Field Theory - Algebrais Extensions - Dummit and Foote Section 13.2

[Math Amateur]

I am trying to clarify my understanding of Proposition 11 of Dummit and Foote Ch13 Field Theory concerning the degree of \alpha over F.

Proposition 11 reads as follows:

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Proposition 11. Let \alpha be algebraic over the field F and let F(\alpha) be the field generated by \alpha over F.

Then F(\alpha) \cong F[x]/(m_{\alpha}(x))

so that in particular [F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) = deg \ \alpha

i.e. the degree of \alpha over F is the degree of the extension it generates over F

-------------------------------------------------------------------------------------------------------------------------------------------------------------- However Corollary 7 (Dummit and Foote page 518) states the following: -----------------------------------------------------------------------------------------------------------------------------------------------------------------

Corollary 7. Suppose in Theorem 6 that p(x) is of degree n. Then

F(\alpha) = \{ a_0 + a_1 {\alpha} + a_2 {\alpha}^2 + ... ... a_{n-1} {\alpha}^{n-1} \ | \ a_0, a_1, ... ... a_{n-1} \in F \} \subseteq K

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Given that F(\alpha) consists of polynomials of degree (n-1) should not the degree of [F(\alpha) \ : \ F] = deg \ m_{\alpha}(x) - 1 = deg \ \alpha -1 - that is the degree of \alpha over F be one less than the degree of the minimal polynomial? Can someone please clarify this situation for me

Peter

[This has also been posted on MHF]

 

[Math Amateur]

I have just been reflecting about my own question above.

It is possible that F(\alpha) viewed as a vector space of deg (n-1) polynomials would have the following as a basis:

1, {\alpha}, {\alpha}^2, {\alpha}^3, ... ... {\alpha}^{n-1}

Then the degree of the space would be n which would equal the degree of the minimal polynomial involved.

If this thought is actually correct, could someone please confirm this as the case.

Peter

 

[caffeinemachine]

I have just been reflecting about my own question above.

It is possible that F(\alpha) viewed as a vector space of deg (n-1) polynomials would have the following as a basis:

1, {\alpha}, {\alpha}^2, {\alpha}^3, ... ... {\alpha}^{n-1}

Then the degree of the space would be n which would equal the degree of the minimal polynomial involved.

If this thought is actually correct, could someone please confirm this as the case.

Peter

Yes Peter, this is absolutely correct.

 

[Deveno]

A basis of any vector space $V$ over $F$ is a subset that both spans $V$ and is linearly independent.

The fact that $\alpha$ satisfies a polynomial of degree $n$ is sufficient to establish that the set $\{1,\alpha,\dots,\alpha^{n-1}\}$ spans $F(\alpha)$ (equivalently, the canonical ring homomorphism $F[x] \to F(\alpha)$ is surjective, with kernel $m_{\alpha}(x)$).

So $\text{dim }_F(F(\alpha)) = [F(\alpha):F] \leq n$.

To establish linear independence we COULD employ the division algorithm to show that any polynomial in $F[x]$ has a UNIQUE remainder of degree $< n$ (or is the 0 polynomial) upon division by $m_{\alpha}(x)$, but I prefer the following:

Suppose we have:

$c_0 + c_1\alpha + \cdots + c_{n-1}\alpha^{n-1} = 0$

If ANY of the $c_j$ are non-zero, then $\alpha$ satisfies a polynomial of degree $< n$ (namely: $c_0 + c_1x + \cdots + c_{n-1}x^{n-1}$) in $F[x]$, contradicting the minimality of $m_{\alpha}(x)$ (which is necessarily irreducible). So it must be that ALL the $c_j = 0$, which proves linear independence.

The standard example used is typically $F = \Bbb Q$, with $\alpha = \sqrt{2}$. Now by Eisenstein, we have:

$x^2 - 2$ irreducible over $\Bbb Q$ (or we could use the fact that $\sqrt{2}$ is irrational, to show that $x^2 - 2$ has no roots in $\Bbb Q$, for if one root of a rational quadratic is rational, so is the other root).

It follows from the above that $\{1,\sqrt{2}\}$ is a basis for $\Bbb Q(\sqrt{2})$, which means that:

$\Bbb Q(\sqrt{2}) = \{x \in \Bbb R: x = a + b\sqrt{2}, a,b \in \Bbb Q\}$

The above can explicitly be shown to be a field, which contains $\Bbb Q$ (since it contains real numbers of the form $a + 0\sqrt{2}$) and $\sqrt{2}$ (as the real number $0 + 1\cdot \sqrt{2}$), moreover the closure axioms for a field ensure that $\Bbb Q(\sqrt{2})$ must contain all $\Bbb Q$-linear combinations of $1$ and $\sqrt{2}$, so this is indeed the smallest subfield of $\Bbb R$ containing both $\Bbb Q$ and $\sqrt{2}$.

 

[Math Amateur]

A basis of any vector space $V$ over $F$ is a subset that both spans $V$ and is linearly independent.

The fact that $\alpha$ satisfies a polynomial of degree $n$ is sufficient to establish that the set $\{1,\alpha,\dots,\alpha^{n-1}\}$ spans $F(\alpha)$ (equivalently, the canonical ring homomorphism $F[x] \to F(\alpha)$ is surjective, with kernel $m_{\alpha}(x)$).

So $\text{dim }_F(F(\alpha)) = [F(\alpha):F] \leq n$.

To establish linear independence we COULD employ the division algorithm to show that any polynomial in $F[x]$ has a UNIQUE remainder of degree $< n$ (or is the 0 polynomial) upon division by $m_{\alpha}(x)$, but I prefer the following:

Suppose we have:

$c_0 + c_1\alpha + \cdots + c_{n-1}\alpha^{n-1} = 0$

If ANY of the $c_j$ are non-zero, then $\alpha$ satisfies a polynomial of degree $< n$ (namely: $c_0 + c_1x + \cdots + c_{n-1}x^{n-1}$) in $F[x]$, contradicting the minimality of $m_{\alpha}(x)$ (which is necessarily irreducible). So it must be that ALL the $c_j = 0$, which proves linear independence.

The standard example used is typically $F = \Bbb Q$, with $\alpha = \sqrt{2}$. Now by Eisenstein, we have:

$x^2 - 2$ irreducible over $\Bbb Q$ (or we could use the fact that $\sqrt{2}$ is irrational, to show that $x^2 - 2$ has no roots in $\Bbb Q$, for if one root of a rational quadratic is rational, so is the other root).

It follows from the above that $\{1,\sqrt{2}\}$ is a basis for $\Bbb Q(\sqrt{2})$, which means that:

$\Bbb Q(\sqrt{2}) = \{x \in \Bbb R: x = a + b\sqrt{2}, a,b \in \Bbb Q\}$

The above can explicitly be shown to be a field, which contains $\Bbb Q$ (since it contains real numbers of the form $a + 0\sqrt{2}$) and $\sqrt{2}$ (as the real number $0 + 1\cdot \sqrt{2}$), moreover the closure axioms for a field ensure that $\Bbb Q(\sqrt{2})$ must contain all $\Bbb Q$-linear combinations of $1$ and $\sqrt{2}$, so this is indeed the smallest subfield of $\Bbb R$ containing both $\Bbb Q$ and $\sqrt{2}$.

Thanks Deveno, the post is most helpful!

Peter