- Thread starter
- #1

- Jun 22, 2012

- 2,918

I am trying to understand the proof of Theorem 6 in Chapter 13 of Dummit and Foote.

Theorem 6 states the following: (see attachment)

=====================================================================================

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

======================================================================================

The proof then begins as follows:

======================================================================================

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

======================================================================================

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]

Theorem 6 states the following: (see attachment)

=====================================================================================

**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]

======================================================================================

The proof then begins as follows:

======================================================================================

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

======================================================================================

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: