Download Field _ extensions

CHAPTER THREE-
Field _ extensions
1 ,if
Galois's original theory was co{i~ed in terms of polynof!}ia1s over the complex field. The modern approach is a_
-/consequence of the methods used, in the 1920's and 1930's,
. to generalize the theory to arbitrary fields. From this viewpoint the central object of study ceases to be a
polynomial,
and becomes instead - a 'field extension' related to a polynomial. Every polynomial f over a field K defines
-another
_ field L containing K (or at any rate a subfield isomorphic to K). There are considerable advantages in setting up
the theory from this field"-theoretic point of view, and ihtroducing polynomials at a later stage.
In this chapter we shall define field extensions and explain the link with polynomials. We shall also classify
certain basic types of extension and give methods by which these may be constructed.
Field extensions
Roughly speaking, a field L is an extension of another field Kif K is a subfield ofL. For technic"al reasons this
definition
- is too restrictive ; we wish to-allow cases where L contains'a subfield ~to K, but not necessarily equal.
Definition. A field extension is a monomorphism i: K ~ L
33
34
GALOIS THEORY
where K and L are fields.
K is the small field, L the large field.
Notice that with a stric~ set~thearetic definitiqn af function, the map i determines bath K and L.
Examples
'1 The inclusian maps i1: Q -+" ~ i2: R -+" C, and i3 : Q ~ C are aU-held extensians.
2 If K is any field, and K(t) is the field ()fratianal expressians aver K, there is a natural manomorphismi: K -4 K(t)
mappirg each elelnent of K to' the carrespanding constant polynamial. This is again a field extensian.
3 Let P be the setaf real numbers of the farm p +
(p
*" qJ2)
-1
q..j2 where p, q E Q.P is a subfield af R, since
= p2!. 2q2 p2:! 2q2J2
if p and q are nan-zerO'. The inclusian map i: Q -+" P is a
field extensian.
'
If i: K -+" L is a field extensian, we can usually identify K with its image i(K), sa that i can be thaught of as an inclusion
x:
map and K can be thaught,ofas a subfield of L. V"hoer these
circumstances we use the natatian
L:K'
for the extension, and say that L is an extension of K'.
In future we shall make the, identification orK and i(K)
whenever this is legitimate.'
,. The next cancept is one whic~pervades much of abstract algebra:
Definition. Let K be a field, X a non-empty s~b~et of K. Thoo the subfield of K generated by X is the intersection of all
subfields of K which contain X.
The reader should convince himself that' this definitian is equivalent to either af the fallowing:
r
FIELD EXTENSIONS
(a} The smallest subfield of K which contains X ..
'4 (b) The set of all elements of K which can be obtained fr()~ elements of X by·a fini te sequence 'of field
operations.
Example., We' shall find the subfield of C gener~ted by , X = ,{ 1, i}. (Whenever we are talking of C the symbol i
will' denote
.J
'7
1,- as usual.) Let L be this subfield. Then L must contain the prime sublleld Q of C; and since L is
closed under the field 'operations it must contain all elements of
, theCform
p+qi
where p, q E Q. Let M be the set of aU such elements. We claimthat'M is a field. Clearly M is closed, under sunlS
and proclQc~further
'/'
p.
q
i
(p + iq) -1 = p2 + q2 - p2 + q ,.,
so tha~ every non-zero element of M has a multiplicative inverse in'M: Hence M is a field, and contains X. Since L
is the smallest subfield containing X, we have L £' M. But M '£ L by definition. Hence L = M, and we have found
"a description of the subfield generated by X, '
In the case of a field extension L: K we are more interested in fields lying between K and L. This means that we can
restrict our attention to'subsets X which contain K; equivalently sets of the form K u Y where Y £ L.
Definition. If L: K is an extension.. and Y is a subset of L, then tbe subfield of L generated by K u Y is written
K(Y)
'and is said to be obtained from K by adjoinihg Y.
Notice that K(Y) is in general'considerably larger than
'Ku~
This notation is open to all sorts of useful abuses: If Y
n
'
35
36
GALOIS THEORY
has a single element y we write K(y) instead of K({y}); and in the same spirit K(Yl, ... , Yn) will replace K({Yl'
... , Yn})·
Examples.
