For two unital rings $R,S$, let $R\sim S$ denote that $R$ and $S$ are Morita equivalent, i.e. the categories $\mathrm{Mod}_R$ and $\mathrm{Mod}_S$ are equivalent. Examples include:

  • Matrix rings: $R\sim M_n(R)$ for any $n\geq 1$,
  • Full corners: $R\sim eRe$ whenever $e$ is a full idempotent in $R$, i.e. $e=e^2$ and $ReR=R$.

The former is an example of the latter, since $R\simeq \varepsilon_{11}M_n(R)\varepsilon_{11}$ where $\varepsilon_{11}\in M_n(R)$ is the idempotent with $1$ in the $(1,1)$ entry and $0$'s everywhere else.

It turns out these are essentially the only ways to construct Morita equivalent rings.

Theorem. $R\sim S$ if and only if $S\simeq eM_n(R)e$ for some $n\geq 1$ and some full idempotent $e\in M_n(R)$.

If $R$ is a unital ring, a proper ideal $P\subseteq R$ is prime if $IJ\subseteq P$ implies $I\subseteq P$ or $J\subseteq P$ for all ideals $I,J$. Let $\mathrm{Spec}(R)$ denote the set of prime ideals of $R$, called the prime spectrum of $R$, with the usual Zariski topology: the closed sets are

$$V(I):=\{ P\in\mathrm{Spec}(R) : P\supseteq I\}$$

for $I$ an ideal of $R$.

My question is: if $R$ and $S$ are Morita equivalent rings, then is $\mathrm{Spec}(R)$ homeomorphic to $\mathrm{Spec}(S)$? I'm also wondering if Morita equivalence determines the poset of ideals. By the above Theorem, we just have to investigate the cases where $S=M_n(R)$ or $S=eRe$.

It's not hard to see that $I\mapsto M_n(I)$ is an order-preserving bijection between ideals of $R$ and ideals of $M_n(R)$, which is multiplicative in the sense that $M_n(IJ)=M_n(I)M_n(J)$, and this multiplicativity implies that it takes prime ideals to prime ideals. It's also easily seen to be continuous when restricted to $\mathrm{Spec}(R)$. So we can conclude that

$$\mathrm{Spec}(R) \simeq \mathrm{Spec}(M_n(R)).$$

Now how about full corners? If $eRe$ is not full then it's easy to see that it's false: for example $S=\mathbf{Z}$ is a corner of $R=\mathbf{Z}\oplus \mathbf{Z}$ where $e=(1,0)\in R$, and yet $\mathrm{Spec}(\mathbf{Z})\not\simeq \mathrm{Spec}(\mathbf{Z}\oplus\mathbf{Z})$. This corner is not full though.

Here's my idea: we have a map $$\Phi:\{\text{ideals of } R\} \rightarrow \{\text{ideals of } eRe\}$$ $$I\longmapsto eIe$$ which is surjective --- a right inverse is given by $\Psi(J):=RJR$.

  • Is $\Phi$ injective if $eRe$ is a full corner?
  • Does $\Phi$ preserve prime ideals? Would it be Zariski-continuous?

EDIT: Clearly order-preserving maps are Zariski-continuous, so $\Phi$ is definitely continuous.


1 Answer 1


I figured it out --- all my dreams were true.

$\Phi$ is injective

Since $e$ is full we have $ReR=R$. Now for any ideal $I$ of $R$ we have

$$I=RIR=(ReR)I(ReR) = (Re)I(eR) = R(eIe)R$$

thus observing that $eIe$ completely determines $I$. So $I\mapsto eIe$ is injective.

$\Phi$ is multiplicative

We want to show that $eIJe =(eIe)(eJe)$. But this is again fullness of $e$.

$$eIJe = (eI)R(Je) = (eI)(ReR)(Je) = eIeJe = (eIe)(eJe).$$

We conclude $\Phi$ restricts to a bijection $\mathrm{Spec}(R)\rightarrow \mathrm{Spec}(eRe)$. But $\Phi$ and $\Phi^{-1}$ are order-preserving, in particular they are Zariski-continuous. So $\Phi$ gives a nice homeomorphism

$$\mathrm{Spec}(R)\simeq \mathrm{Spec}(eRe).$$

  • 1
    $\begingroup$ A natural followup question is to write down a definition of the prime spectrum solely in terms of $\text{Mod}(R)$. I think such a definition is known but I'm not personally familiar with it. $\endgroup$ Oct 7, 2015 at 21:46

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