Isomorphism of $k$-Algebras

Question

Let $k$ be a field an $A$ a finitely generated $k$-algebra (think of the polynomial ring). Suppose I have an isomorphism of $k$-algebras $$ A \cong A/I $$

for some ideal $I$. If $A$ were a finitely generated $k$-vector space we can conclude $I=0$, as opposed to the non finitely generated case. So can we still get $I=0$, even though $A$ is not f.g. as a vector space over $k$, only as an algebra?

If you mean finitely generated as a $k$-algebra, you can reduce the problem to finding ideals $I$ in $A=k[x_1,\ldots, x_n]$ such that $A\cong A/I$. — Nishant, Mar 24 '16 at 23:31

score 4 · Answer 1 · edited Mar 25 '16 at 00:18

4

Your question is usually phrased as follows:

is a finitely generated algebra Hopfian?

A well-known theorem of Maltsev states that every presentable finitely generated algebra is Hopfian —presentable means here that the algebra embeds in a matrix algebra $M_n(R)$ for some commutative ring $C$. This condition holds for an algebra $A$ if one can separate elements in $A$ with maps to algebras of finite dimension, for example (One says in this case that the algebra $A$ is residually finite dimensional)

But not all finitely generated algebras are Hopfian. For example, there exist finitely generated groups $G$ with surjections $G\to G$ which are not injective, and the induced morphisms of group algebras $k G\to k G$ are surjective algebra maps which are not injective. The classical example is the Baumslag-Solitar group

edited Mar 25 '16 at 00:18

Pedro

122,002

answered Mar 24 '16 at 23:34

Mariano Suárez-Álvarez

135,076

But every such group is non-abelian. I assume the OP meant to ask about commutative $k$-algebras, given the (commutative-algebra) tag on this question... – Alex Kruckman Mar 24 '16 at 23:38
1

A (left or right) noetherian ring is hopfian, so nothing interesting happens there. – Mariano Suárez-Álvarez Mar 24 '16 at 23:42
(That is proved here for groups, and the same argument, mutatis mutandi, works here) – Mariano Suárez-Álvarez Mar 24 '16 at 23:44
Right, and I think this statement, together with a sketch of the easy proof, would be a better answer to the OP's question. – Alex Kruckman Mar 24 '16 at 23:45
You are free to provide that. – Mariano Suárez-Álvarez Mar 24 '16 at 23:46

Alex Kruckman · Answer 2 · 2016-04-16T02:55:58.763

The answer is yes for finitely generated commutative $k$-algebras (Mariano's answer deals with the non-commutative case). More generally, if $A$ is a Noetherian ring and there is an isomorphism $f\colon A/I \to A$, then $I = (0)$.

Why? Let $\pi\colon A\to A/I$ be the quotient map. If $I\neq (0)$, then $f^{-1}[I]$ is a nonzero ideal in $A/I$, so $\pi^{-1}[f^{-1}[I]]$ is an ideal in $A$ which properly contains $I$. Repeating, we can build an infinite ascending chain of ideals.

Edit: Here are a few more details. We start with $I_0 = (0)$ in $A$, and we set $I_{i+1} = \pi^{-1}[f^{-1}[I_i]]$ for all $i$. Let's check by induction that we get a proper increasing sequence $I_0 \subsetneq I_1 \subsetneq \dots$ of ideals in $A$. In the base case, $I_1 = \pi^{-1}[f^{-1}[(0)]] = \pi^{-1}[(0)] = I$, and we've assumed that $(0) \subsetneq I$. And given the inductive hypothesis $I_{i-1} \subsetneq I_i$, we have $I_i = \pi^{-1}[f^{-1}[I_{i-1}]] \subsetneq \pi^{-1}[f^{-1}[I_{i}]] = I_{i+1}$.

Either there is a flaw or I'm overlooking something: in the "repeating" part, we get a sequence of ideals all containing $I$, but not necessarily their predecessor, as $\pi$ is mapping into the fixed quotient ring $A/I$. — TheHumanHighway, Apr 16 '16 at 01:48

Isomorphism of $k$-Algebras

2 Answers2