Radicals

On one hand, for any $X ∈ M_n (J(R))$, $(I - XA)$ has a right inverse $(\det (I - XA))^{-1} ⋅ \mathrm{adj}(I - XA)$. Here $\det (I - XA) ∈ 1 + J(R)$ is a unit in $R$. On the other hand, $J(M_n (R))$ is a two-sided ideal of a full matrix ring, thus (consider entry $(1,1)$) it takes the form $M_n (I)$ for some two-sided ideal $I ⊆ R$. Taking diagonal matrices, we see $I ⊆ J(R)$, thus $I = J(R)$.

We show for cyclic modules $(x) ≃ R / I$. Note that the submodules corresponds to ideals containing $I$. Zorn’s lemma complete the proof. Assume we show every $k$-generated module has maximal submodules. For every $k+1$-generated module $⟨x_1, \ldots ,x_{k+1}⟩$, there exists a maximal submodule of $\frac{⟨x_1, \ldots ,x_{k+1}⟩}{⟨x_1, \ldots ,x_{k}⟩} ≃ \text{some cyclic module}$. This is recovers maximal submodules of $⟨x_1, \ldots ,x_{k+1}⟩$ containing $⟨x_1, \ldots ,x_{k}⟩$. Induction completes the proof.

Let $K = \frac{M}{N}$ be f.g.. It suffices to show $(KJ(R) = K) → (K = 0)$. Let $K'$ be any maximal submodule of $K$, then $(K/K')(R) = 0$, i.e., $KJ(R) ⊆ K'$. A contradiciton.

We take $K = \frac{M}{N}$. It suffices to show $(KI = K) → (K = 0)$. Now $K = KI^{n} = K0$ for some $n$.

Let $𝐦$ be the row vector of generating elements of $M$. There is $I$-square matrix $𝐀$ s.t. $𝐦𝐀 = 𝐦$. By characteristic polynomial, $(𝐈 - 𝐀) ⋅ \mathrm{adj}(𝐈 - 𝐀) = \det (𝐈 - 𝐀) ⋅ \mathrm{id}$. Note that $\det (𝐈 - 𝐀) = 1 - x$ for some $x ∈ I$. We see $(1 - x)$ annihilates the generating set of $M$, thus $M(1-x) = 0$.

(1). Recall that $J(M)$ is the intersection of morphism of the type $φ : M → S$ ($S$ is simple). We show $f(J(M)) ⊆ J(N)$.
(2). Is $f$ is an epimorphism with superfluous kernel, then $f$ and $f^{-1}$ shows a $1:1$ correspondence of maximal submodules of $M$ and $N$. The rest is clear. We show such corrrespondence:

• If $Y ⊆ N$ is maximal, then $f^{-1}(Y) ⊆ M$ is proper. Assume $f^{-1}(Y)$ is not maximal, then there is $x ∉ f^{-1}(Y)$ s.t. $(x) + f^{-1}(Y)$ is proper. Since $Y$ is maximal, $f^{-1}f((x) + f^{-1}(Y)) = f^{-1}(N) = M$. Note that $f^{-1}f((x) + f^{-1}(Y)) = (x) + f^{-1}(Y) + \ker f ≠ M$, a contradiction.
• If $X ⊆ M$ is maximal, then $f(X)$ is proper as $\ker f ⊆ X$. Assume $f(X)$ is not maximal, then there is $y ∉ f(X)$ s.t. $(y) + f(X)$ is proper. We take $x ∈ f^{-1}(y)$. Note that $x ∉ X$ and $(x) + X$ is proper, a contradiction.

(3). When there is $f : M ↠ N$, we see $\frac{M}{J(M)} ↠ \frac{N}{J(N)}$ by cokernel. Conversely, if $\frac{M}{J(M)} ↠ \frac{N}{J(N)}$, then $f(M) + J(N) = N$. Since $J(N) ⊆ J$ is superfluous, we obtain $f(M) = N$.
(4). If $\frac{M}{J(M)}$ has a superfluous submodule $\frac{X}{J(M)}$, then $X ⊆ M$ is superfluous. We must have $X = J(M)$.
(5). $J(X_∙ ) ⊆ J(∐ (X_∙ ))$ shows $∐ J(X_∙ ) ⊆ J(∐ X_∙ )$. Conversely, for any maximal submodule $X_i' ⊆ X_i$, $X_i' ⊕ ∐ _{≠ i}X_∙$ is a maximal submodule of $∐ X_∙$. Hence $∐ J(X_∙ )$ is an intersection of some maximal submodules of $∐ X_∙$. We obtain $J(∐ X_∙ ) ⊆ ∐ J(X_∙ )$.
(6). By previous analysis, $J(M) = 0$ if $M$ is a submodule of a product of simple modules. Conversely, if $J(M) = 0$, then any non-zero element $x ∈ M$ survives in some $M → S$. Let $S$ be the product of all simple modules, then there is monomorphism

$$\begin{equation} φ ∈ (M, ∏ _{(M,S)} S) ≃ ∏_{(M,S)}(M,S) ∋ 1_{(M,S)}. \end{equation}$$

Let $R$ be Artinian. There is a descending chain $J(R) ⊇ J(R)^2 ⊇ \cdots$ which stabilises at some $n$. We show $J^n(R) = 0$. By Artinian, there is some minimal non-zero ideal $I$ s.t. $J^n(R) ⋅ I ≠ 0$. Clearly, $J^n (R) ⋅ I = I$ and $I = (x)$ is principal. We take $r ∈ J^n (R)$ s.t. $rx = x$. Since $(1-r) x = 0$, we see $x = 0$, a contradiction!

