Extending a morphism to a finite algebraic field extension

Asked Aug 12 '14 at 22:49

Active Aug 27 '14 at 21:47

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Theorem 5.21

I am trying to understand the proof of Theorem 5.21 in Introduction to Commutative Algebra, and am stuck on the portion underlined in red (note that $$\Sigma := \{(A,f) \mid A \text{ is a subring of $K$ and $f$ is a homomorphism of $A$ into $\Omega$} \}$$ is a partially ordered set: $(A,f) \leq (A', f') \iff A \subset A' \text{ and } f'\mid A=f$).

edited Aug 27 '14 at 21:47

user26857

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asked Aug 12 '14 at 22:49

kfriend

1

If youy ask about the result. See the proposition$1.1$ in this document. – Hamou Aug 12 '14 at 23:03
I understand the result, but in this specific proof, I am not understanding why the underlined statement holds. – kfriend Aug 13 '14 at 01:15
This statement is a theorem to extending a morphism of fields . – Hamou Aug 13 '14 at 14:42
So how would one define $\bar{g}$'(x+m'), for x+m' ∈ k' ? – kfriend Aug 14 '14 at 20:15

Extending a morphism to a finite algebraic field extension

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