# Jacobson radical of a finitely generated algebra (A-M 5.22)

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?

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

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 $$\Omega$$.)
• 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! :) May 20 at 12:09