let $R$ be a local ring, $\varphi\in \End(R_{R}^{2})$, $\overline{\varphi}\in \End(\overline{R}_{\overline{R}}^{2})$, $\overline{R} =R/J(R)$ , $J(R)$= Jacobson radical $R$. where neither $\varphi$ nor $1-\varphi$ is invertible. Why neither $\overline{\varphi}$ nor 1- $\overline{\varphi}$ is invertible in $\End(\overline{R}_{\overline{R}}^{2})$ ?

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    $\begingroup$ I had a go at cleaning up the LaTeX, but am not sure it still makes sense $\endgroup$ – Yemon Choi Jul 10 '11 at 20:19
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    $\begingroup$ I did a little more LaTeX cleaning up. I guess $R^2_R$ is the $R$-module $R^2$. $\endgroup$ – Mark Sapir Jul 10 '11 at 20:24
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    $\begingroup$ This is a trivial exercise. An endomorphisms $\phi\in \mathrm{End} R^n$ is invertible if and only if $\det \phi$ is if and only if $\overline{\det\phi}$ is if and only if $\overline{\phi}$ is. $\endgroup$ – Mohan Jul 10 '11 at 21:44
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    $\begingroup$ @Mohan: There is nothing in the original post (look at the history!) that tells us $R$ is commutative. $\endgroup$ – darij grinberg Jul 10 '11 at 22:27
  • $\begingroup$ To support Darijs remark: In commutative algebra, no one speeks of the jacobson radical of a local ring; it's just the maximal ideal. $\endgroup$ – Martin Brandenburg Jul 11 '11 at 7:35

Clearly, it is enough to prove that if $\overline{\varphi}$ is invertible, then so is $\varphi$. (In fact, showing the same thing for $1-\varphi$ is analogous.)

Also it is clear that $\mathrm{End}\left(R_R^2\right)\cong \mathrm{M}_2\left(R\right)$ (canonically in $R$), and thus we can WLOG consider $\varphi$ as a $2\times 2$ matrix over $R$, with $\overline{\varphi}$ being its projection onto $\mathrm{M}_2\left(\overline{R}\right)$.

So what we have to prove now is the following:

Proposition. Let $R$ be a (not necessarily commutative) ring, and let $J\left(R\right)$ denote the Jacobson radical of $R$. Let $n\in\mathbb N$, and let $A\in \mathrm{M}_n\left(R\right)$. Then, $A$ is invertible if and only if the matrix $\overline{A}\in\mathrm{M}_n\left(R/J\left(R\right)\right)$ is invertible. Here, $\overline{A}$ denotes the image of $A$ under the canonical projection $\mathrm{M}_n\left(R\right)\to \mathrm{M}_n\left(R/J\left(R\right)\right)$ which is induced by the canonical projection $R\to R/J\left(R\right)$.

Before we prove this, a lemma:

Lemma. Let $R$ be a (not necessarily commutative) ring, and let $J\left(R\right)$ denote the Jacobson radical of $R$. Let $n\in\mathbb N$. Then, $\mathrm{M}_n\left(J\left(R\right)\right)$ (the subset of $\mathrm{M}_n\left(R\right)$ formed by all matrices each of whose entries lies in $J\left(R\right)$) is contained in the Jacobson radical $J\left(\mathrm{M}_n\left(R\right)\right)$.

Actually it is not just contained but equal to $J\left(\mathrm{M}_n\left(R\right)\right)$, as proven in the first absatz of this text (and probably many texts about noncommutative rings), but we only need the "contained" part. Still let me give a proof of this Lemma in the Appendix at the end of this reply. But first let us prove the Proposition using the Lemma:

Proof of the Proposition. The $\Longrightarrow$ direction is trivial, so let us now prove the $\Longleftarrow$ direction.

Assume that $\overline{A}$ is invertible. Then, there exists some matrix $U\in \mathrm{M}_n\left(R/J\left(R\right)\right)$ such that $\overline{A}U=U\overline{A}=I_n$. Clearly, there exists some matrix $B\in \mathrm{M}_n\left(R\right)$ such that $U=\overline{B}$ (because the canonical projection $\mathrm{M}_n\left(R\right)\to \mathrm{M}_n\left(R/J\left(R\right)\right)$ is surjective). Now, $\overline{AB}=\overline{A}\overline{B}=\overline{A}U$ (since $\overline{B}=U$), so that $\overline{AB}=\overline{A}U=I_n=\overline{I_n}$. Hence, $AB-I_n$ lies in the kernel of the canonical projection $\mathrm{M}_n\left(R\right)\to \mathrm{M}_n\left(R/J\left(R\right)\right)$. But this kernel is $\mathrm{M}_n\left(J\left(R\right)\right)$. Thus, $AB-I_n\in \mathrm{M}_n\left(J\left(R\right)\right)\subseteq J\left(\mathrm{M}_n\left(R\right)\right)$ (due to the Lemma). Hence, $I_n-AB=-\left(AB-I_n\right)$ lies in $ J\left(\mathrm{M}_n\left(R\right)\right)$ as well, so that (by the definition of $J\left(\mathrm{M}_n\left(R\right)\right)$) we see that $I_n-\left(I_n-AB\right)$ is invertible. In other words, $AB$ is invertible. That is, there exists some $C\in \mathrm{M}_n\left(R\right)$ such that $CAB=ABC=I_n$. Now, $ABC=I_n$ shows that $A$ is right-invertible. Similarly, $A$ is left-invertible, and we conclude that $A$ is invertible. This proves the $\Longleftarrow$ direction of the Proposition, qed.

There might be a more interesting algebraic undercurrent flowing beneath this argument; for example, does the Lemma have anything to do with the Morita equivalence between $R$ and $\mathrm{M}_n\left(R\right)$ ? I don't know. Some noncommutative algebraist could enlighten me.

Appendix: I don't feel like bumping up this thread for the only reason to fix a single typo, so let's add an alternative proof for the lemma. It is long (though the proof linked above isn't too short either when detailed), but I am somewhat proud of the penultimate paragraph.

Proof of the Lemma. Let $D\in\mathrm{M}_n\left(J\left(R\right)\right)$ be arbitrary. We want to prove that the matrix $I_n-DE$ is invertible for every $E\in\mathrm{M}_n\left(R\right)$.

So let $E\in\mathrm{M}_n\left(R\right)$ be arbitrary. Since $D\in\mathrm{M}_n\left(J\left(R\right)\right)$, every entry of the matrix $D$ lies in $J\left(R\right)$.

Now, the entries of the matrix $DE$ are sums of products of entries of $D$ with entries of $E$. Since the entries of $D$ all lie in $J\left(R\right)$, and since $J\left(R\right)$ is an ideal, this yields that all entries of $DE$ lie in $J\left(R\right)$.

The (noncommutative) Nakayama lemma says that if $M$ is a finitely generated right $R$-module such that $M\cdot J\left(R\right)=M$, then $M=0$. Now, let $R^n$ be the right $R$-module of column vectors of length $n$ over $R$. Then, $\mathrm{End}_R\left(R^n\right)$ can be canonically identified with the matrix ring $\mathrm{M}_n\left(R\right)$, acting from the left(!) on $R^n$. Let $\left(e_1,e_2,...,e_n\right)$ be the standard basis of the right $R$-module $R^n$. Now, let $M$ be the quotient module $R^n / \left(\left(I_n-DE\right)R^n\right)$. Then, $M$ is a finitely generated right $R$-module (being a quotient of $R^n$) with generating system $\left(\overline{e_1},\overline{e_2},...,\overline{e_n}\right)$, where $\overline x$ denotes the residue class of a vector $x\in R^n$ modulo the submodule $\left(I_n-DE\right)R^n$. For every $i\in\left\lbrace 1,2,...,n\right\rbrace$, we have $e_i-DEe_i=\left(I_n-DE\right)e_i\in \left(I_n-DE\right)R^n$, so that $\overline{e_i}=\overline{DEe_i}$. Now, the vector $DEe_i$ is the $i$-th column of the matrix $DE$ (because $e_i$ is the $i$-th basis vector of the standard basis $\left(e_1,e_2,...,e_n\right)$ of $R^n$). Thus, all entries of the vector $DEe_i$ are entries of the matrix $DE$, and thus lie in $J\left(R\right)$ (since all entries of the matrix $DE$ lie in $J\left(R\right)$). Hence, $DEe_i \in R^n\cdot J\left(R\right)$, so that $\overline{DEe_i} \in \underbrace{\left(R^n / \left(\left(I_n-DE\right)R^n\right)\right)}_{=M} \cdot J\left(R\right) = M \cdot J\left(R\right)$. We have thus shown that $\overline{e_i} = \overline{DEe_i} \in M \cdot J\left(R\right)$ for every $i$. Thus, $\overline{e_i}\cdot R \subseteq M \cdot \underbrace{J\left(R\right)\cdot R}_{\subseteq J\left(R\right)\text{ (since }J\left(R\right)\text{ is an ideal)}} \subseteq M\cdot J\left(R\right)$ for every $i$.

But since $\left(\overline{e_1},\overline{e_2},...,\overline{e_n}\right)$ is a generating system of $M$, we have $M = \sum_{i=1}^n \underbrace{\overline{e_i}\cdot R}_{\subseteq M\cdot J\left(R\right)} \subseteq \sum_{i=1}^n M\cdot J\left(R\right) \subseteq M\cdot J\left(R\right)$ (since $M\cdot J\left(R\right)$ is a right $R$-module). Combined with the obvious inclusion $M \cdot J\left(R\right)\subseteq M$, this yields $M = M \cdot J\left(R\right)$. Hence, the Nakayama lemma (quoted above) yields that $M=0$. Since $M = R^n / \left(\left(I_n-DE\right)R^n\right)$, this shows that $\left(I_n-DE\right)R^n = R^n$. In other words, the left action of the matrix $I_n-DE$ on the right $R$-module $R^n$ (seen as a right $R$-module map $R^n\to R^n$) is surjective. Since every surjective right $R$-module map $R^n\to R^n$ is right-invertible, this yields that the left action of the matrix $I_n-DE$ on the right $R$-module $R^n$ (seen as a right $R$-module map $R^n\to R^n$) is right-invertible. In other words, the matrix $I_n-DE$ is right-invertible.

But now let us look again at the fact that the left action of the matrix $I_n-DE$ on the right $R$-module $R^n$ (seen as a right $R$-module map $R^n\to R^n$) is right-invertible. By applying the same argument to the left $R$-module $R^n$ instead of the right $R$-module $R^n$, we can show that the right action of the matrix $I_n-DE$ on the left $R$-module $R^n$ (seen as a left $R$-module map $R^n\to R^n$) is right-invertible. But this time, this means that the matrix $I_n-DE$ is left-invertible. (The reason for this is the following: If $X$ and $Y$ are two $n\times n$ matrices over $R$, then the right action of the matrix $XY$ on the left $R$-module $R^n$ is the composition $\left(\text{the right action of the matrix }Y\text{ on the left }R\text{-module }R^n\right)\circ \left(\text{the right action of the matrix }X\text{ on the left }R\text{-module }R^n\right)$, and not the other way round. Thus, an $n\times n$ matrix is left-invertible if and only if its right action on the left $R$-module $R^n$ is right-invertible.).

Thus we now know that the matrix $I_n-DE$ is left-invertible and right-invertible. This yields that the matrix $I_n-DE$ is invertible. Since this is proven for every $E\in\mathrm{M}_n\left(R\right)$, this yields that $D\in J\left(\mathrm{M}_n\left(R\right)\right)$ (by the definition of the Jacobson radical). Since we have proven this for every $D\in\mathrm{M}_n\left(J\left(R\right)\right)$, we thus have shown that $\mathrm{M}_n\left(J\left(R\right)\right) \subseteq J\left(\mathrm{M}_n\left(R\right)\right)$. The Lemma is proven.

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    $\begingroup$ Yes, this result does follow from Morita theory. If R and S are Morita equivalent rings, then there is an isomorphism between the lattices of (two-sided) ideals of R and of S. This isomorphism preserves the anninilators of modules. Since the Jacobson radical is the intersection of all (say) right primitive ideals of a ring, this correspondence preserves the radical of a ring. The last thing that you need to know is that the correspondence between R and M_n(R) is the obvious one: it sends an ideal I of R to M_n(I). [Reference: T.Y. Lam, Lectures on Modules and Rings, Cor. 18.50.] $\endgroup$ – Manny Reyes Sep 1 '11 at 19:58
  • $\begingroup$ "Anninilators": they send every module element to something nilpotent? ;) Thanks, a very informative comment. $\endgroup$ – darij grinberg Sep 1 '11 at 20:25

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