# Field Theory - Dummit and Foote - Chapter 13 - Theorem 6

#### Peter

##### Well-known member
MHB Site Helper
I am trying to understand the proof of Theorem 6 in Chapter 13 of Dummit and Foote.

Theorem 6 states the following: (see attachment)

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Theorem 6. Let F be a field and let [TEX] p(x) \in F[x] [/TEX] be an irreducible polynomial. Suppose K is an extension field of F containing a root [TEX] \alpha [/TEX] of [TEX] p(x): \ p( \alpha ) = 0 [/TEX]. Let [TEX] F ( \alpha ) [/TEX] denote the subfield of K generated over F by [TEX] \alpha [/TEX],

Then [TEX] F( \alpha ) \cong F[x] / (p(x)) [/TEX]

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The proof then begins as follows:

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Proof: There is a natural homomorphism

[TEX] \phi : F[x] \longrightarrow F( \alpha ) \subseteq K [/TEX]

[TEX] a(x) \longmapsto a( \alpha ) [/TEX]

obtained by mapping F to F by the identity map and sending x to [TEX] \alpha [/TEX] and then extending so that the map is a ring homomorphism ( i.e. the polynomial a(x) in x maps to the polynomial [TEX] a( \alpha ) [/TEX] in [TEX] \alpha [/TEX] )

Since [TEX] p( \alpha ) = 0 [/TEX] by assumption, the element p(x) is in the kernel of [TEX] \phi [/TEX], so we obtain an induced homomorphism ( also denoted by [TEX] \phi [/TEX] ):

[TEX] \phi : F[x]/(p(x)) \longrightarrow F( \alpha ) [/TEX] ... ... ... (1)

etc etc

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My problem is in understanding the last sentence above - how exactly is the homomorphism shown in (1) induced by what comes before it - and anyway why it it the same as the natural homomorphism defined earlier (since it is also called [TEX] \phi [/TEX] I am assuming the two defined homomorphisms are the same).

Also it is subsequently shown that (1) above is an isomorphism - but how can it be a bijection when there are cosets on the left but polynomial evaluations on the right???

Can someone please clarify this situation for me ... perhaps using an example to make the explanation tangible?

Peter

[Note: This has also been posted on MHF]

Last edited:

#### Deveno

##### Well-known member
MHB Math Scholar
OK, it should be clear that the map:

$$\displaystyle \phi:F[x] \to F(\alpha)$$

given by:$$\displaystyle \phi(f(x)) = f(\alpha)$$ is a ring-homomorphism.

Moreover, since $$\displaystyle p(x)$$ is irreducible, it follows that any element of $$\displaystyle F(\alpha)$$ is of the form:

$$\displaystyle c_0 + c_1\alpha + \cdots + c_{n-1}a^{n-1}$$

where $$\displaystyle \text{deg}(p(x)) = n$$.

(Since $$\displaystyle p(\alpha) = 0$$ all higher powers of $$\displaystyle \alpha$$ can be expressed in terms of these. Also, these powers of $$\displaystyle \alpha$$ must be linearly independent over $$\displaystyle F$$ or else $$\displaystyle \alpha$$ satisfies a non-zero polynomial of degree less than deg(p(x)) in $$\displaystyle F[x]$$ contradicting the irreducibilty of p(x)).

Thus any element of $$\displaystyle F(\alpha)$$ has the pre-image under $$\displaystyle \phi$$ of $$\displaystyle c_0 + c_1x + \cdots + c_{n_1}x^{n-1}$$. So $$\displaystyle \phi$$ is surjective.

By the fundamental isomorphism theorem for rings:

$$\displaystyle F(\alpha) \cong F[x]/\text{ker}(\phi)$$.

Well, what is $$\displaystyle \text{ker}(\phi)$$? Clearly, any element of the ideal generated by p(x) is in the kernel, since $$\displaystyle \phi$$ annihilates all elements of this ideal. But $$\displaystyle F[x]$$ is a principal ideal domain (because F is a field), so we have:

$$\displaystyle \text{ker}(\phi) = (g(x))$$ for some $$\displaystyle g(x) \in F[x]$$.

Thus g(x)|p(x), so p(x) = h(x)g(x), for some h(x) in F[x]. Since p(x) is irreducible, we conclude that h(x) is a unit, that is, h is in F*. Thus:

g(x) = (1/h)(p(x)), which is in (p(x)), showing that $$\displaystyle \text{ker}(\phi) \subseteq (p(x))$$, so we must have:

$$\displaystyle \text{ker}(\phi) = (p(x))$$.

