305 -Extension fields revisited (2)

Most of our work from now on depends on the following simple, but very useful observation:

Theorem 1 Let {{mathbb F}:{mathbb K}} be a field extension. Then {{mathbb F}} with its usual addition is a vector space over {{mathbb K},} where multiplication of elements of {{mathbb F}} by elements of {{mathbb K}} is just the usual product of {{mathbb F}}.

Proof: Remember that the statement means that the elements of {{mathbb F}} (the vectors) can be added in a way that:

  • Addition is commutative: {v+w=w+v} for all {v,win{mathbb F}.}
  • Addition is associative: {v+(w+x)=(v+w)+x} for all {v,w,xin{mathbb F}.}
  • There is an additive identity {0:} {v+0=0+v=v} for all {vin{mathbb F}.}
  • All vectors admit additive inverses: For any {v} there is a {-v} such that {v+(-v)=-v+v=0.}

All of these are immediate consequences of {{mathbb F}} being a field. The statement also means that we can multiply vectors by elements of {{mathbb K}} (the scalars) in such a way that:

  • {a(v+w)=av+aw} for all {ain{mathbb K}} and {v,win{mathbb F}.}
  • {(a+b)v=av+bv} for all {a,bin{mathbb K}} and {vin{mathbb F}.}
  • {a(bv)=(ab)v} for all {a,bin{mathbb K}} and all {vin{mathbb F}.}
  • {1v=v} for all {vin{mathbb F}.}

Again, these are all immediate consequences of {{mathbb F}} being a field, and in fact hold for all {a,b,v,w} in {{mathbb F}.} Box

From linear algebra we know that if {{mathbb F}:{mathbb K},} then the dimension of {{mathbb F}} as a vector space over {{mathbb K}} is well-defined, and it is the size of any basis. Recall that a basis is a set {S} of vectors such that for any {vin {mathbb F}} there are vectors {alpha_1,dots,alpha_nin S} and scalars {a_1,dots,a_n} such that

displaystyle  v=a_1alpha_1+dots+a_nalpha_n,

(we say that {S} spans {{mathbb F}}), and this representation is unique (we say that {S} is linearly independent over {{mathbb K}}). Another way of stating linear independency is by saying that if {alpha_1,dots,alpha_nin S,} {a_1,dots,a_nin{mathbb K},} and

displaystyle  a_1alpha_1+dots a_nalpha_n=0,

then {a_1=dots=a_n=0.}

Definition 2 If {{mathbb F}:{mathbb K},} then {[{mathbb F}:{mathbb K}]} is the dimension of {{mathbb F}} as a vector space over {{mathbb K}}.

We will be interested in finite-dimensional extensions. They correspond to extensions by algebraic numbers. Recall:

Definition 3 If {{mathbb F}:{mathbb K}} and {alphain{mathbb F},} we say that {alpha} is algebraic over {{mathbb K}} iff there is a nonzero polynomial {pin{mathbb K}[x]} such that {p(alpha)=0.}

(If {{mathbb K}={mathbb Q},} we simply say that {alpha} is algebraic.)

If {alpha} is not algebraic over {{mathbb K},} we say that it is transcendental over {{mathbb K}.}

For example, {pi} is transcendental over {{mathbb Q}} but algebraic over {{mathbb R},} as witnessed by the polynomial {p(x)=x-piin{mathbb R}[x].} (That {pi} is transcendental is a deep theorem of Lindemann that we won’t prove here.)

{sqrt2} is algebraic over {{mathbb Q},} as witnessed by the polynomial {x^2-2.} This is also witnessed by the polynomial {2x^4-8,} although this is not particularly surprising, since {x^2-2|2x^4-8.}

This is an instance of a general phenomenon:

Theorem 4 Let {{mathbb F}:{mathbb K},} and suppose that {alphain{mathbb F}} is algebraic over {{mathbb K}.} Then there is a polynomial {p_alphain{mathbb K}[x]} with the following properties:

  1. {p_alpha} is nonzero.
  2. {p_alpha} is monic.
  3. {p_alpha(alpha)=0.}
  4. {p_alpha} is irreducible over {{mathbb K}[x].}

Moreover, {p_alpha} satisfies that whenever {qin{mathbb K}[x]} and {q(alpha)=0,} then {p_alpha|q,} and {p_alpha} is the unique polynomial in {{mathbb K}[x]} with these properties.