1 The subfield R(i) of C must contain all elements·.t + iy where x, Y E R. So C = R(i).
2 Let K be any field, K(t) the field of rational expressiotlSin t over K. This notation would appear to be
ambiguous, In that K(t) also denotes the subfield generated by K u ,{t}. But this subfield, since it is closed
under the field operations, must contain all rational expressions in t; hence it is the whole field of rational
expressions. Thus K(t) has'the same meaning under ,either interpretation. ,
3 The subfield of R consisting of all elements p + qJ2 where p, q E Q is easily seen to be Q(J2). :
4 It is not always true th~lt a field of the form K(a) consists of all elements of the form j + ka where j; k E K.'
It certainly contains, all such elements, but they need, not for~ a field. For example, in R: Q let a be the real
cube root of 2,f,lnd consider Q(a). This. is ip. fact tbe set of all elements of R of ~he formp'+qa+ra2, where
p,q, r. E Q. To 'show thi's, we prove that the set of such elements ,is a field. The only diffi, culty is finding a multiplicative, inverse; the re~der should'
work out·the details.
'
,
Simple extens.ions.
We define a special kind of field extension .
. ~. D~finition. A simple extension is an extension L:K having the property that L = K(a) f9{ some IX EL.
The examples discussed above are all simple extensions.
On the other hand R: Q is. not a simple extension, (see, Exercise 3.6). Note that a simple extension need not
be presented in a form which makes its simplicity ·obvious. For example, Q(fl,.J3): Q is a simple extension,
.even though two elements are adjoined, not Just one. For consider the
FIELD EXTENSIONS
37
J3
Y1>2
fl.' eld Q. ~. (. 2+'.
. 3). This contains the elements (J].. 2 +
= 5.+2 6;,henceJ6;hence~(~+J3)
= 2J3+3J2; hence 2andJ3. Thus Q(J2, ..j3)=. Q(.J2 +)3).
Our aim for the rest of this chapter is to "classify all'pos .. ~ible simple extensions. We first formulate the
concept of isomol"phISmo1 ext~sions, then develop "techniques .Jor constructing simple extensions, and
flnallyshow that up. to
. isomorphism" we have' constructe'd all possible .' simple
extensions.·/
.
Definitio9/Ari isomorphism. between; two field extensions i : K -+ K*, j : L
isomorphisms A:K -+ L, jl :K* -+ L*,such that for alLk eK
-+ L
*is a pair (A, j,l) of field.
j(2(k» = jl(i(k)) .
. Another, more pictorial, way -ofputti:l1g this is to say that the diagram
KlK*
A~
~j1
L -;+ L* .J
commutes - that is, the two paths from K to L * give the same map.
The reason for making the definition this way is that as well ~. the field structure being preserved b-y
i~omorphism~ ~eembedding of the small field in the large Qnelsalso' pfeserved.----..~-·- ----·---v anous Identifications may be made. Ifwe identify K and' i(K), and ~ and j(L),then i and j are inclusions,
and the cO:r,1mutativlty. condition now becomes .
"
j,lIK = A
where JlIK denotes the restriction of Jl to K. If we further identify K and J-. then A becomes the identity, and
so JlIKiS. the identity. In what follows we shall attempt to use these 'identified' conditions wherever possible.
Butona few occasions (notably Theorem .8.3) we shall need the'· full;
. generality of the first definition ...
. 38
GALOIS .THEORY
ConstntctiDg simple extensionS
Any attempt to develop' a theory of simple extensions l~ads rapidly to a fundamental dichotomy.
,
I·
Definition. Let !,(a)': K be a simple ~xtension. If the~e exists' a non-zero polynomial p over K such t~at p(a)
= 0 then a is an algebraic element pver K ~nd the extension is a simple algebraic extension. Other,wise a is
transcendental over K arid K(a):K is a simple transc~ndental extension.
The next result.' gives a way of corist~ctii1g a simple transcendental' extension' of any field.
3.1 Theorem. Thefield o/rational expressions K(t) is a simple . . transcendental extension of the field K.
 Proof. Clearly it is a simple extension. If p is a polynomial over K such that p(t) = 0 then p = 0 by
definition of K(t).
The construction of siInple algebraic extensions is a much more delicate problem .. First we need a technical
definition .
. Definition. A polynomial
f(t) =·aO+a·1t+···+ant" over a field K is monic if an = 1.
Clearly every polynomial is 'a copstant multiple of some' monic polynomial; and for a non-zero polynomial
this mOhic polynomial is unique. Further, the product of two monic polynomials is again monic ....
