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 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 ${{\mathbb Q}.}$

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