A nilpotent element is annihilated by some polynomial $t^n$. Conversely, we take monic polymonial $p$ s.t. $p(x) = 0$ and assume $p$ is not monic. Assume the least term in $p(x)$ has degree $k$, then $(x^k) ⋅ (x) = (x^k)$. By Nakayama lemma, $x^k = 0$.

We show any $x ∈ J(R)$ is nilpotent (= algebraic). Note that the collection $\{(λ - x)^{-1} : λ ∈ (k ∖ \{0\})\}$ is linearly dependent, as $\dim R < |k|$. Clearing the denominators in a vanishing linear combination, we see $x$ is algebraic.

Well-known.

For any matrix ring $M_n (R)$, one has $𝐦𝐨𝐝 _{M_n (R)} = 𝐦𝐨𝐝 _{R}$. Note that $R$ is semisimple iff $𝐦𝐨𝐝 _R$ is semisimple category.

(1 ↔ 3) is due to compact-$\varinjlim^{fil}$ argument. (1 → 2) is clear. We show (2 → 3). Note that the identity $𝐩𝐫𝐨𝐣 = 𝐦𝐨𝐝$ is Morita invariant, and every $M ∈ 𝐦𝐨𝐝 _R$ is the quotient ring of some $M_n (R)$ by a f.g. ideal.
(2 → 4). Any principal ideal $(x)$ is generated by an idempotent element $e$. We write $x = ez$ and $e = xy$. Now

$$\begin{equation} x = ez = e(ez) = xyx. \end{equation}$$

(4 → 2). Clearly, principal ideals are projective. By induction, it suffices to show that any $2$-generated ideal $(x,y)$ is principal. We assume

• $x^2 = x$, since $(x)$ is projective;
• $xy = 0$, since $(x,y) = (x, (1-x)y)$.

We take $\overline y$ s.t. $y \overline y y = y$, and consider the idempotent element

$$\begin{equation} e := y \overline y (1-x) ,\quad e^2 = e. \end{equation}$$

We show $(e + x) = (x,y)$. $e,x ∈ (x,y)$ is clear. Conversely,

• $y = (e+x) y = ey = y \overline y y = y$, and
• $x = (e+x)x = ex + x = 0 + x = x$.

The final definition is two-sided. The identity $𝐩𝐫𝐨𝐣 = 𝐦𝐨𝐝$ is Morita invariant.

For any $x ∈ J(R)$, we take $y$ s.t. $xyx = x$. Now $1 - xy$ is not a unit, a contradiciton.

Suppose $R / I$ has no non-zero divisor. For $[x] ∈ R / I$, there is some $y$ s.t. $[x][y][x] = [x]$. Hence $[y]$ is a two-sided inverse of $[x]$ when $[x] ≠ 0$. Hence $R / I$ is a division ring.

Semisimple rings are Noetherian and vNR. Conversely, any ideal in a Noetherian vNR ring is projective, thus the ring is semisimple.

By Hahn-Banach theorem, for any $x,y ∈ (X ∖ 0)$ there is $φ ∈ \mathcal{B}(X)$ mapping $x$ to $y$. The $\mathcal{B}(X)$-module $X$ is generated by any non-zero element, thus it is simple. By definition, $\mathrm{ann}(X) = 0$. Hence $J(\mathcal{B}(X)) = 0$.
The ring $\mathcal{B}(X)$ is not vNR for most $X$. When $\mathcal{B}(X)$ is vNR, then all $T$ has a generalized inverse $S$ s.t. $TST = T$. Note that $\mathrm{Range}(T) = \mathrm{Range}(TS) = \ker (I - TS)$ is a closed subspace. However, not all continuous operators have closed range.

The local ring has idempotent $\{0,1\}$. Conversely,

Note that $I = Ie ⊇ IRe ∩ eRee = I ∩ eRe = I$. Hence the equality holds.

Note that

$$\begin{equation} J(eRe) = ⋂_{𝔪 ⊆ eRe} 𝔪 = eRe ∩ ⋂_{𝔪 ⊆ eRe} 𝔪R ⊆ eRe ∩ J(R) ⊆ eJ(R)e. \end{equation}$$

Conversely, for any $x ∈ J(R)$, we show $exe ∈ J(eRe)$. Note that any $e - exeeye = e(1-x(eye))$. We take $z$ s.t. $(1-xeye)z$ is a unit. Now $e - exeeye$ has right inverse $eze$.

We take the homomorphism of $eRe$-bimodules, and obtain

$$\begin{equation} eRe ↠ e' R' e',\quad exe ↦ e'x'e'. \end{equation}$$

The kernel is $eJ(R)e = J(eRe)$. Hence $eRe / J(eRe) ≃ e' R' e'$. The ring $eRe$ is local iff $e' R' e'$ is a division ring. We show $e' R' e'$ is a division ring iff $e' R'$ is a minimal (=simple) right ideal.

1. When $e'R'$ is simple, then $e'R'e'$ is a division ring by Schur’s lemma.
2. When $e'R'e'$ is a division ring, we show $e'R'$ is a simple right $R'$-module, i.e., $e'R'$ is cyclicly generated by arbitrary non-zero element $e'r'$. Note that $e's' = e'r'e (e'r'e')^{-1}(e's')$.

Let $0 = X_0 ⊆ \cdots ⊆ X_m = M$ and $0 ⊆ Y_0 ⊆ \cdots ⊆ Y_n = M$ be two composition series. By Zassenhaus lemma, we define

$$\begin{equation} M_{i,j}^→ := \frac{X_{i+1} ∩ (Y_{j+1} + X_i)}{X_{i+1} ∩ (Y_j + X_i)} ≃ \frac{Y_{j+1} ∩ (X_{i+1} + Y_j)}{Y_{j+1} ∩ (X_i + Y_j)} =: M_{i,j}^↓. \end{equation}$$

The tessellation $(M_{i,j})$ has precisely one simple factor in each row and column. Hence $m = n$ and the simple factors are the same.

Note that $\{\ker (f^n)\}_{n ≥ 1}$ stables at some $n$. We show $\ker f^n ∩ \operatorname{im} f^n = 0$.

• Suppose $x = f^n (y) ∈ \ker f^n ∩ \operatorname{im} f^n$. Then $y ∈ \ker f^{2n} = \ker f^n$. Hence $x = f^n (y) = 0$.

We also show that $\ker f^n + \operatorname{im} f^n = M$.

• For any $x ∈ M$, there is $y$ s.t. $f^{2n}(y) = f^n (x)$. Now $x = (x-f^n(y)) + f^n (y)$ is the desired decomposition.

By finite length, any direct sum decomposition lies in a maximal one. For any two maximal decompositions $M = ∐ _i X_i = ∐_j Y_j$. We show $X_1 ≃ Y_j$ for some $j$. We define $x_i : M \overset {π_i} ↠ X_i \overset {ι _i} ↪ M$ and $y_j$ (similarly). Note that all $x_i$ and $y_j$ are idempotents. Since $x_1 ∑ y_j = x_1$, there is some $j$ s.t. $x_1 y_j ≠ 0$. Note that $(x_1y_j)|_{Y_j} ∘ (y_jx_1)|_{X_1} ∈ \mathrm{End}(X_1)$ is a non-zero idempotent, hence a unit. We see $X_1$ is a direct summand of $Y_j$. The rest of the proof is clear.

Fitting lemma shows that non-units in $\mathrm{End}(M)$ are nilpotent. By Nakayama lemma, multiplying a nilpotent strictly decreases the length of a f.g. module. Hence, the multiplication of $n$ nilpotent endomorphisms is $0$. It remains to show $f$ is nilpotent iff $f(M)$ is a superfluous submodule.

• Suppose $f^n(M) = 0$ for some minimal $n ≥ 1$, and $f(M) + N = M$ for proper submodule $N$. We see $0 ≠ f^{n-1}(M) = f^{n-1}(N)$. Nakayama lemma shows $f^{n-1}(N) = 0$, a contradiction.
• Suppose $f(M) ⊆ M ≠ 0$ is superfluous, then it is a proper submodule by existence of maximal submodules. Note that the surjection never increase the length, hence there is no $g$ s.t. $gf(M) = M$. We see $f$ is a non-unit, thus nilpotent.

We take $π : A ⊕ B ↠ C$. Note that $π ((a,b)) = π ((0,b))$. Hence $π |_B$ is epic. If $π ((0,b)) = 0$, then $b ∈ A$. Hence $b = 0$, yields $π |_B$ is also monic. We see $π |_B$ is an isomorphism.