,Now suppose that K(a): K is a simple·a:igebraic ex'tenslo,n.
There is a polynomialp over K such that p(a) == '0. We may suppose that p is monic. There is at least one
monic polynomial of smallest degree of which ais a zero. If p, q are two ~uch, then p(a}....:.·q(a) = o. If p =1=
q some constant ,multipl~ 'oCp - q is a monic polynomial of which a is a zero, .contrary
FIELD EXTENSIONS
39
to the definition. Hence,thereisa unique monic polynomial, pof smallest degree such that p(ex) = O ..
Definition. ,Let' L : K be a field extension, and suppose that' ' ex E L is algebraic over K. Then the ,minimum
a
polvnohliar of over· K is the unique 'monic .polynomial 11'2 over/( of
smallest degree such that 11'2(0:) = O.
'
"
For example, i E C is, algebraic ov~r' Rt ,HWe let" , m(t) = t2 + 1 then 11'2(0 = O. Clearly 11'2 \ is, mopifThe qnly
monic polynomials over R of smalle(degree\ ~re' those pf the form t + r (r E R) or 1. But i cannot be a zero of any
,of these, or' else 'we would have iE R. Hence the minimum polynomial,ofi over R is t2 + 1.
It is natural to ask' ~hich p01ynomials can occuras-, minimum polynomial§. The next lemma provides informa.; " \
<,tion on this question.
3.2 Lemma.
,r
ex is an" algebraic element over the field K, " then the minimum polynomial of ex. aver K is
irreducible over K. It divides every polynomial of which ex is a zero ..
-
Proof. Suppose that the minimum polynomial 11'2 ,of ex over
K is reducible, so that 11'2 =fg wbete f and g are of'smaIier degree. We mayasStime, f "nd g are monic. Since
11'2(0:) ,;'" 0 we have f(ex)g(cx) = 0 so. either f(cx) = 0 or g(cx) = O. But
~ this contradicts-,thedefinition of m. Hence, 11'2 ,is irreducible over K.
Now, suppose that p 'is a polynomial over K such that 'p(cx) =0. By the division algorithm there exist polynomials'
q and rover K such' that p == mq + rand 8r <
em. Then o = p(ex) =,0 + r(ex). If r =1= 0 a suitable constant
multiple' of
)t is moni,c,which contradicts the definition of m. Therefon~ r = 0 so 'that p divides 11'2,
Remark for the sophisticate. Using highet..:powereq masN- , nery we can. express ourselves' more succinctly.
The .ring· KIt] is a princiJ?al ideal domain, and {p :p(Ct),= Q}.; is<,an
40 GALOIS THEORY 'ideal~8}
principal ideaL Its uniq~¢' monic geperator is m. For details, consult Hartley and
Hawkes [141 pp. 59 ff.
9iven any tield K and any irreducible monic polynomial mover K we shall construct an extension K(a) : K such
that IX has minimum polynomial m ovet: K. We need two
·,diminary lemmas.
V 3.3 Lemma. 'If ¢"is a ring hom~morphismfromafield K into
I a ring R and 4>
=1= 0
then 4> is a
monomorphism.
Proof. The kernel of 4> is an ideal of K. But K, being a field, has no ideals other than 0 and K itself. Since 4> =1=
0 the kernel is not K, so must be O. Therefore 4> is amonomor..;
phism,
Note that this lemma fails if K is just a ring: the natural map Z -+ Z2 is neither zero nor a monomorphism.
3.4 Lemma. If m is an irreduCible polynomial over the field K, and I is the ideal of K[t] consisting of all
multiples afm, then the quotient ring K[t]/1 is a field.
,
Proof. Let theQt 1 +fbe ~ nonz~ro.elementofS = K[t]jt Since m is irr~ibl~ m and f are coprime. Hence by 1. J
2
.
I
there exist polynomials a, b over K-such that
af+bm := 1.
" Then
(1 +a)(1 +f)+'(1 +0)(1 +m) = 1 + 1.
But 1+ m .= 1 is the zero element of S, and 1 + 1 i~ the iden.;. tityelement, so that
(1+a)(I+f)·.= 1+1
and ·1,+ f has inve!se I + a. Hence S is a ~eld.
