CG )
57
MapC X. AIX = Map IX. At )
Def . Let A be a ring .Then A is called a field
if A is commutative and At = A - Eo }.
( Thus. every non
- zero element in A is a unit . )
e. g .
Q.
IR, E . Xp
Caution.
Do not forget the commutativity condition !
c. e . Ht .
* Modules.
Def .Let ( R
,t ,
. ) be a ring with unity and I 14 .tlbe an
abelian group . Then M is called an R - module if
there exists a map• : Rx M - M
,
usually denoted . Crim ) = r . m,
such that
- for all r ER and m , n EM,
r - ( men ) =r .mx r . n,
- for all r .SER and MEM
,( res ) . m= r . Mt s . m
,
- for all r. SER and me 14
, Crs ) . m = r - Cs - m ),
- for all me M ,I 'M
-
- M -
1/11/16e. g .
1) If A is a ring then
A is itself an A - module -
2) If 14,
N are A - modules.
then
MXN is an A - module, usually denoted
by MAN.
( direct sum of modules ).
By iteration .An i= AAAA . . . AA is an
A - module,
( free module )
3) Every abelian group is a 2- module-
4) If A is a ring ,
then A"
is a
Matnxn CA ) - module -
Def . An A - module M is called an A - vector spaceif A is a field .
e. g . Q "
, R"
,Eh , 21pm . . . .
Lemma.
I ) Hm C- M, o . m = o .
2)ht
a E A, a - o = o .
3) hta C- A ,
ttmEM,
C- a ) - on = a . Gm ) = - am
4) Ka EA ,Vin EM ,
C - a) C - m ) = Am .
pf . Similar to the lemma for rings .
¥ Algebra .
Def. Let ( A. t
.
. ) ,( B. t
.
. ) be rings and
suppose that A is commutative.
Then
B is called an A - algebra if B is an
A - module and for ta EA, VI.YEB ,
they satisfy acxy ) = Cady -_ xcay ) .
In other words.
"
( Algebra ) = ( Ring ) -1 C Module )"
We will give examples later.
14.
Sub objects . Homomorphisms .
Ideals.
and Quotients ( Ch 18.19 )
* Sub objects .
Here we define sub objects of rings .modules
,
and
algebras which are equivalent to subgroups ofgroups .
Def . Let A be a ring .Then BCA is called
a subring of A if B is a ring with respectto t and . inherited from A ,
and
B contains I of A.
Def .
Let A be a ring and te be an
A - module.
Then Nc M is called an
A- submodule of M if N is an
A - module with respect to addition and
Scalar multiplication inherited from 14 .
Def .Let A be a commutative ring and
B be an A - algebra .
Then CCB is called
an A - subalgebra of B if C is an
A - algebra with respect to the algebrastructure inherited from B and
C contains I of B .
How to check something is a sub object ?
- A
CR, R is a ring . Then A is a
subring of R if and only if
I 7V-a.BE
A,
Atb.
ab EA -C closed )
2) o EA,
C additive identity )
37 Ha EA , - a E A ( additive inverse )
4) LEA C unity )
- A a ring ,µ an A - module ,
Note.
Then
N is an A - submodule of 14 if NIX and✓
a EA ,
htx. YEN , ax , xty EN C closed )
Q.
Where are the other conditions ?
- A a ring , B an A - algebra .CCB .
C is an A - subalgebra of B if
I ) ta EA , V-a.ge C, ax
, xxy.xy.EC .
Caked )
2) I E C .
Kulig
* HomomorphismsWe define homomorphisms for rings .
modules,
and
algebras similar to group homomorphisms .
Def .
Let A, B be rings .
Then the function
§ : A → B is called a homomorphism if
I ) tx. YEA , Olney ) = Xcx ) e- cfcy )
2) V-x.ge A , ¢cxy3= daddy ,
3) 0/43 = I.
Def . Let M , N be A - modules.
Then the function
of : M → N is called a homomorphism if
1) txcyc M, cfcxty ) = ok x Ky )
2) Hae A ,txEM , flax ) = a 04×3
.
Def.
Let B ,C be A - algebras . Then the function
¢ : B→ C is calleda homomorphism if
I ) tX. ye B , Cfcxty ) = olcxstcfcy )
2) U-X.yEB.co/cxy3--o/cx7o/cy )
3 ) ht at A , AXE B , cfcax ) = acfcx )
47 0/43=1.