Let $M$ and $M'$ be $|I|$-exchangable. Suppose $\text{all} = M ⊕ M' ⊕ N = ∐ _I X_∙$. We obtain

$$\begin{equation} M ⊕ (M'⊕ N) = M ⊕ (∐ _I Y_∙),\quad φ : (M'⊕ N) ≃ ∐ _I Y_∙. \end{equation}$$

By assumption, $∐ _I Y_∙ = φ (M)' ⊕ ∐ _I Z_∙$. Hence

$$\begin{equation} M ⊕ M' ⊕ N = (M ⊕ M') ⊕ ∐ _I φ ^{-1}(Z_∙). \end{equation}$$

Let $M = M_1 ⊕ M_2$ be $|I|$-exchangable. Suppose $M_1 ⊕ N = ∐ _I X_∙$, then $M ⊕ N = (M_2 ⊕ X_0) ⊕ ∐ _{≠ 0} X_∙$. We take $Y_0 ⊆ (M_2 ⊕ X_0)$ and $Y_{≠ 0} ⊆ X_{≠ 0}$ s.t. $M ⊕ N = M ⊕ ∐ Y_∙$. Let $π : M ⊕ N ↠ M_1 ⊕ N$ be canonical projection. It straightforward to check that $M_1 ⊕ ∐ _I π (Y_∙) = M_1 ⊕ N$.

Suppose $M$ is $|I|$-exchangable. For any given $(f_i)$, there is a retract $M \xrightarrow {(f_∙ )} ∐ _I M \xrightarrow{Σ } M$. We take

$$\begin{equation} M ⊕ ∐ _I M_∙ ^s = ∐ _I M_∙ ,\quad M_∙ = M_∙ ^s ⊕ M_∙ ^c. \end{equation}$$

Let $τ : ∐ _I M_∙ ^c ≃ M$ denote the isomorphism by canonical projection. We take

$$\begin{equation} e_i : M \xrightarrow {(f_∙ )} \underset{\text{induced by } \ τ }{\underbrace{∐ _I M_∙ ↠ M_i ^c ↪ ∐ _I M_∙}} \xrightarrow {Σ } M. \end{equation}$$

Conversely, suppose there is identity $M ⊕ N = ∐_I X_∙$, we take the composition

$$\begin{equation} f_i := M ↪ M ⊕ N ↠ X_i ↪ ∐ _I X_∙ ↠ M. \end{equation}$$

This extends to $(f_∙ ) ∈ (M, ∐ M)$ which summing up to $1$. By assumption, there are orthogonal idempotents $\{g_i ∘ f_i\}_I$ summing up to $1$. Now $Y_i := \operatorname{im}(M ↪ M ⊕ N ↠ X_i )$ is the desired summand of $X_i$.

We show $M$ is $(k+1)$-exchangable when it is $k$-exchangable ($k ≥ 2$). We take $(f_i)_{i=0}^k$ summing up to $1$. We apply $2$-exchangable property to $f_0 + (∑ _{0<∙ ≤ k}f_∙)$, hence WLOG we assume $f_0$ is idenpotent. Set $e := (1-f_0)$. Now $∑ _{0< ∙ ≤ k}ef_∙ e = e$. Note $Me$ is $k$-exchangable with endomorphism ring $e \mathrm{End}(M)e$. Now there is some idempotent decomposition in $e\mathrm{End}(M)e$:

$$\begin{equation} ∑ _{0 < i ≤ k} e g_i e f_i e = e,\quad (e g_i e f_i e)(e g_j e f_j e) = δ ^{i,j} (e g_i e f_i e). \end{equation}$$

We set $F_i := e g_i e f_i e g_i e f_i$ for $0 < i ≤ k$, and $F_0 := 1 - ∑ _{0<∙ ≤ k}F_∙$.

• $F_i^2 = (e g_i e f_i e)^3 g_i e f_i = F_i$;
• $F_i F_j = e g_i e f_i (e g_i e f_i e g_j e f_j e) g_j e f_j = 0$ when $i ≠ j$;
• $F_0 F_{≠ 0} = F_{≠ 0} F_0 = 0$ and $F_0^2 = F_0$ follow from the construction;
• $F_0 f_0 = F_0 - F_0e = F_0$. Hence $F_0 = ? ∘ f_0$.

(1 → 2). Endolocal modules are indecomposable by checking idempotents. When $\mathrm{End}(M)$ is local, any $x+y = 1$ implies $x$ or $y$ is a unit. WLOG we take $x^{-1}x = 1$ and $0y = 0$, hence $M$ is $2$-exchangable. By previous analysis, $M$ is finite exchangable.
(2 → 3). We show $M$ is $|I|$ exchangable when it is finite exchangable. When $M ⊕ N = ∐ _I X_∙$, we take non-zero $m ∈ M$ which lies in a minimal finite summand $∐ _{I_0} X_∙$. Now we consider $M ⊕ N = ∐ _{I_0} X_∙ ⊕ (∐ _{I ∖ I_0}X_∙ ) =: ⨁ _{t=0}^{|I_0|} L_t$. By finite exchangable property, there is $L_t := Y_t ⊕ Z_t$ s.t. $M ⊕ ∐_{t=0} Y_t = M ⊕ N$. Hence $M ≃ ⨁_{t=0}^{|I_0|} Z_t$ by cancellable property. Since $M$ is indecomposable, there is precisely one $t_0$ s.t. $Z_{t_0} ≃ M$. We see $t_0 ≠ 0$, as $X_{i_0} ∩ M ≠ 0$ for any $i _0 ∈ I_0$. In this case,

$$\begin{equation} M ⊕ N = M ⊕ L_{t_0} ⊕ (∐ _{t ≠ t_0} X_t) ⊕ (∐ _{I ∖ I_0}X_∙ ). \end{equation}$$

Hence $M$ is $|I|$-exchangable.
(3 → 1). We take $κ =2$ and show that $x + y = 1$ implies $x$ or $y$ is a unit.