We n'ow have what we need to prove:
\IIELDEXTENSIONS' , " 41
V 3.5 Theorem. IfK is any field and m is any irreducible mo~ic polynomial 0v.er K,
then there exists an extension. K(a): K such that a has minimum polynomial mover K:
Proof. There is a,natural monomorphism ~:K ~ K~Let rObe the ideal of K[f] consisting of all multipl~s, oUm,.Jet
S = K[t]/I, and let v be the natural homomorphism K[t] '-. S. By Lemma 3.4 S is a field, and by Lemma 3.3the
composite map vi is a monomorphism. )dentify K with its imagev(i(K», ~! a = 1+ t. Clearly S = K(a). Since mE
Iwe have m(a) = J, and I is the zero etem-el1t of S. Since m is irreducible and monic it must be the minimum
polynomialofa. For if p is the minimum polynomial, then by 3.2 p/m. Therefore p = m.
There is another way to look at this construction. From each coset T+ rwe can select a unique polynomial of
om.
degree smaller than
The field structure on S induces operations on the set of representative polynomials as
follows: addition is as usual. Multiplication is as usual except that after multiplying it is necessary to take the
remainder on division bY"m. It is possible jo define K(a) in this way.
Classifying simple extensions
We shall now demonstrate that the above methods suffice for the construc,tion of all possible simple extensions,
(up to isomorphism). Again transcendental extensi09s are easily dealt wi tho
v
3.6' Theor~m. \Every simple transcendental extensionK(a) :K is isomorphic to the extension K(t) :
Kofrational expressions in t -over .f(. The isomorphism can be chosen to' carry t into a'.
Prorif:Define a map ¢ : K(t) -. K(a) by ¢(f(t)/g(t») = f(a)/g(a'.
If g =1= 0 then g(a) i= 0 (since a ~s transcendental) so this
42
GALOIS THEOR Y
definition makes sense. It is clearly a homomorphism, and is a monomorphism by 3.3. It is clearly onto; and so is an
isomorphism. Further, <l>IK is the identity, so that <I> defines
'. an isomorphism of extensions. And <I>{t) = ex.
To deal with algebraic extensions we must establish a standard form for elements of the large field~
3.7 Lemnta. Let K{ex):K be a simple algebraic extension, whEre ex has minimum polynomial m overK.
The~ any element of K(a) has a unique .expression in the form pea) wher.e p is a.
'. polynomial over K and op < om.
(
Proof. Every element of K{a) can be expressed in the form . f(a)/g{a) where f, 9 E K[t] and g{a) =1= 0 (since the set of
all
. such elements' is a field, contains K and a, and lies inside K(a)) .. Since g(a) =1= 0, m doe~ hot· divide g; and since m
is irreducible m and 9 are coprime. By 1.12 there exist' polynomials a, b over K such that ag+ bm = O. Hence a(a)g(a)
= 1, so that f(a)/g(a).= f(a)a(a)' = h(a) for some polynomial hover K. Let r be the remainder on dividing h by m. Then
rea) = h(a): Since or < Qm, the existence of such an expression is proved.
·Now we show uniqueness. Suppose that [(a) = g{a) where of,og < om~ ~f e .. = f- g then e{a) ::::= 0 and oe < am ..
By definition ofm we have e.= ~, so thatf = g. The le'mma
is proved ..
"
ExampJe. Suppose K = R, m(t) = ·t2+t+1. According t9 the above lemma, if a ha"s mipimum polyrtomial .mover K
then every element of K (a). is a "polynomial in a of degree < 2. ConsIder. the element (3a2 +'2)/(a + 4). Observe that
1 = /1(~2 + t+ 1)+t~3(t+4)
" so that
·1/(a.+4) = (a-3)/11.
PI E L D', EX T,E N S ION S Therefore'
(3a2 + 2)j({t +4) = 1\ (3a2 + 2)(a -"- 3)
= '1\(3( -a--:J)+2)(a~3)
l'
2 --
'
= n( - 3a +8a+3) ,
= /1(11a+6)
=a-t161'
We can now prove a preliminary version of the result that' K and. m between, them' determine the extension K(a).
43
a
3.8 Theorem. $uppose K{a):Kand K{f3):K are simple al~~., braic extensiQns" such thai
and f3 have the
same minimum' polynomialm over K .. Then the two. extensions aridsomorphic, and the isomorphism of the
large fields can be taken to map
, ato f3.
