## 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 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}] 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}] 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 ${{mathbb Q}.}$

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