Such decomposition exists by definition. We show any indecomposable summand is endolocal.

• We take $L ⊕ N = ∐ M_∙$. For any non-zero $x ∈ L$, there is a minimal finite summand $∐ _{I_0} M_∙$ containing $x$. Note that $∐ _{I_0}X_∙ ⊕ ∐ _{I ∖ I_0}X_∙ = L ⊕ N$ where $∐ _{I_0}X_∙$ is finite exchangable. Now $L ⊕ N = ∐ _{I_0}X_∙ ⊕ L_0 ⊕ N_0$ for $L_0 ⊆ L$ and $N_0 ⊆ N$ are summands. $L_0 = 0$ as $L ∩ (∐ _{I_0}X_∙ ) ≠ 0$. Hence $L$ is isomorphic to a summand of finite exchangable module $∐_{I ∖ I_0}X_∙$, thus endolocal.

To see the uniqueness, we take $∐ M_∙ = ∐ N_∙$. Clearly $N_∙$’s are endolocal. We fix an endolocal module $E$ and take $G = \{i ∣ M_i ≃ E\}$ and $H = \{j ∣ N_j ≃ E\}$. We show $|G| = |H|$.
($|G| < ∞$ iff $|H| < ∞$). Suppose $|G| < ∞$. There is finite $J_0 ⊆ J$ s.t. $∐ _G M_∙ ⊆ ∐ _{I_0} N_∙$. Clearly $H ∩ (I ∖ I_0) = ∅$; otherwise $∐ _{I ∖ G}M_∙$ has a summand isomorphic to $E$, and exchangable property shows contradiction. A symmetric analysis shows $|G| < ∞$ iff $|G| < ∞$.
($|G| = |H|$ for finite case). There are finite sets $I_0$ and $J_0$ s.t.

$$\begin{equation} ∐ _GM_∙ ⊆ ∐ _{H ∪ J_0} N_∙ ⊆ ∐ _{G ∪ I_0} M_∙. \end{equation}$$

Hence, there is $∐ _GM_∙ ⊕ ? = ∐ _{H ∪ J_0} N_∙$. By exchangable property, there are summands $N_∙ '$ s.t. $∐ _{H ∪ J_0} N_∙ ' = ∐ _GM_∙$. Hence $|H| ≥ |G|$. A symmetric analysis shows $|G| = |H|$.
($|G|$ and $|H|$ are infinite). For each $t ∈ G$, there are finite sets $H_t$, $I_t$ and $J_t$ s.t.

$$\begin{equation} M_t ⊆ ∐ _{H_t ∪ J_t} N_∙ ⊆ ∐ _{G ∪ I_t} M_∙. \end{equation}$$

Hence $∐ _G M_∙ ⊆ ∐ _{⋃ H_t ∪ ⋃ J_t} N_∙ ⊆ ∐ _{G ∪ ⋃ I_t} M_∙$. By exchangable property, $∐ _G M_∙ ⊆ ∐ _{⋃ H_t ∪ ⋃ J_t} N_∙$ is a summand. Since $∐ _{J ∖ (⋃ H_t ∪ ⋃ J_t)} N_∙$ is isomorphic to a summand of $∐ _{I ∖ G} M_∙$, it contains no summand isomorphic to $E$. Hence $⋃ H_t = H$. Now $|G| ≤ ω |H| = |H|$. A symmetric analysis shows $|G| = |H|$.

Let $e$ be arbitrary preimage of $[x]$. There is $k$ s.t. $(e(1-e))^k = 0$. There is a polynomial $p = p(e)$ s.t. $e^k + (1-e)^k + p = 1$. Clearly $p ∈ I$, we take $q := (1-p)^{-1}$ (also a polynomial in $e$). Now

• $e^kq$ is a preimage of $[x]$;
• $e^kq + (1-e)^kq = (1-p)^{-1}q = 1$;
• $e^kq (1-e)^kq = 0$;
• $(e^kq)^2 = e^kq (1-(1-e^k)q) = e^kq$.

Let $e$ and $e'$ be liftings of $[x]$. When the ring is commutative, we conclude from $(e-e')^3 = (e-e')$ and $(e-e')^N = 0$ that $e = e'$. For non-commutative rings, the lifting is not unique in general. For instance, we take $M_2(ℤ / 4ℤ) ↠ \frac{M_2(ℤ / 4ℤ)}{J(M_2(ℤ / 4ℤ))} ≃ M_2(𝔽 _2)$, then $\mathrm{id}$ has distinct liftings $\mathrm{id}$ and $\binom{1 \ 2}{0 \ 1}$.