Also we use the terms
monomorphic m . epi morphism . isomorphism .
endomorphism , automorphismsimilar to
group homomorphisms .
* Another definition of algebras .
Let A be a commutative ring and B be
an A - algebra .
Also, let us write
2433 : -_ { a c- BI ay -_ yx tyEB3 .
called the center of B -
Caution.
This notion is ( similar but ) different from
a center of a group . Here,
B is a priorian abelian group .
thus its center as a groupis B itself .
The center we defined here is
with respect to multiplication .not addition .
Now let f : A - B : a - a. I.
Then it is
easy to See that f is a ring homomorphism .and
from the definition of algebras we have
ta EA,
tbEB ,La.Db -
- a.b= acts . D= bca . I ).
Thus FCA ) CZCB ).
i - e,
the image of f- is
contained in the center of B .
Conversely ,if A
, B are rings .
A is commutative,
and f : A - B is a homomorphism et .
FCA ) CZCB ) , then the mapA x B → B : LA,b) 1- fcaib gives
a well-defined scalar multiplication structure,
Thus we may regard B as an A - algebra .
In Sum,
if A Ts a commutative ring we have"
A - algebras"
=
" f : A → B homomorphismset
. FCA ) CZCBJ"
- M ,
* Quotient objects .
Similarly to the definition of quotient groups .
we want to define quotient rings ,modules
.and
algebras .
Let us start with modules.
which are
easier than others -
Let A be a ring , M bean A- module
,and
Nc M be an A - submodule.
we want to givethe " quotient group
"
MIN an A - module structure.
i. e,
ta. YEN ,
tae A, we want to have
It-
y = Fey ,i.e . (xtN)tlytN ) = XtytN
AI = AT, i. e
,a ( Xx N ) = axtN .
But it is always true !
Def ,Let A be a ring ,
14 be an A- nodule,
and
N is an A - submodule of M.
Then an A - module
141N is well-defined ,called the quotient of
14 by N . Also,
the natural mapµ → MIN : M 1- me N is a homomorphism .
This time let us consider rings . Similarly .let
A be a ring and I CA be a subgroup of A .
We want to define a natural ring structure on
A II,
in other words,
for a. BE A ,
Itb = atb , i. e , Cattle ( bit ) = atbtI,
Jib - Tb , i.e.
Latte C BTI ) = abt I,
E. I = I . I = I, i.e ,
( att ) (1+1)=(1+2) Catt ) = ATI.
The first part is clear since I is a normal
subgroup of A .( Recall that A is an abelian
group)
.
But the second and third ones are not trivial .
Indeed , we should have
abt I = ( atI7CbtI ) = abi-Ib-a-L-I.TL,
ATI = ( 1+27 ( att ) = at Ia -11+1 . I
= ( att ) I Itt ) = at It AIT I - I.
Thus, for these statements to hold we should
have that for any at A .at and Ia
are subsets of 2 , ice ,at . Ia CI
.
( From this condition I - ICI is automatic . )
Def . Let A be a ring and ICA be a
subset of A .Then I is called an ideal
of A ifI ) ( I . t ) is a subgroup of A
.
2) KAGA, a. I CI and I - a CI
.
In this case the natural ring structureon Ali is well - defined . which is
called the quotient of A by I.
Also ,the natural map A → At : an att
is a ring homomorphism .
How about algebras ? Let A be a commutative
ring . B be a ring . and f : A → Bbe
a homomorphism such that FCA ) C ZCB ).
In other words . B is an A - algebra .
Since' '
algebra"
=
"
ring" & "
module "
, if we
want to define BIZ for some ICB
then I must be at least an ideal of B.
We claim that then I is also an
A - submodule ; indeed,
for taEA ,
A . I = fca )I or La.1) I CI by the
definition of ideals .Thus
.
in this case
we can also define B12 as an A - algebraSimilarly to the ring case .
Caution.
Let I be an ideal of a ring A.
Then,
I is not necessarily a subring of A,
that is we do not require that LEI.
Indeed , we have
Lem.
I -01 if and only if I -_ A.
pf , ⇐ is obvious. Conversely .
for any AEA
I sat. thus Iaa .1=a
.
~ > ACL.
④
Thus in our convention . Subring and ideals
are ( slightly ) different .