Here is an example:

Let $$\displaystyle F = \Bbb Z_2 = \{0,1\}$$. Consider the polynomial:

$$\displaystyle p(x) = x^2 + x + 1 \in \Bbb Z_2[x]$$.

This is clearly irreducible over $$\displaystyle \Bbb Z_2$$, because we have no linear factors (roots) in $$\displaystyle \Bbb Z_2$$.

Convince yourself that if $$\displaystyle u$$ is a root of p(x) in some extension field of F, the map:

$$\displaystyle \phi:x + (x^2 + x + 1) \to u$$

gives an isomorphism of $$\displaystyle \Bbb Z_2[x]/(x^2 + x + 1)$$ with $$\displaystyle \Bbb Z_2(u)$$. Some things to answer:

1) How many elements do each of these fields have?
2) What are the explicit addition and multiplication tables for each field?

#### Peter

##### Well-known member
MHB Site Helper
OK, it should be clear that the map:

$$\displaystyle \phi:F[x] \to F(\alpha)$$

given by:$$\displaystyle \phi(f(x)) = f(\alpha)$$ is a ring-homomorphism.

Moreover, since $$\displaystyle p(x)$$ is irreducible, it follows that any element of $$\displaystyle F(\alpha)$$ is of the form:

$$\displaystyle c_0 + c_1\alpha + \cdots + c_{n-1}a^{n-1}$$

where $$\displaystyle \text{deg}(p(x)) = n$$.

(Since $$\displaystyle p(\alpha) = 0$$ all higher powers of $$\displaystyle \alpha$$ can be expressed in terms of these. Also, these powers of $$\displaystyle \alpha$$ must be linearly independent over $$\displaystyle F$$ or else $$\displaystyle \alpha$$ satisfies a non-zero polynomial of degree less than deg(p(x)) in $$\displaystyle F[x]$$ contradicting the irreducibilty of p(x)).

Thus any element of $$\displaystyle F(\alpha)$$ has the pre-image under $$\displaystyle \phi$$ of $$\displaystyle c_0 + c_1x + \cdots + c_{n_1}x^{n-1}$$. So $$\displaystyle \phi$$ is surjective.

By the fundamental isomorphism theorem for rings:

$$\displaystyle F(\alpha) \cong F[x]/\text{ker}(\phi)$$.

Well, what is $$\displaystyle \text{ker}(\phi)$$? Clearly, any element of the ideal generated by p(x) is in the kernel, since $$\displaystyle \phi$$ annihilates all elements of this ideal. But $$\displaystyle F[x]$$ is a principal ideal domain (because F is a field), so we have:

$$\displaystyle \text{ker}(\phi) = (g(x))$$ for some $$\displaystyle g(x) \in F[x]$$.

Thus g(x)|p(x), so p(x) = h(x)g(x), for some h(x) in F[x]. Since p(x) is irreducible, we conclude that h(x) is a unit, that is, h is in F*. Thus:

g(x) = (1/h)(p(x)), which is in (p(x)), showing that $$\displaystyle \text{ker}(\phi) \subseteq (p(x))$$, so we must have:

$$\displaystyle \text{ker}(\phi) = (p(x))$$.

Here is an example:

Let $$\displaystyle F = \Bbb Z_2 = \{0,1\}$$. Consider the polynomial:

$$\displaystyle p(x) = x^2 + x + 1 \in \Bbb Z_2[x]$$.

This is clearly irreducible over $$\displaystyle \Bbb Z_2$$, because we have no linear factors (roots) in $$\displaystyle \Bbb Z_2$$.

Convince yourself that if $$\displaystyle u$$ is a root of p(x) in some extension field of F, the map:

$$\displaystyle \phi:x + (x^2 + x + 1) \to u$$

gives an isomorphism of $$\displaystyle \Bbb Z_2[x]/(x^2 + x + 1)$$ with $$\displaystyle \Bbb Z_2(u)$$. Some things to answer:

1) How many elements do each of these fields have?
2) What are the explicit addition and multiplication tables for each field?

Deveno,

This is really helpful and will assist me to make progress with Field Theory.

I will work carefully through what you have said.

Thanks,

Peter