By induction. Since being an othogonal idempotent set is verified by collection of finite sums, the induction extends to countable sets.

Let $I$ be such nil ideal. By Noetherian conditions, there is a maximal nilpotent ideal $N ⊆ I$. Suppose $N ≠ I$, then $\{\mathrm{ann}(x) : x ∈ (I ∖ N)\}$ has a maximal element $\mathrm{ann}(x_0)$. For any $r ∈ I$, there is $n$ s.t. $(x_0 r)^n = 0$. One has either $x_0rx_0 = 0$ or $\mathrm{ann}(x_0rx_0) = \mathrm{ann}(x_0)$ by maximality of $\mathrm{ann}(x_0)$. By induction, $x_0 r x_0 = 0$. Suppose $N^k = 0$, then $(N + (x_0))^{2k}$ is generated by elements of the types

1. the monomial consisting of at least two $x_0$, or
2. the monomial with letters in $N^k$.

Hence $(N + (x_0))$ is nilpotent, a contradiction!

If $P ∈ 𝐌𝐨𝐝 _R$ is projective, then $P/PI ∈ 𝐌𝐨𝐝 _{R/I}$ is projective.

• (Surjective). Any projective module $Q$ over $R/I$ takes the form $[e] (R/I)^n$. By idempotent lifting, $Q = (eR^n)/I$.
• (Injective). For projective $R$ modules $P$ and $P'$ s.t. $P/PI ≃ P'/P'I$. We take $P ↠ P/PI$ and $P' ↠ P'/P'I$. The lifting property yields $P → P'$. Note that $\operatorname{im}(P → P') + P'I = P'$, and $P' I ⊆ P' J(R)$ is superfluous (by Nakayama lemma). Hence $P → P'$ is surjective. By symmetric analysis, $P ≃ P'$.

(1 → 2). When $R$ is semi-perfect, then $R / J(R) ≃ ∏ M_{n}(D)$ by Wedderburn-Artin theorem. It is clear to find a complete set of primitive orthogonal idempotents $\{e_i'\}_{i=1}^n ⊆ R / J(R)$ which lifts to $\{e_i\}_{i=1}^n ⊆ R$.

1. If $n = 1$, then $R ≃ \mathrm{End}(R)$ is local. We complete the proof.
2. We show $n$ from $n - 1$. Let $e$ be any lifting of $e_1'$. By induction, there is $\{(1-e)e_i(1-e)\}_{i=2}^2 ⊆ (1-e)R(1-e)$ primitive orthogonal which lifts $(1-e_1')e_i'(1-e_1') = e_i$. We define $e_1 := (1-∑ _{i=2}^n e_i)$ serving as a lifting of $e_1'$. Clearly, each $\frac{\mathrm{End}(e_i R)}{J(\mathrm{End}(e_i R))}$ is a division ring (= diagonal entries of each $M_n(D)$).

(2 → 3). Take $M_i := e_i R$.
(3 → 1). We arrange $M_∙$’s by isomorphism classes. By fitting lemma, $\frac{\mathrm{End}(⨁ M_∙ )}{J(\mathrm{End}(⨁ M_∙ ))} ≃ ∏ M_{n_i} (\frac{\mathrm{End}(M_i)}{J(\mathrm{End}(M_i))})$, where $n_i$ is the multiplicity of the isomorphism class of $M_i$. Hence $\frac{R}{J(R)}$ is semi-simple. This also determined a finite complete set of orthogonal idempotents in $R / J(R)$. The lifting property of idempotents follows from the finite exchangable property of $⨁ M_∙$.

Note that semi-simple rings are precisely $\mathrm{End}(⨁ _{\text{finite}} \ \text{endolocal modules})$.

(1 → 2). $J(R)$ is transfinite nilpotent by previous proposition. $J(R/J(R)) = 0$ is again transfinite nilpotent.
(2 → 3). We take arbitrary non-zero $[x] ∈ R / J(R)$ and show that $([x]) ⊕ \mathrm{ann}([x]) = R / J(R)$. WLOG We assume $J(R) = 0$. We consider the flat module $F:= \operatorname{cok}(R^{(ω )}\xrightarrow {1- x} R^{(ω)})$.

• If $F = 0$, then $\varinjlim (x^k) = 0$. By non-existence of nilpotent, $x = 0$.

If $F ≠ 0$, we take a maximal submodule $M ⊊ F$.

Let $[e_n]$ (we assume $R^{(ω )} = ∐_{n ≥ 1} R e_n$) be the least coordinal idempotent $∉ M$. We take the combination

$$\begin{equation} ([e_n]) + M = F,\quad r[e_n] + m = [e_{2n}], \end{equation}$$

and obtain

$$\begin{equation} [e_{2n}] = x^n [e_n] = r[e_n] + m \end{equation}$$

We show $J(R)$ is right transfinite nilpotent.

• Any left module $X ≠ 0$ has a simple submodule $(x)$. By radical property, $J(R) x = 0$.

Now we show $R' := \frac{R}{J(R)}$ is semi-simple.

• By assumption, any left ideal contains a simple left ideal $S$. By $J(R') = 0$, there is a maximal ideal $𝔪 ∩ S = 0$. Hence $S ⊕ 𝔪 = R$. Note that the procedure of taking simple summands of $R'$ terminates within finite steps. $R'$ has to be semi-simple.

(1 → 2-(3)). Clearly, $Q → M$ factors through $p$ by the lifting property of projectives. We show that $p ∘ α = p$ implies $α$ is an automorphism. We construct the pullback square of monomorphisms:

By $p ∘ α = p$, the morphism $p|_{\operatorname{im}α }$ is epic, thus the first row is a short exact sequence. By exercise, the subobject $\operatorname{im}(α ) ↪ P$ is an isomorphism. This shows $α$ is epic. We show $α$ is also monic. Let $j$ be the right inverse of $α$ (by the lifting property of projectives). Then $p ∘ j = p ∘ α ∘ j = p$. Hence $j$ is epic. Since $j$ is also split monic, $j$ is an isomorphism. Therefore $α$ is an automorphism.
(2(1) → 1). $p$ is epic, since any epimorphism $Q ↠ M$ factors through $p$. We show $\ker p$ is small. If not, there exists a proper subobject $N ↪ P$ such that $N + \ker p = P$. Therefore we have the pullback pushout square of short exact sequences. We introduce $l$ by the lifting property of projectives, then $α : λ ∘ l$ is a non-epimorphism such that $p ∘ α = p$, a contradiction.

We show 1 → 2(3) → 2(1,2), and 2(1) → 1. It remains to show 2(2) → 2(3). If there is a strict monomorphism $α$ s.t. $p ∘ α = p$, then there is an induced monomorphism $α|_{\ker p} : \ker p → \ker p$.

In the above diagram, the left square is pullback-pushout. Hence $\operatorname{im}α + \ker p = P$. A contradiction.

(1 ↔ 2) follows from the definition. In the case of $1$, a $\mathcal{P}$-cover must be a summand of $p$. Since there is no differences between $\mathcal{P}$-cover and $𝐚𝐝𝐝 (P)$-cover for $M$, we show (1 ↔ 3). (3 → 4) is given by the special Yoneda lemma. It remains to show (4 → 1).
When $p_∗ : (P, P)_{\mathrm{End}(P)} ↠ (P, M)_{\mathrm{End}(P)}$ admits a projective cover, it takes the form

$$\begin{equation} (P, P)_{\mathrm{End}(P)} ↠ Q ↠ (P, M)_{\mathrm{End}(P)}. \end{equation}$$

Note that $Q ≃ (P, P_0)_{\mathrm{End}(P)}$ for $P_0 ∈ 𝐒𝐦𝐝 (P)$ by projectivisation. With the special Yoneda lemma, we may write the mapping sequence as

$$\begin{equation} (P, P)_{\mathrm{End}(P)} \overset {a_∗ } ↠ (P, P_0)_{\mathrm{End}(P)} \overset {b_∗ } ↠ (P, M)_{\mathrm{End}(P)}. \end{equation}$$

We show that $b : P_0 → M$ is a $\mathcal{P}$-cover. By special Yoneda lemma, $(ba)_∗ = p _∗$ yields $ba = p$. As $a$ is split epic, $b$ and $p$ factors through each other, which shows that $b$ is a $\mathcal{P}$-precover. If there exists $b ∘ γ = b$, we have $b_∗ ∘ γ _∗ = b_∗$. Now $γ _∗$ is an isomorphism as $b_∗$ is a projective cover. By projectivisation, $γ$ is an automorphism. Hence $b$ is a $\mathcal{P}$-cover.

(1 → 5). When $M$ is f.g., then $\frac{M}{MJ}$ is semi-simple with finite length. By idempotent lifting, we take

$$\begin{equation} ⨁ _{i= 1}^n e_i R \overset p → M ↠ \frac{M}{MJ} ≃ ⨁ _{i= 1}^n \frac{e_i R}{e_i J}. \end{equation}$$

Note that $\operatorname{im} p ↠ \frac{M}{MJ}$. Hence $\operatorname{im} p + MJ = M$. Nakayama lemma shows $\operatorname{im} p = M$. Hence $p$ is surjective. $\ker p ⊆ ⨁ _{i= 1}^n e_i J$ is superfluous. Hence $p$ is a projective cover.
(5 → 4). Clear.
(4 → 1). We show idempotent lifting property and semi-simplicity of $\frac{R}{J}$ in turns.