* kernel of homomorphisms .
In the case of groups .
the kernel of a homomorphismis a normal subgroup . Conversely . any normal subgroupcan be considered as the kernel of some homomorphism .
Here we observe that similar thing happens for
rings and modules L thus also algebras ).
Lemma.
Let A be a ring .Then for a subset
Ic A , I is an ideal of A if and only if
there exists a homomorphism f : A - B such that
Ker f -
- I.
pf .⇒ If I is an ideal of A . then it is
the kernel of a natural homomorphism A - At.
⇐ If f : A → B is a homomorphism such that
kerf =L.
then for any AEL and XIYEIwe have
f- City ) =fcxkfcy ) = o .
'
. Xxy EI
flax ) -_ ftafcx ) = o , fcxat-fcxlfca-o.i-ax.NET.
Thus I is an ideal of A .
Lemma.
Let A be a ring and 14 be an A - module.
Then for a subset NCM, N is an
A - submodule of 14 if and only ifthere exists a homomorphism f -
- M → P
such that kerf .
- N .
The proof is similar to above.
I
* fundamental homomorphism theorem.
We also have versions of the fundamental homomorphismtheorem for rings .
nodules.
and algebras .
Them.
Let f i A → B a homomorphism of rings .
Then
im f is a subring of B and the mapI : A / kerf → imf : at kerf - tea )
is an isomorphism of ringsThen . Let A be a ring and f : M - N be
a homomorphism of A - modules.
Then
imf is an A - submodule of N and the map
I : 141 kerf → Imf : me kerf c- fan )
is an isomorphism of A- modules.
The proofs are similar to the group case. 1/4/26
.
* ideals and submodule generated by a subset .
Similar to thegroup case , one can define ideals
and submodule generated by certain subsets.
Def . Let A be a ring and SCA be a subset.
Then we define ( S ) to be the smallest ideal
containing S, called the ideal generated by S
.
Def . For an ideal ICA . we say that z
is principal if I -_ (a) for some AEA . i. e.
I is generated by a single element .
e. g .I ) Co ) = 63
.
G) = A
2) NEZ ,
(a)= [ multiples of a } .
Prop .Let A be a ring and S CA be a subset
.
Then ( S ) is equal to
{ as , bit . . - + Akskbkl KEIN, Ai . . . . .ae
,
be, . . .
. bkEA.si.
. . . SKI }The proof is also similar to the group case .
Def. Let A be a ring ,
14 be an A- module.
and
SCM be a subset . Then we define LS >
to be the smallest A- submodule of 14
containing S,
called the A- submodule generated by S.
Def . Let A be a ring ,14 be an A - module
,
and
NC 14 be an A - submodule. Then Nis called
cyclic if D= ( m > for some MEN.
i. e,
N is
generated by one element .
e.g .
Vector subspace generated by vectors.
# intersection and sums .
Lem.
Let A be a ring and I.
TCA be ideals
of A.
Then It ], INT are ideals of A .
pf . Clearly It 'S. I are ( normal ) subgroups of A .
Also forany
AE A, we have
a C Its ) c aItaJ CJ -1J .
and
a CI AT ) CAI CI,
act ) ca ] CJ .thus
act ) c I AT .
④
* prime and maximal ideals.
Here . we will only consider commutative rings .
Def . Let A be a ( commutative ) ring and ICA
be an ideal. Then I is called a maximal ideal
if I # A and for any ideal JCA .
if To I then J - I or J= A.
In other words.
I is a maximal ideal if it is
maximal among theproper ideals of A with respect
to containment -
Prop .
Let A be a commutative ring and ICA be
an ideal of A. Then At is a field
if and only if I is maximal .
pf . ⇐ For a EA, suppose that of a- Eats .
Then
a # I .
Therefore. It (a) DI
,and by maximally
we have Itoh = A .In particular, IE It Cal
.
Thus F BEA such that ab - I EI.
us a- I- T = o E AIL
,i.e
.a- I -
- T -
- I C- At.
⇒ Let J CA be an ideal at . IET .
Then FAE J - I. Since AIL is a field ,
I be A sit . IT -
- T, i.e
,ab - LEI .
. : To - ( ab - 1) tab =L . ice , f- A.
Cor.
Let A be a commutative ring .
Then A is
a field if and only if to ) CA is maximal.
i. e,
there are only two ideals Sos.
