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This is regarding exercise 5.22 in Atiyah-MacDonald :

$A$ is a subring of an integral domain $B$ such that $B$ is finitely generated over $A$. Prove that if the Jacobson radical of $A$ is zero, then so is the Jacobson radical of $B$.

From a previous exercise (5.20) we have the following result:

There exist $y_1, \dots, y_r\in B$ algebraically independent over $A$ and an $s\in A\setminus 0$ such that $B_s$ is integral over $B_s'$ where $B' = A[y_1, \dots, y_r]$.

Here's my solution. With $y_1, \dots, y_r, s$ as above, I claim that any non constant $f\in B'$ is not a unit in $B$. Suppose not, say $f\in B'$ has an inverse $g\in B$, then $g/1$ satisfies a monic polynomial over $B_s'$ of some degree $m$. Multiplying such a relation by $f^{m-1}$ gives a relation of the form $$g-\frac{p}{s^k} = 0, p\in B', k\geq 0.$$ Multiplying again by $f$ and rearranging, $$s^k = pf.$$ Since $f$ is a non constant polynomial, and $p\neq 0$ we have an algebraic relation in $B'$ which is impossible.

Next, let $b\in B$ be a nonzero non unit and let $b/1$ satisfy a monic polynomial over $B_s'$. Clearing the denominators and rearranging we get an equation of the form $$bx+f = 0, x\in B, f\in B'.$$ Note that $f$ is a nonzero non unit in $B$. First, suppose $r\geq 1$, then multiplying the above by $y_1$ and adding $1$ we get $bxy_1+(fy_1+1) = 1$ and $fy_1+1$ is a non constant polynomial in $B'$, hence not a unit in $B$. This means that $b$ is not in a maximal ideal containing $fy_1+1$.

So we are left with the case when $r = 0$. This means that there is an $s\in A\setminus 0$ such that $B_s$ is integral over $A_s$. From the arguments above, it suffices to prove that for every $0\neq f\in A$ there is a maximal ideal in $B$ not containing $f$. If $f$ is a unit in $A$ this is obvious, so assume $f$ is not a unit, in which case there's a maximal ideal $\mathfrak{m}$ of $A$ not containing $sf$, hence not containing $s$ and $f$.

Push the maximal ideal to $A_s$, use integrality to obtain a maximal ideal $\mathfrak{m}'$ lying over $\mathfrak{m}$ in $B_s$ and then pull it back to $B$. This is a maximal ideal not containing $f$ as required.

My answer seems correct, am I missing something? Anyway, this is the hint in the book

To take any $v\neq o$ in $B$, localize at $v$, then $B_v$ is finitely generated over $A$. Obtain $s$ as above and a maximal ideal $\mathfrak{m}\not\ni s$ in $A$, and consider the map $A\to A/\mathfrak{m} = k$. Extend this homomorphism to $g\colon B_v\to\Omega = \bar{k}$. Then $g(v) \neq 0$ and $ker(g)\cap B$ is maximal in $B$ not containing $v$.

I don't see how the kernel is maximal in $B$, could someone explain that?

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    $\begingroup$ Here's a previous solution where the user followed the hint. $\endgroup$
    – rschwieb
    Nov 9, 2020 at 15:03
  • $\begingroup$ Thanks for the link @rschwieb, that works. Is my solution correct? $\endgroup$ Nov 9, 2020 at 16:07
  • $\begingroup$ sorry, don’t have time to check. $\endgroup$
    – rschwieb
    Nov 9, 2020 at 17:27

1 Answer 1

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It is clear that given the homomorphism $ g \colon B_v \rightarrow \Omega$ we have $g(v) \neq 0$, since if $g(v)=0$ then $g(1)=g(v v^{-1})=g(v)g(v^{-1})=0$, which leads to a contradiction. Furthermore, it is clear that $ker(g) \cap B$ is a maximal ideal in $B$; note that $ker(g) \cap B \cap A = \mathfrak{m}$ and since $B$ is integral over $A$, this shows that $ker(g) \cap B$ is maximal in $B$ (by Corollary 5.8 in Atiyah-Macdonald). Be careful that when applying (5.8), one must check that $ker(g) \cap B$ is prime in $B$ (this is clear since $B/ker(g)$ is isomorphic to a subring of $\Omega$.)

Hope this helps!

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    $\begingroup$ Sorry for the late comment. But could you explain more on why $B$ is integral over $A$, since in the exercise, we only know that $B$ is finitely generated $A$-algebra, which may not be a finitely generated $A$-module, if I haven't got that wrong. Thank you! :) $\endgroup$
    – Hetong Xu
    May 20 at 12:09

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