Proof: Let {I={qin{mathbb K}[x]:q(alpha)=0}.}

We claim that {I} is an ideal. To see this, note that {Ine0,} since {0in I.} If {p,qin I,} then {p-qin I,} since {(p-q)(alpha)=p(alpha)-q(alpha)=0-0=0.} If {pin I} and {qin{mathbb K}[x],} then {pqin I,} since {(pq)(alpha)=p(alpha)q(alpha)=0q(alpha)=0.}

As shown last lecture, {I} is principal. Let {pin{mathbb K}[x]} be such that {I=(p).}

Note first that {pne0.} This is because {alpha} is algebraic, so there is at least one nonzero polynomial {qin{mathbb K}[x]} such that {q(alpha)=0.} This shows that {qin I,} so {Ine(0).}

If {a} is the leading coefficient of {p,} then {ain{mathbb K}} and {ane0,} so {a^{-1}in{mathbb K}} and if {hat p=a^{-1}p,} then {hat pin I} and {hat p} is monic. Moreover, {I=(hat p).} This is because if {qin I,} then there is some {rin{mathbb K}[x]} such that {q=rp.} But then {q=arhat p,} so {qin(hat p).} (This shows that {Isubseteq(hat p).} On the other hand, since {hat pin I,} then {(hat p)subseteq I.})

Let {p_alpha=hat p.} Then {p_alphain{mathbb K}[x]} is monic and nonzero, and {p_alpha(alpha)=0.} Moreover, whenever {q(alpha)=0} and {qin{mathbb K}[x],} then {p_alpha|q.} We claim that {p_alpha} is irreducible over {{mathbb K}[x].} To see this, suppose that {p_alpha=rq} for some polynomials {r,qin{mathbb K}[x].} Since {p_alpha(alpha)=0,} we must have {r(alpha)=0} or {q(alpha)=0,} so one of {r,q} is also in {I,} and therefore {p_alpha|r} or {p_alpha|q,} but then (since {r|p_alpha} and {q|p_alpha}), {{rm deg}(p_alpha)={rm deg}(r)} or {{rm deg}(p_alpha)=q.}

Finally, we argue that {p_alpha} is the unique polynomial in {{mathbb K}[x]} that is monic (and therefore nonzero), irreducible, and such that {p_alpha(alpha)=0.} For suppose {q(alpha)=0} and {qin{mathbb K}[x]} is monic and irreducible. Then {qin I} so {p_alpha|q.} Since {q} is irreducible, we must have that {q=p_alpha a} for some unit {a.} The only units in {{mathbb K}[x]} are the nonzero elements of {{mathbb K},} so {{rm deg}(p_alpha)={rm deg}(q).} Since both are monic, we must have {a=1,} so {p_alpha=q.} Box

Definition 5 Suppose that {alpha} is algebraic over {{mathbb K}.} The minimal polynomial of {alpha} over {{mathbb K}} is the unique monic irreducible polynomial {p_alphain{mathbb K}[x]} such that {p_alpha(alpha)=0.}

The main connection between this concept and the idea of field extensions as vector spaces is the following result:

Theorem 6 Suppose that {{mathbb F}:{mathbb K}.}

  1. If {{mathbb F}} contains a transcendental over {{mathbb K},} then {[{mathbb F}:{mathbb K}]=infty.}
  2. If {{mathbb F}={mathbb K}(alpha)} and {alpha} is algebraic over {{mathbb K},} then {[{mathbb F}:{mathbb K}]={rm deg}(p_alpha).}

Proof: Suppose first that {betain{mathbb F}} is transcendental over {{mathbb K}.} We claim that the numbers {1,beta,beta^2,beta^3,dots} are linearly independent over {{mathbb K}.} This is because for any finitely many integers {0le n_0<n_1<dots<n_k,} if {a_0,dots,a_kin{mathbb K}} and

displaystyle  a_0beta^{n_0}+a_1beta^{n_1}+dots+a_kbeta^{n_k}=0,

then {beta} is a root of the polynomial

displaystyle  p(x)=a_0x^{n_0}+a_1x^{n_1}+dots+a_kx^{n_k}in{mathbb K}[x].

