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I don’t understand the part of the proof of prop5.23 of Atiyah-Macdonald, Introduction to commutative algebra. The author separates the proof into two cases, the case that x is transcendental or the case that x is algebraic over A. About the former case, I don’t find where the author uses the hypothesis that x is transcendental.

proof of the book

nat-cat
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If $x$ is algebraic over $A$ and $P(t) \in A[t]$ is a (possibly non-monic) polynomial with $P(x) = 0$, then in the proof of (i), the element $\xi$ has to be a root of the non-zero polynomial $f(P)(t) \in \Omega[t]$ (where $f(P)(t)$ is the polynomial we obtain by applying $f$ to the coefficients of $P(t)$).

But then there are only finitely many possibilities for $\xi$, and so we cannot guarantee that for $v = a_0x^n + a_1x^{n-1} + \dotsb + a_n$ as in the proof, we have that $f(a_0)\xi^n + f(a_1)\xi^{n-1} + \dotsb + f(a_n) \neq 0$.

One thing to note here is that $f$ may have non-trivial kernel, and hence, $f(a_0) \xi^n + \dotsb + f(a_n) = 0$ for all roots $\xi$ of $f(P)(t)$ does not imply that $v = 0$.

Claudius
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  • Thank you for your response. I understand that there may be complications when $x$ is algebraic. I am concerned about whether it is always possible to extend $f$ to $g$ so that $g(x) = ξ$ when x is transcendental. Does this mean that if $x$ is transcendental, we can freely choose the image because there are no restricting relationships for $x$? – nat-cat Feb 21 '24 at 04:18
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    You are right. If $x$ is transcendental over $A$, then $A[x]$ is the polynomial ring in one variable. The fact that for any ring morphism $f\colon A\to \Omega$ and any $\xi\in \Omega$ there exists a unique ring morphism $g\colon A[x]\to \Omega$ with $g\big|_{A} = f$ and $g(x) = \xi$ is known as the universal property of the polynomial ring. – Claudius Feb 21 '24 at 12:00
  • Thank you very much. I did not realize that it satisfies the universal property of the polynomial ring. My understanding has deepened. – nat-cat Feb 21 '24 at 12:10