A of A .
pf . Apply the above proposition to the case I - 63.
Def.
Let A be a commutative ring and ICA
be an ideal of A . Then I is called
a prime ideal of A if forany a. BEA
,
if abt I then either a EI or BEI.
It maybe regarded as a
"
partial converse" of
the " absorbing rule "
in the definition of ideals .
Prop .
Let A be a commutative ring and ICA be an ideal
of A. Then AIL is an integral domain if
and only if I is a prime ideal of A .
pf . ⇐ Let a. be A be sit . IT -5=0. Then
ab EI,
thus a C- I or BEI,
i.e .
a- =o onto.
⇒ Let a. BEA be sit . ab EI.
Then aT=o,
thus a- =o or b- =o, i.e
.
AEI or BEI.
Cor. Let A be a commutative ring .
Then it is an
integral domain if and only if { o ) is a prime ideal.
Pf . Apply the proposition above to 1=103 .
Con. Every maximal ideal is prime .
pf . ICA is maximal ⇒ Ali is a field⇒ AIL is an integral domain⇒ I is prime .
Dk
15 . Fields and vector spacesC Ch 28 )
* Properties of fields.
Lem. Let F be a field . Then there are only two
ideals of F, namely l o ) and F itself
Rank.
Note that { o ) C zero ring ) is not a feed :
we require that FX = F -103.
but I of -9-3=14 .
Pf . It is because Co ) is maximal.
④
Cor. Let X : E → A be a ring homomorphism .
If
At Eo ) ,
then X is injective .
pf . As A ¥103,
In -40A.
Thus XCIF ) -
- IA t OA.
which means that I # Ker X .
Since kerfis an ideal of F
,we have Kerry = I -3
.Dk
* Vector spaces .
Def . Let F be a field and V be an F- module.
Then we say that V is finite-dimensional Cor
finitely generated ) if there exists a finite set
Sc V such that V -
- LS >. 1/11/30
We want to classify all the finite - dimentional
F-modules.
Let V be such a module and
choose S CV such that ISI is minimum .
and let S= { vi. Vz, . . . .
Vu } for some NEW .
Than.
The morphismX : F
"
→ U :(Ac,
As.
. . .
,Au ) ↳ Aivitdsvzt c- - tank
is an isomorphism of F- modules.
In other words,
all the finitely generated F- modules
are isomorphic to F"
for some a E IN .
pf . Y is definitely an F- module homomorphism ,and
also surjective by definition - Now suppose that
XC Ai , Az. . . . . An ) =o for Some Al .az .
. . . . An C- A,
Assume that Aito for some LEE Eu. WLOG
,
wemay assume that a , -4 o .
Then we have
A. Vit Azvzt a .. C- AuVu=o , thus
Uct Aza ? Vat . .
. + Quai'
Vu -
- o . In other words,
Vi E ( Va .V3 , . . .
, Un >,
thus( V2
, Vy,
.- . . Un > = L vi. Vz
, - . .
.V = V .
But it contradicts the minimal .
- of 1St.
Thus
A , = As = . . . = An =o, i.e .
Ker =o. T
Prop . Let min EIN be at . men and
X : F"
→ Fm is an F- module homomorphism .
If X is injective , then men and Xis an isomorphism .
pf .We proceed by induction on n
. If n=o . then
we -
- n -
- o and X : 90 } → 9-3 is obviously an
isomorphism .
Now suppose that the statement is
true for n Ekt. If n - K
. i . e,
we have
a moaomorphism Y : Ek → FM.
then consider
the composition Fk - ' ↳Fk 4-Fm where
¢ : ( ai .As
. . . . . Ace ? 1- ( As .az,
. . . . Ae - i. o ) . ¢ is clearlya mono morphism . thus so is X. of .
First we suppose that mtk,
ice .ME Kt
.
Then by induction assumption in -
-Kt and
Xo of is bijective .But it means that X is
surjective .thus X is also bijective .
Thus,
of = X- to C X . of ) is also bijective .
But
im of I C o.O , . . .
. o,
I ),
which is a contradiction.
Thus M=k.
Now we supposethat X is not
surjective .
Let { e. .eu
. . .
. ek 's be a standard basis of
Fk , then there exists kick s 't . imf # ee .
WLOG wemay assume that election X .