Since {beta} is transcendental, we necessarily have {p=0,} i.e., {a_0=dots=a_k=0.}

This shows that if {{mathbb F}} contains any transcendentals over {{mathbb K},} then {[{mathbb F}:{mathbb K}]=infty.}

Conversely, suppose {{mathbb F}={mathbb K}(alpha)} and that {alpha} is algebraic over {{mathbb K}.} Let {n={rm deg}(p_alpha).} Then {1,alpha,dots,alpha^n} are {n+1} elements of {{mathbb F}} that are linearly dependent over {{mathbb K},} since {p_alpha(alpha)=0} but {p_alphane0.} This shows that {[{mathbb K}(alpha):{mathbb K}]le n.}

If {1,dots,alpha^{n-1}} are not linearly independent over {{mathbb K},} then there are numbers {a_0,dots,a_{n-1}in{mathbb K},} not all of them zero, such that

displaystyle  a_0+a_1alpha+dots+a_{n-1}alpha^{n-1}=0.

But then {p(x)=a_0+a_1x+dots+a_{n-1}x^{n-1}in{mathbb K}[x]} is nonzero and {p(alpha)=0.} This contradicts that {{rm deg}(p_alpha)=n,} since {p_alpha|p} and {0le{rm deg}(p)le n-1.} It follows that {1,dots,alpha^{n-1}} are linearly independent, and therefore {[{mathbb K}(alpha):{mathbb K}]ge n.} Box

Corollary 7 If {alpha} is algebraic over {{mathbb K}} and {betain{mathbb K}(alpha),} then {beta} is algebraic over {{mathbb K}.} Moreover, {{rm deg}(p_beta)le{rm deg}(p_alpha).}

Proof: By the above, {[{mathbb K}(alpha):{mathbb K}]<infty.} Since {betain {mathbb K}(alpha),} then {beta} is algebraic.

Moreover, since {{mathbb K}(alpha):{mathbb K}(beta):{mathbb K},} we have {[{mathbb K}(beta):{mathbb K}]le[{mathbb K}(alpha):{mathbb K}],} as {{mathbb K}(beta)} is a vector subspace of {{mathbb K}(alpha).} But {{rm deg}(p_beta)=[{mathbb K}(beta):{mathbb K}]} and {[{mathbb K}(alpha):{mathbb K}]={rm deg}(p_alpha).} Box

Corollary 8 If {alpha} is algebraic over {{mathbb K}} and {[{mathbb K}(alpha):{mathbb K}]=n,} then

displaystyle  {mathbb K}(alpha)={a_0+a_1alpha+dots+a_{n-1}alpha^{n-1} : a_0,dots,a_{n-1}in{mathbb K}};

moreover, if {betain{mathbb K}(alpha),} then there is exactly one way to write {beta=a_0+a_1alpha+dots+a_{n-1}alpha^{n-1}} with {a_0,dots,a_{n-1}in{mathbb K}.}

Proof: This is because the argument of Theorem 6 shows that {1,alpha,dots,alpha^{n-1}} is a basis of {{mathbb K}(alpha)} over {{mathbb K}.} Box

We knew that Corollary 8 holds in a few particular cases, for example for {{mathbb Q}(sqrt n)} for {nin{mathbb Z}.} See also lecture 4.5. The argument above is much more general, and reduces significantly the amount of computations involved in each case.

Lemma 9 If {{mathbb F}:{mathbb H}:{mathbb K}} then {[{mathbb F}:{mathbb K}]=[{mathbb F}:{mathbb H}][{mathbb H}:{mathbb K}].}

Proof: Let {S} be a basis of {{mathbb F}} over {{mathbb H}} and {T} be a basis of {{mathbb H}} over {{mathbb K}.} It is enough to show the following two facts:

  • {ST:={vw:vin S,win T}} is a basis of {{mathbb F}} over {{mathbb K},} and
  • if {v,v'in S} and {w,w'in T,} then {vw=v'w'} implies {v=v'} and {w=w'.}

The second fact shows that {|ST|=|S||T|.} This together with the first fact implies the lemma.

