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,w\in{\mathbb F}.}
  • Addition is associative: {v+(w+x)=(v+w)+x} for all {v,w,x\in{\mathbb F}.}
  • There is an additive identity {0:} {v+0=0+v=v} for all {v\in{\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 {a\in{\mathbb K}} and {v,w\in{\mathbb F}.}
  • {(a+b)v=av+bv} for all {a,b\in{\mathbb K}} and {v\in{\mathbb F}.}
  • {a(bv)=(ab)v} for all {a,b\in{\mathbb K}} and all {v\in{\mathbb F}.}
  • {1v=v} for all {v\in{\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 {v\in {\mathbb F}} there are vectors {\alpha_1,\dots,\alpha_n\in S} and scalars {a_1,\dots,a_n} such that

\displaystyle v=a_1\alpha_1+\dots+a_n\alpha_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 independence is by saying that if {\alpha_1,\dots,\alpha_n\in S,} {a_1,\dots,a_n\in{\mathbb K},} and

\displaystyle a_1\alpha_1+\dots a_n\alpha_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 {\alpha\in{\mathbb F},} we say that {\alpha} is algebraic over {{\mathbb K}} iff there is a nonzero polynomial {p\in{\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-\pi\in{\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\mid 2x^4-8.}

This is an instance of a general phenomenon:

Theorem 4 Let {{\mathbb F}:{\mathbb K},} and suppose that {\alpha\in{\mathbb F}} is algebraic over {{\mathbb K}.} Then there is a polynomial {p_\alpha\in{\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 {q\in{\mathbb K}[x]} and {q(\alpha)=0,} then {p_\alpha\mid q,} and {p_\alpha} is the unique polynomial in {{\mathbb K}[x]} with these properties.

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

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

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

Note first that {p\ne0.} This is because {\alpha} is algebraic, so there is at least one nonzero polynomial {q\in{\mathbb K}[x]} such that {q(\alpha)=0.} This shows that {q\in I,} so {I\ne(0).}

If {a} is the leading coefficient of {p,} then {a\in{\mathbb K}} and {a\ne0,} so {a^{-1}\in{\mathbb K}} and if {\hat p=a^{-1}p,} then {\hat p\in I} and {\hat p} is monic. Moreover, {I=(\hat p).} This is because if {q\in I,} then there is some {r\in{\mathbb K}[x]} such that {q=rp.} But then {q=ar\hat p,} so {q\in(\hat p).} (This shows that {I\subseteq(\hat p).} On the other hand, since {\hat p\in I,} then {(\hat p)\subseteq I.})

Let {p_\alpha=\hat p.} Then {p_\alpha\in{\mathbb K}[x]} is monic and nonzero, and {p_\alpha(\alpha)=0.} Moreover, whenever {q(\alpha)=0} and {q\in{\mathbb K}[x],} then {p_\alpha\mid q.} We claim that {p_\alpha} is irreducible over {{\mathbb K}[x].} To see this, suppose that {p_\alpha=rq} for some polynomials {r,q\in{\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\mid r} or {p_\alpha \mid q,} but then (since {r\mid p_\alpha} and {q\mid 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 {q\in{\mathbb K}[x]} is monic and irreducible. Then {q\in I} so {p_\alpha\mid 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_\alpha\in{\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 {\beta\in{\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 {0\le n_0<n_1<\dots<n_k,} if {a_0,\dots,a_k\in{\mathbb K}} and

\displaystyle a_0\beta^{n_0}+a_1\beta^{n_1}+\dots+a_k\beta^{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_\alpha\ne0.} 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_1\alpha+\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\mid p} and {0\le{\rm deg}(p)\le n-1.} It follows that {1,\alpha,\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 {\beta\in{\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 {\beta\in {\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_1\alpha+\dots+a_{n-1}\alpha^{n-1} : a_0,\dots,a_{n-1}\in{\mathbb K}\};

moreover, if {\beta\in{\mathbb K}(\alpha),} then there is exactly one way to write {\beta=a_0+a_1\alpha+\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 already knew that Corollary 8 holds in a few particular cases, for example for {{\mathbb Q}(\sqrt n)} for {n\in{\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:v\in S,w\in 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 {v\ne 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 {v\in S} and {S} is a basis, {v\ne0.} 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 {\alpha\in{\mathbb F}.} Since {S} is a basis of {{\mathbb F}} over {{\mathbb H},} there are vectors {v_1,\dots,v_n\in S} and scalars {a_1,\dots,a_n\in{\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,} {1\le i\le 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_k\in ST} and {c_1,\dots,c_k\in{\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_i\in S,} {w_i\in T.} For each {v\in S,} let {I_v=\{i:v_i=v\}.} Then we can rewrite the expression above as

\displaystyle 0=\sum_{v\in S}\left(\sum_{i\in I_v}c_iw_i\right)v.

Note that {I_v=\emptyset} for all but finitely many {v\in S;} for these {v,} we use the convention that empty sums are zero, that is, {\sum_{i\in I_v}c_iw_i=0.}

For any {v\in S,} {\sum_{i\in I_v}c_iw_i\in {\mathbb H}.} Since {S} is a basis for {{\mathbb F}} over {{\mathbb H},} it follows that {\sum_{i\in I_v}c_iw_i=0} for all {v\in S.} Since {c_i\in{\mathbb K}} for all {i\in I_v} and {w_i\in T} for all {i\in I_v,} and {T} is a basis of {{\mathbb H}} over {{\mathbb K},} then {c_i=0} for all {i\in I_v.} This holds for all {v\in 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,\beta\in{\mathbb F}} are algebraic over {{\mathbb K},} then so are {-\alpha,} {\alpha+\beta} and {\alpha\beta;} and if {\beta\ne0,} 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,\alpha\beta,\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.


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 )

Facebook photo

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

Connecting to %s

%d bloggers like this: