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If I have two $K$-algebras $A$ and $B$ (associative, with identity) and an algebra homomorphism $f\colon A\to B$, is it true that $f(\operatorname{rad}A)\subseteq\operatorname{rad}B$, where $\operatorname{rad}$ denotes the Jacobsen radical, the intersection of all maximal right ideals?

I can think of two proofs in the case that $f$ is surjective, but both depend on this surjectivity in a crucial way. The first uses the formulation of $\operatorname{rad}A$ as the set of $a\in A$ such that $1-ab$ is invertible for all $b\in A$, and the second treats an algebra as a module over itself, and uses the fact that the radical of $A$ as a module agrees with the radical of $A$ as an algebra, and is the intersection of kernels of maps onto simple modules; here the surjectivity is needed to make $B$ into an $A$-module in such a way that the radical of $B$ as an $A$-module is contained in the radical of $B$ as a $B$-module.

If a counter example exists, $A$ will have to be infinite-dimensional, as in the finite dimensional case all elements of $\operatorname{rad}A$ are nilpotent, and (I think, although I don't remember a proof right now, so maybe I'm wrong) that the radical always contains every nilpotent element.

This is my first question on here, so let me know if I should have done anything differently!

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Dear Matt, welcome to this site. Your question is extremely clear and shows you have thought about it carefully.The prediction that a counterexample will be infinite dimensional has an absolutely correct proof . So, no: definitely don't do anything differently! –  Georges Elencwajg Feb 17 '12 at 12:08
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That the radical contains all nilpotents follows from the pleasantly analogous characterization (in the commutative case) of the set of nilpotents as the intersection of all prime ideals, which is obviously smaller than the Jacobson radical, where you intersect only the maximal ideals. –  Georges Elencwajg Feb 17 '12 at 12:22
    
Typesetting exercise: \operatorname{rad} versus \mathrm{rad}: $\operatorname{rad} A$ versus $\mathrm{rad} A$. A difference should be visible. –  Michael Hardy Feb 17 '12 at 17:03
    
@Georges: alternately, note that by Schur's lemma any element of a ring that acts nontrivially on a simple module acts invertibly, but a nilpotent element can never act invertibly. –  Qiaochu Yuan Feb 17 '12 at 17:06
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"the radical always contains every nilpotent element": That's not true (unless I'm missing a commutativity assumption in the question). For instance $\mathbb{M}_n(K)$ where $n>1$ contains many nonzero nilpotent elements but has zero radical. –  Cihan Apr 23 '12 at 9:48

2 Answers 2

up vote 8 down vote accepted

No, it is false that the Jacobson radical is sent to the Jacobson radical.

Take $A=K[X]_{(X)}$ [localization at $(X)$], $B=K(X)$ and the inclusion $f:K[X]_{(X)}\hookrightarrow K(X)$
Then $f(Rad(A))=f(XK[X]_{(X)})=XK[X]_{(X)} \nsubseteq Rad(B)=Rad(K(X))=(0)$

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Thanks, that's great. :-) –  Matt Pressland Feb 17 '12 at 13:16
    
I like this example Georges. It also serves to show that for a dominant map of irreducible $K$-schemes $X\rightarrow Y$, without finiteness hypotheses, $\dim(X)$ need not be greater than or equal to $\dim(Y)$. I think I will always remember $K[X]_{(X)}$ now! –  Keenan Kidwell Feb 17 '12 at 19:30
    
Ah, but I hadn't thought of this interesting consequence/interpretation: thanks a lot, @Keenan! –  Georges Elencwajg Feb 17 '12 at 19:49

This is pretty obviously wrong: Take $A$ to be the ring of upper-triangular $2 \times 2 $-matrices $ \left[ {\begin{array}{*{20}c} {K } & {K} \\ {0 } & {K } \\ \end{array}} \right]$, $B$ to be the ring of all $2\times 2$ -matrices $ \left[ {\begin{array}{*{20}c} {K } & {K} \\ {K } & {K } \\ \end{array}} \right]$, and $f$ to be the canonical inclusion. Then $radA=\left[ {\begin{array}{*{20}c} {0 } & {K} \\ {0 } & {0 } \\ \end{array}} \right]$ and $radB=0$. However, when $f$ is a surjective homomorphism of $K$-algebras, it is right. In this case, the inclusion $f(radA)\subseteq radB$ is clear since, for any maximal left ideal $m$ of $B$, the inverse image $f^{-1}(m)$ is a maximal left ideal of $A$.

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To give an example of strict inclusion, let $A=Z$ and let $f$ be the natural projection of $A$ onto $B=Z/4Z$. Here, $f(radA)=f(0)=0$ but $ radB=2Z/4Z\neq0$ –  Aimin Xu Nov 2 '12 at 23:54

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