To see the second fact, note that if {vw=v'w'} then

displaystyle  wv-w'v'=0,

and use that {v,v'in S} and {w,w'in {mathbb H}:} Recall that {S} is linearly independent over {{mathbb H}.} If {vne v',} then it follows that {w,w'=0,} contradicting that {{mathbb T}} is a basis so, in particular, it consists of nonzero elements. Then {v=v'.} Since {vin S} and {S} is a basis, {vne0.} Then {wv-w'v'=wv-w'v=(w-w')v} and we must have {w-w'=0,} so {w=w'.} This proves the second fact.

To prove the first fact, we need to show that {ST} spans {{mathbb F}} over {{mathbb K},} and that {ST} is linearly independent.

{ST} spans. Let {alphain{mathbb F}.} Since {S} is a basis of {{mathbb F}} over {{mathbb H},} there are vectors {v_1,dots,v_nin S} and scalars {a_1,dots,a_nin{mathbb H}} such that

displaystyle  alpha=a_1v_1+dots+a_nv_n.

Since {T} is a basis of {{mathbb H}} over {{mathbb K},} for each {i,} {1le ile n,} there are {w_{1,i}dots,w_{m_i,i}in T} and scalars {b_{1,i},dots,b_{m_i,i}in{mathbb K}} such that

displaystyle  a_i=b_{1,i}w_{1,i}+dots+b_{m_i,i}w_{m_i,i}.

Replacing each {a_i} with the corresponding expression above in the displayed representation of {alpha,} and expanding, gives as a representation of {alpha} as a linear combination of elements of {ST.}

{ST} is independent. If {y_1,dots,y_kin ST} and {c_1,dots,c_kin{mathbb K}} are such that

displaystyle  c_1y_1+dots+c_ky_k=0,

we need to prove that {c_1=dots=c_k=0.} To this end, note that each {y_i=v_iw_i} for some {v_iin S,} {w_iin T.} For each {vin S,} let {I_v={i:v_i=v}.} Then we can rewrite the expression above as

displaystyle  0=sum_{vin S}(sum_{iin I_v}c_iw_i)v.

Note that {I_v=emptyset} for all but finitely many {vin S;} for these {v,} we understand {sum_{iin I_v}c_iw_i=0.}

For any {vin S,} {sum_{iin I_v}c_iw_iin {mathbb H}.} Since {S} is a basis for {{mathbb F}} over {{mathbb H},} it follows that {sum_{iin I_v}c_iw_i=0} for all {vin S.} Since {c_iin{mathbb K}} for all {iin I_v} and {w_iin T} for all {iin I_v,} and {T} is a basis of {{mathbb H}} over {{mathbb K},} then {c_i=0} for all {iin I_v.} This holds for all {vin S,} and we are done. Box

The argument above is written in a way that proves the result even if the extensions are infinite-dimensional. However, all our applications will be restricted to finite dimensional extensions, in which case the write up above can be simplified slightly.

Corollary 10 If {{mathbb F}:{mathbb K}} and {alpha,betain{mathbb F}} are algebraic over {{mathbb K},} then so are {-alpha,} {alpha+beta} and {alphabeta;} and if {betane0,} then so is {beta^{-1}.}

Proof: Note that if {beta} is algebraic over {{mathbb K},} then it is also algebraic over {{mathbb K}(alpha),} since {{mathbb K}[x]subseteq {mathbb K}(alpha)[x].}

Thus, {[{mathbb K}(alpha,beta):{mathbb K}]=[{mathbb K}(alpha)(beta):{mathbb K}(alpha)][{mathbb K}(alpha):{mathbb K}]<infty.} Since {-alpha,alpha+beta,alphabeta,beta^{-1}in{mathbb K}(alpha,beta),} the result follows from Theorem 6. Box

Corollary 11 The set {bar{mathbb Q}subseteq{mathbb C}} of algebraic numbers is a field.

Proof: This is immediate from the corollary above. Box

Note that a simple counting argument shows that {bar{mathbb Q}} is countable, which means that almost every complex number is transcendental over {{mathbb Q}.}

Typeset using LaTeX2WP. Here is a printable version of this post.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: