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- Jun 22, 2012

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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 \(\displaystyle f \ : \ R \ \to \ T \); and let \(\displaystyle \alpha \in T \).

Then there is a unique ring homomorphism \(\displaystyle f_1 \ : \ R[X] \ \to \ T \) which extends \(\displaystyle f \) (that is, is such that \(\displaystyle f_{1|R} \)) and satisfies \(\displaystyle f_1(X) = \alpha \)

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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 \(\displaystyle f_1 \) would have to satisfy

\(\displaystyle f_1(rX^i) = f(r) {\alpha}^i \)

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