Proof. ByJ.? every element X-E K(a) is uniquely expressible in the form
x = xo+Xta+····+xnan , (Xl' ""Xn E K)
,
,
where n= om-:- L Define'a map 4) :K(a) -+ K(f3) by"
..•.
4>(x) = XO+Xlf3+"·+Xnpn.
By3.? 4> is onto and 1-1. Obviously ¢(x + y)= 4>(x)+¢(y): . We shall show fhat 4>(xy) == 4>(x)¢(y), for any
x, .y E /((a). Let x = f(a), y= g(a)" xy = h(a); where f, g, h are polynomials overK of degree <am. Then
f{a)g(a) - h(a) =xy :"""xy = O.
By 3.2 In divides [g - h, so there exists a polynomial q over. K s~ch tqatfg = mq + h. Sinc~ oh < am it follows'
thath is;: 'the remaInder on dividing fg by m.By the same reasoning ,
. we must havef(p)g([3) = h([3). Thus ¢(xy) = h(P) = f(/J)g(P) = 4> (x)¢(y), so ¢ is an \somorphi&m. Since¢ is
the identity
~ .........................................
,,-
-
.
-"
:.: ........................:,'
on K the' two extensions· a're isomorphic. An<i clearly.,
04>(&)'= {3.
44
GALOIS THEORY
For certain l,\ter applications we need a slightly stronger version of the previous- theorem, to cover
extensions of' isomorphi~ (rather than identical) fields. Before we can state
. the more general theorem we need the following:
Definitif!n. Let i: K ~ L be a field monomorphism. There . · is a monomorphism i:K[t] ~ L[t] defined by
i(ko+k1t+ ... +kntn) = i(ko)+.i(k~)t+· .. + i(kn)~
!:
(ko, ... ; kn E K). If i is an isomorphism so is i.
-
.
The hat is unnecessary and may be dispensed with; and' in future we shall use the same symbol i for the map
between fields and for its extension .to the polynomial rings. This will not cause confusion since i(k) = i(k) for
any k E K.
3.9 Theorem. Suppose K and L· are fields and i : K -:-+ Lisan isomorphism. Let K(a), L{{3) be simple
algebraic extensions of K and L respectively, such that a has minimum polynomial mIX(t) over K and {3 has
minimum polynomial mp(t) over L .. Suppose further that mp(t) = i(mIX(t»!Then 'there exists an
isomorphism j: K(a) ~ L(fJ) such that j/K = j and j(a) = {3 .
.
· Proof. We have the diagram
K ~ K(a) .
i~
~j
L ~ LCfJ)
~ (where the dotted arrow indicates that j has not. yet been
 found). Using the proof of 3.8 as a guide, define j as follows: every element of K(a) is of the form pea) for
a ·polynqmial p over K of degree < oma.. 'Let j(p(a» = (i(p»(fJ) where i{p) , is defined as above. The'
detailed proof-follows the lines of 3.8 and is left to the reader.
The point of this theorem is that the given mapi can be extended to a map j between the larger fields. Such
e~tensiol1 tkeel'ems, saying that under suitable conditions maps
FI ELD ·EXTENSIONS
45
between. sub-objects can be extended to maps between objects, constitute important weapons in the
mathematician's armoury. Using them we can extend our knowledge from small structures to large ones in a
sequence of simple steps. Not surprisingly, extension theorems are usually hard to come by!
Theorem 3.9 implies that u~der the given hypotheses the extensions K(a): K and L(f3) : L are isomorphic.
This allows us to identify K with Land K(a) with L([3), via the maps i andj .
. Theorems 3.5 and 3.8 together give a complete characterization of simple algebraic extensions in terms of
polynomials. To each extension corresponds an irreducible monic polynomial; and given the smalt' ·field and
this _ ,. Ilolynomial we can reconstruct the' extension.
Note that the correspondence is not strictly 1-1: isomorphic extensions may give different polynomials, since.
there is latitude of choice for a. No difficulties arise because of this ..
Exercises
3.i Prove that isomorphism of field extensions IS an
equivalence relation.
3.2 Find the subfields of (: generated by: (a) {O, I}
0
(b) { 0 }
(c) {O, 1, i}
,. (d) {i,J2} (e) {..[2,
(g) R u {i}.
J3} (f) R-
3.3 Describe the subfields of C of the form
.