Consider the composition map Fk
IsFke- > Ftl
where f i ( Ai , Az.
. . . . Ak ) 1- ( Ac ,Az
.. . .
, Ok - e ).
Then
Ker f = { Colo , a . . , o , Ak ) I AKE F } = Lek > . Also ,
Ker e. y = { ve # I ecxcv ) ) = o }= I VEE
'sI Xu ) C- Ker e }
= 2/-1 ( Ker = 2/4 (
Leks ) However .
Xlv ) taekfor any Afo . Thus
Ker fo 4=0 , i. e . foy is injective . But
it is impossible by the argument above.
Thus Y is bijective .
Cor . If Fm = Eh , then man .
pf .
obvious.
Cor . A finite - dimensional vector spaceis completely
determined by the minimum of the size ofa generating set
.
pf . Combine the results above.
16.
Z and Fcx ]. ( Ch . 21
. 22,24 .25,26 )
* I - the ring of integersIt is a very special ring , in a sense that .
..
Thin . Let R beany ring .
Then there exists a unique
homomorphism X : 21 → R .
pf .We define Xcn )i= n . IR
.
Then it is clearly a
homomorphism ( exercise ! ). Also . if 0 : 21 → R is another
homomorphism . then since Xcel ) = I, thus
n To,
Cf Ln ) = & Clt . . . ti ) = ¢ C ht . . -+0/47 = It . - . c- I -_ n . L,
n Lo, cfcn ) = - 01C - u ) = - ( - n . I ) = n - I
.
h = o , ¢ Co ) = o = o - I.
Thus 4=8 . Da
Q.
Canyou
find a ring R such that forany
ringS there exists a unique homomorphism
s - R ?
What are the ideals of TL ? We know for example( n ) = { multiples of us
.
Are there other ideals ?
Than . Every ideal of 21 is principal .
pf . Suppose that I CTL is an ideal.
If 1=63.
then I = ( o ) thus it is principal .
Thussuppose
otherwise,
ice .I - 903¥ of .
Since I is an
abeliangroup .
it means that In E In 21 > o.
Choose an n such that it is the minimum
amongthe elements in Intl > o .
Then clearly( n ) CI
.
We will show that in fact Cn ) =L .
For the sake of contradiction.
assume that I ? Cu ).
Then F X E I - Cu ).
Now we use"
division of
integers"
to find of EZ, o Er Ln such that
X = qntr .Since X. n E I
, we have
r -
- X - qn E I.
But r cannot be nonzero,
since
otherwise it contradicts the minimality of n. Thus
r=o,
which means X=qn ,i.e
,X E Ca )
.
This is again contradiction since X E I - Cn ).
Exercise,
For any a. BE 2. show that
(a) t (b) = ( god C a. b ) ),
(a) Acb ) = Clear C a. b ) ).
What are prime and maximal ideals of I ?
Prop .
Let ( n ) E Z be an ideal of 21.
Then,
I ) Ln ) is prime if and only if n=o or
n is a prime integer .
2) C n ) is maximal if and only if n is a prime
integer .
pf . First suppose that onto.
If n-
- ab for some
a. BEZ such that K lat, Ibl Llnl
,then
ab E Cn ) but a ¢ Cn ),
b € Cn ),
Thus cu )
is not a prime ideal. Similarly .
if a is
prime .then for
any a BEZ such that
ab ELM, we have n lab thusn/aorn lb ,
which means that a E Ca ) orbe Cn ) .
Thus C u ) is a prime ideal.
Now we still suppose that into and this time assume
that Cu ) is maximal.
Then Cut is prime , thus
n is prime . Conversely suppose that n is prime .
If Cn ) is not maximal .then Fa ETL at .
Ln I I Ca ) E Z . In. particular ,
n = ab for some BEZ.
As n is prime ,either A= It or A= In
.
However,
+ ten (a) = 21 or (a) = Cn ),
which is contradiction.
It remains to check the case n=o . But we
know that Iko ) I Z is an integral domain but
not a field .
Thus Co2 is prime but not maximal.
* FCXT : a polynomial ring .
( F : field )
Indeed ,TL and FCXT share
many properties . First ,
Thur.
Let A be an F - algebra .
Then forany
a EA,
there exists a unique F - algebra
homomorphism 4 : FCXT → A such that Xcx )=a.
pf . Clearly X : F → A : fix ) 1- fca ) is an
F - algebra homomorphism .
