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I am reading R.Y. Sharp: Steps in Commutative Algebra.
Lemma 1.13 on page 7 (see attachment) reads as follows:
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1.13 LEMMA. let R be a commutative ring, and let X be an indeterminate; let T be a commutative R-algebra with structural ring homomorphism $$ f \ : \ R \ \to \ T $$; and let $$ \alpha \in T $$.
Then there is a unique ring homomorphism $$ f_1 \ : \ R[X] \ \to \ T $$ which extends $$ f $$ (that is, is such that $$ f_{1|R} $$) and satisfies $$ f_1(X) = \alpha $$
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I must confess that while I have a vague and general idea of what is going on here (very vague :-( ) I do not fully comprehend this Lemma.
Could someone give me a simple example showing explicitly what is going on here - and if possible, explain the situation?
Is this about the evaluation of polynomials? In the proof (see attachment) Sharp mentions that $$ f_1 $$ would have to satisfy
$$ f_1(rX^i) = f(r) {\alpha}^i $$
This appears to be a generalisation of the evaluation homomorphism (in the evaluation homomorphism we would have f(r) = r) - is that correct?
Would appreciate someone explaining the situation using an example.
Peter
Lemma 1.13 on page 7 (see attachment) reads as follows:
--------------------------------------------------------------------------------------
1.13 LEMMA. let R be a commutative ring, and let X be an indeterminate; let T be a commutative R-algebra with structural ring homomorphism $$ f \ : \ R \ \to \ T $$; and let $$ \alpha \in T $$.
Then there is a unique ring homomorphism $$ f_1 \ : \ R[X] \ \to \ T $$ which extends $$ f $$ (that is, is such that $$ f_{1|R} $$) and satisfies $$ f_1(X) = \alpha $$
--------------------------------------------------------------------------------------
I must confess that while I have a vague and general idea of what is going on here (very vague :-( ) I do not fully comprehend this Lemma.
Could someone give me a simple example showing explicitly what is going on here - and if possible, explain the situation?
Is this about the evaluation of polynomials? In the proof (see attachment) Sharp mentions that $$ f_1 $$ would have to satisfy
$$ f_1(rX^i) = f(r) {\alpha}^i $$
This appears to be a generalisation of the evaluation homomorphism (in the evaluation homomorphism we would have f(r) = r) - is that correct?
Would appreciate someone explaining the situation using an example.
Peter