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I don't understand the first sentence in the proof of Proposition 5.23 of Atiyah, Macdonald, Introduction to Commutative Algebra.

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Assume we know the result true for all $A$-algebras that can be generated by $n$ elements. Suppose $x_1,\dots,x_{n+1}\in B$ generate $B$ as an $A$-algebra, and let $v\in B$ be non-zero. But it might be the case that the only $S\subset\{x_1,\dots,x_{n+1}\}=X$ such that $v\in A[S]$ is $X$ itself, so I don't see how to exploit the induction hypothesis.

Alternatively, from the induction hypothesis one could obtain: there exists $u_0\neq 0$ in $A$ such that any homomorphism $f$ of $A$ into an algebraically closed field $\Omega$ with $f(u_0)\neq 0$ can be extended to a homomorphism $g$ of $A[v]$ into $\Omega$ with $g(v)\neq 0$. But now, how can we extend $g$ to all $B$?

Elías Guisado Villalgordo
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2 Answers2

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Assume the cases $n=1$ and $n=m$ have been proved and let us prove the case $n=m+1,$ so let $v$ be a non-zero element of $B=A[x_1,\dots,x_{m+1}].$

  • By the case $n=m,$ there exists a non-zero element $u_1\in A[x_1]$ such that any homomorphism $f_1$ from $A[x_1]$ to an algebraically closed field $\Omega$ such that $f_1(u_1)\ne0$ extends to a homomorphism $g:B\to\Omega$ such that $g(v)\ne0.$
  • By the case $n=1,$ there exists a non-zero element $u\in A$ such that any homomorphism $f$ from $A$ to an algebraically closed field $\Omega$ such that $f(u)\ne0$ extends to a homomorphism $f_1:A[x_1]\to\Omega$ such that $f_1(u_1)\ne0.$

Putting things together, we get the desired result.

(It goes without saying that you could alternatively apply first the case $n=1$ and then $n=m,$ the intermediate algebra $A[x_1]$ being then replaced by, say, $A[x_1,\dots,x_m]$.)

Anne Bauval
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Using your notation, let $A'=A[x_1,\ldots,x_n]$. Then $B=A'[x_{n+1}]$. By the $n=1$ case, which is explained in the text, there is $u'\in A'$ such that any homomorphism $f':A'\to \Omega$ with $f(u')\neq 0$ can be extended to a homomorphism $g:B\to \Omega$ and $g(v)\neq0$.

By induction hypothesis (applied to the finitely generated algebra $A\to A'$), for $u'\in A'$, there is $u\in A$ such that any homomorphism $f:A\to \Omega$ with $f(u)\neq0$ can be extended to $f':A'\to \Omega$ with $f'(u')\neq 0$.

Thus, the required element in $A$ is $u$. Given any homomorphism $f:A\to \Omega$ with $f(u)\neq 0$ we can first extend to a homomorphism $f':A'\to \Omega$ with $f'(u')\neq 0$ and then to $g:B\to \Omega$ with $g(v)\neq 0$.

user5389
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