"
evaluation at a"
.
Also, it is the unique homomorphism satisfying
the desired property as X. generates
the F- algebra FCXT.
What are the ideals of FCX ] ?
Prop . Every ideal of FCX ] is principal .
Pf . Suppose I C FCXT is an ideal.
If 1=903.
then
I -_ Co ).
Thus suppose otherwise,
i.e . IF 903 .
Choose a polynomial f E I - lo ) of minimum degree .
We claim that I -_ Cf ).
For the sake ofcontradiction
. suppose otherwise and let GEL - Cf ).
Then by" division of polynomials
"
,
I f. RE FIX ]
sit . g = offer ,r -
- o or degr C deg f . Byminimality of degf , we have r=o since
r = of - off E I.
However . then of - off E Cf ),
which contradicts the fact that g E I - Cf ).
④
Exercise .
Forany text . gcx ) E FGS .
show that( fcxi ) + C gem ) = ( gcdcfcxi.ge/D .
and
( fcx ) ) n C gem ) = ( lcmcfcx ), gcx
!) ) .
What are prime and maximal ideals of FCXT ?
Prop .
Let (f) CF Cts be an ideal.
Then.
I ) Cf ) is prime if and only if f- =o or
f- is an irreducible polynomial .
2) Cf ) is maximal if and only if f is an
irreducible polynomial .
Pf . First suppose that f- to.
Then,
if f=ghfor some g.
HE FCXT such that deg g. deghzts ,
then gh E Lf ) but g. h # Cf ). Conversely ,
if
f is an irreducible polynomial then For any
g. HE FED such that gh E Cf ) . we have
f- I gh ⇒ fly or fth ⇒ g E Cf ) or he Cf ).
Thus f is a prime ideal.
Now suppose that f is irreducible.
If Lf ) is
not maximal ,
then A g C- FED at . Cf ) Ecg ) FED.
In particular .The FAD at . gh=f .
Since
f- is irreducible, g
-
- of for some CEFCXT or
g EF .
But then (g) = FCX ] or Cgl = Cfl.
which is a contradiction. Conversely if Cf ) is
maximal ,then if ) is prime ,
thus f is irreducible.
It remains to consider the case when f=o . But
Falco ) = FCX ] is an integral domain but not a
field . Thus Co ) is prime but not maximal. ME
* Why does this similarity happen ?
Def .
Let R be an integral domain.
Then it
is called an Euclidean domain if there is
a function d : R - 903 → IN at .
is dca ) Ed Cab ) forany a. BE R - I -3
.
and
② for all a ER and be R - fo }. Fq .
rER
such that a=bqtr and either r=o ordcrkdcb )
.
Lem. Z and FCXT are ED
.
pf .For 21
,
D= absolute value.
For Ex ] .
d- - deg .
Def . Let A be an integral domain . Then it
is called a principal ideal domain if everyideal of A is principal .
Them. Let A be an ED .
Then A is a PID.
pf .Let I be an ideal and choose of AEI such
that dca ) is the minimum.
We show that I -- la )
.
To this end. suppose BEI
,Then F g. r EA
such that b = Agerand either r -
- o
ordersL dca) . Also, r= b- age EI
.
Thus
by minimality of dca ) ,der ) cannot be
strictly smaller than dca ),
which means r=o
Thus b -- age E Ca ) .
In other words.
I -- Ca ) . to
Thus. both TL and FAD arePIPS .
( Compare precious proofs with the above ! )
Def .Let A be an integral domain
. Then AEA
is called a prime element of A if a # o ,
a is not a unit ,and la ) CA is a prime
ideal,
ice.
for anyb. c EA if
alba then alb or alc .
e. g . prime numbers in Z .
Def . Let A be an integral domain.
Then AEAis called an irreducible element of A if
ato, a is not a unit ,
and for anyb. a e A
s .t . be a .
then b or c is a unit .
e. g .irreducible polynomials in FH ]
.
14%Lem
. Every prime element is irreducible.
Pf . If a is prime and bea.
then
a Ibc,
thus alb or ale by assumption .
Since bc=a. it means that b or c
is a unit .TO
In general.the converse is not true
.
But .. .
Them.
Let A be a PID and p-to.pe/AX.ThenTFAE :
CDCp ) is a prime ideal. i.e . p is a prime element
.