(a) Q(..[2)
(b) Q(i)
(c) Q(a) where a is the.real cube root 0£2
46
GALOIS THEOR Y
(d) Q(~; j7) (e) Q(iJIT) ..
3.4 Let K =. Zz.' Describ~ the subfields ofK(t) of the fO~IIl:
~)K(t~
..
(bf K(t+1): '(c) K(t5) '.
(d) K(e + 1) ..
3.5 Of the extensions. defined in Exercises 3.3 a:nd~.4" which are simple algebraic? Which are simple tr~ns/
cendental?
\,
~.
3.6 Show that R is not a sitpple extensicm qf Q as'
follows:
fa) Q is countable ...
a.
(b) Any simple extension of
countable' field 1s
countable. . (c) R is not countable.
3.7 Prove a tninscenden!al version 'of Theorem 3.9,. modelled on Theorems ~.9 and 3.1.
_ 3.8 Find the minimum polynomials over the~rriall field . of t@ following elements in the following
extensions·:
(a) i in C : Q ,"
.
. (b) i'>' C: R
. (c) ~in R:Q
. (d)
.
.
<.v1 +.1)/2 in C : Q
(e) (iJ3~1)/2 in C·:Q~
'. '" ....................... .
(f) a in K:P where K is the field 'of ExerCise 1.6 and P is its prime subfield· .... '. ' ..
. (g) a in Z3(t)(a): Z3(t) where tis indeterminate ~rid a2 == t + 1.
3.9 Show that if a has minirnumpolynomiaf t2 ~2 .over Q andp~ hasinin~mnm poly~omialt2 :....At+ 2· over,
•
_,.
-0
FIELD EXTE'NSION~
Q, then the extensions Q(a): Q and. Q(J1): Q are isomorphic.
3.10 For which of the following values ofm(t) do there exist extensions K(a) of K for which a has minImum pOlynomial
m(t)?
(a) m(t)' == t~'- 4,K = R (b) m(t) = t2 + 1, K = Z3 (c) m(t) = t2+ 1, K ;=, Zs
(d) m(t) = t7 - 3t6 + 4t3 - t - L K = R.
3.11 Let 1< be any fieid of characteristic =1= 2 and m(t) a' quadratic polynomial over K (i.e. am = 2). Show that
m(t) has both zeros i.n an extension field K(a) of K where a2 =k E K. Thus allowing 'square roots'
Jk
enables us tQ
solve all quadratici equations over K.
3.12 S'how that· for fields of characteristic' 2 there' exist quadratic equations which cartnot be solved by adjoin
ingsql:lCire r()ots of elements in the 'field. (Hint: try
Z2)'
3.13 Show that we can solve quadratic equations over a fie1q of cha~act~iistic 2 if as well as square roots we adjoin
elements ~k defined to be solutions of the equation
(~kf +~!k = k.
3.14 Show [hat the two zeros of t2 + t - k = 0 in the previous question are~;I and 1 + ~/k.
3.15 Let K = Z3' Find all irreducible quadratics over K, and construct all possible extensions of K by an element with
quadratic minimum polynomial. Into how, many isomorphism classes do these extensions fall? How many elements do
they have?
47
48
GALOIS TH.EORY
3.16 Construct extensions Q(a): Q where a has the following minimuln polynomials over Q:
(a) t2 - 5
.
I
4 3 2
3
(b) t +t +t +t+ 1 (c) t + 2.
3.17 Find a field with 8 elements.
3.18 Is Q(J2, 13, 15): Q a simple extension?
3.19 Suppose m(t) is irreducible over K, and ahas ~inimum polynomial m(t) ov~r. K. Doesm(t) necessarily fac
torize over K(a) into linear (degree 1) polynpmials? Hint: try K= Q, a = the real cube root of 2 ..
3.20, Mark the following true or false.
(a) Every field has n~m-trivialextensions,
(b) Every field has non-trivial algebraic extensions,· (c) Every simple' extension is algebraic, '
. (d) Every extension is simple, ...
(e) All simple algebraic. extensions are isomorphic. (f). An simple-transcendental extensionsota giv.en
.
field are isomorphic ..
(g) Every minimum polynomial is monic .
. (h) Monic polynomials are always irreducible.
(i) Every polynomial is a cori~tant multiple of an
irreducible polynomial.
.
(j) It is always safe to identify isomorphic fields ..