(2) p is an irreducible element .
G ) Lp ) is a maximal ideal.
pf .G) ⇒ Css is the lemma above
,and
(3) ⇒ Cl ) is what we proved a couple of lectures
ago. Thus it suffices to
show
(2) ⇒ (3) .
Thus let A be a PID and o # PEA be an
irreducible element .We prove that Cp) is maximal .
Thussuppose
J Z Cp ) and J= C q ).
Then
(f) a
p . thus p-
- ga for some AEA .
But it means that q or a is a unit .
If a is a unit,
then Cp ) -
- Cq ) which is
a contradiction . thus of is a unit . But then
J = (g) = A . In other words.
there is no
ideal " between Cp ) and A"
,Thus Cp> is max
.
Thus in particular . prime = irreducible in PID.
e . g . prime numbers EZ .irreducible polys in FAD
.
Rink.
Let
2/[53-2]:= E
atbrila.BE23 be a subring
of Cl,
We first show that 2 is irreducible in Zaki ].
Suppose that 2 = C atbri ) C Ctdri ). By
taking absolute values.
we have
4=1217latbrifilctdrif = Case+362 ) ( of +3£ ).
But 8+362 ¥2 and chest f- 2.
thus
either a- t 362=1 or Et3d=l, i. e
,
( at 3fbi ) = It or ( at 3rd -21=11.
In other
Words.
either at 3fbi or Ct 3rd i is a unit .
Thus 2 is irreducible.
On the other hand,
2 is not prime since 214 -_ CHP Ct Bi ) but
4+53 12 ¢ Zaki ] and CI - Bill # 21053.
i.e.
2X tri and 2T I - Fsi .
* Unique factorization domain.
Def . Let A be an integral domain.
Then it is called
a unique factorization domain ifI ) for any
of XEA can be expressed as
X= upipz -. - Pr where u EAX is a unit and
p . . . . . . Pr EA are irreducible elements of A,
2) Such an expression is"
unique"
in a sense that if
up , .
. . Pr = v go. . - Gs for some u.VE At , r.SE/N
.
Phi..
. prig , . . .
. qs EA irreducible elements . then r=S
and we may reorder of , . . .
.
. qs such that
for anylEiEr=S
. pi and qi are associates ,
i.e. there exists Ui EA ' such that pi
-
- wifi .
We will not prove the following theorem,
but it is
useful to understand structure of certain rings .
Them. Every PID is a UFD .
Thou.
If A is a UFD .
then Act is a OED.
e - g .
Since FAD is a PID . it is a UFD .
Or.
since F is a UFD , FAD is a OED .
→ unique factorization of polynomials into irreducible s
* Adjoining elements to a ring .
Suppose that A is a commutative ring and
consider an A . algebra of ( multivariate ) polynomialsin Xi , Xz ,
. . .
, Xu with coefficients in A.
We denote such an algebra by Alexi.
Xz.
. .. . Xu ] .
Now forany commutative A - algebra B and for
anybi
.be
, . . . .
bn E B, we define the
"
evaluation
homorphism"
X : ACX, ,
. ..
, Xu ] → B i fix .
. .. ,Xn ) 1- fcbi
,. - . . bn )
,
which is a unique A - algebra homomorphismsatisfying Xcx ,
) - b . . Xan = bz,
. . .
. Xcxn ) = bn.
Def.
Let B be a commutative ring and A be
a subring of B such that A CB.
( Thus in particular B is an A - algebra . )
For be,
. .
. , bn E B , we write Acb , .bz . .. .
. bn ]
to be the A - subalgebra of B . defined
as the image of the evaluation homomorphismACX . , . . . . XD → B ! fcx , .
. . . . Xu ) tsfcbi.
. . .. bn ) .
Exercise.
Show that Acb . . . . .
.ba ] CB is the smallest
A- subalgebra of B containing bi.ba.
. . .
. bn. i. e
. it is the
A - subalgebra.
' generated by bi.
.
. . .bn
"
.
e.
g.DZ/CmtJ=ETn-kEQ/keIN.nEZ
),
Z [ Ii IT = ZET ]
2) Ici ] := E attic f / a. BEZ }"
ring of Gaussian integers"
37 . Crs ) : = E atbr ER I a. BE Q 3 .
4) Rci ] = Cl.