1. (Idempotent lifting). For any $e' ∈ \frac{R}{J}$ and $f' := 1' - e'$, there are projective converse $P → \frac{eR}{eJ}$ and $Q → \frac{fR}{fJ}$. Taking direct sums, WLOG we assume $P ⊕ Q = R$, and write $P = xR$ and $Q = yR$ for $x$ and $y$ idempotents. Note that $x' + y' = 1' = e' + f'$ in $\frac{R}{J}$. Taking projection shows $x' = e'$ and $y' = f'$.
2. (Semi-simplicity). For cyclic $\frac{R}{J}$-module $([x])$, there is a projective cover $p : P ↠ (x)$. Clearly $p' := \frac{P}{PJ} ↠ ([x])$ is epic as $[x]$ is mapped from $P$. Note that $\ker p' = \frac{\ker p}{PJ}$, which is superfluous in $\frac{P}{PJ}$. Hence $p'$ is also a projective cover. Since $J(\frac{P}{PJ}) = 0$, $\ker p' ⊆ J = 0$. Hence $p'$ is an isomorphism. It yields that any cyclic modules are semi-simple, thus $\frac{R}{J}$ is semi-simple.

(1 → 3). The proof is similar to the semi-perfect case. We summarise the procedure:

1. for any $M$, take complete set of orthogonal idempotents of the semi-simple module $\frac{M}{MJ}$;
2. lifting idempotents to $R$, we obtain a projective module;
3. prove the projective cover map, wherein $MJ ↪ M$ is always superfluous as $J$ is transfinite nilpotent.

(3 → 1). $R$ is semi-perfect. Since arbitrary $M$ has a projective cover, the $1:1$ correspondence of maximal submodules yields $J(M) ≠ 0$. Hence $MJ = M$ yields $M = 0$, i.e., $J$ is right transfinite nilpotent.

(←). When $R$ is right perfect, we show $J(M) = MJ$.

• $M/MJ$ is semi-simple as $R / J(R)$ is. Now $J(M / MJ) = 0$. Hence, $J(M) ⊆ MJ$. The other inclusion is clear.

checkcheck

Set $\ker p = : K ⊆ P$. We aim to construct a defect morphism:

• for any $x ∈ K ⊆ P$, there is some $φ _x : P → K$ preserving $x$.

Now $\operatorname{im}φ _x ⊆ P$ is superfluous. We claim $φ _x ∈ J(\mathrm{End}(P))$.

• It suffices to show $φ _x$ lies in every maximal right ideal. If not, then there is $(φ _x) + 𝔪 = R$. We take $φ _x r + m = 1$. Since $\operatorname{im} (φ _x r) + \operatorname{im} (m) = P$ and $\operatorname{im}(φ _x r)$ is superfluous, we see $m$ is surjective and thus has a right inverse. A contradiction!

Hence $(1 - φ _x)$ is a unit while $(1 - φ _x)(x) = 0$. Hence, $x = 0$.
$φ _x$ is constructed by purity. We show $KI = K ∩ XI$ for any ideal $I$:

• $\ker [K ↪ P ↠ P/PI] = K ∩ XI$ by definition;
• $K/KI ≃ K ⊗ R/I ↪ P ⊗ R/I ≃ P/PI$ is monic as $F$ is flat.

WLOG assume $P$ is free by adding idempotents. Set any $x = (c_λ)_{λ ∈ Λ _0} ⊆ P$. We take $I$ as the right ideal generated by $\{c_λ\}$. Since $x ∈ K ∩ PI = KI$, we write $x = ∑ c_λ k_λ$ for $k_λ ∈ K$. We define $φ _x : P → K$ by $1_λ ↦ k_λ$ when $λ ∈ Λ _0$, and $1_λ ↦ 0$ otherwise. This is the desired morphism.

Note that $F$ is countably presented by $F = \ker [∐ R^{ω } \overset {φ }↪ ∐ R^{ω}]$, thus is a filtered colimit of a cotower. By Lazard’s theorem, $F$ is flat.
Consider $φ =(1-r) : (x_0,x_1,\ldots) ↦ (x_0 - x_1r_0, x_1 - x_2 r_1, x_2 - x_3r_2, \ldots)$. When it is split monic, we denote the right inverse by

$$\begin{equation} ψ : (y_0,y_1,\ldots) ↦ (∑ y_∙ c_{∙, 0}, ∑ y_∙ c_{∙, 1}, \ldots). \end{equation}$$

Note that for each fixed $k$, $c_{k, ∙}$ are almost zero. We take the maximal $n$ s.t. $c_{0, n - 1} ≠ 0$. We obtain

$$\begin{equation} c_{k,n} - c_{k+ 1,n}r_k = δ _{k, n}. \end{equation}$$

Hence, $c_{n,n} r_{n-1} \cdots r_0 = 0$. We take $c_{n,n} =c_{n+1, n}r_n + 1$, then

$$\begin{equation} c_{n+1, n} r_n \cdots r_0 = - r_{n-1} \cdots r_0. \end{equation}$$

We see the ideal stablise when multiplying $r_{≥ n}$. The converse is clear. Once such limit exists, it corresponds to the following countable composition stabilises:

$$\begin{equation} R \xrightarrow {r_0} (r_0) \xrightarrow {r_1} (r_1 r_0) \xrightarrow {r_2} (r_2 r_1 r_0) \xrightarrow {r_3} \cdots. \end{equation}$$

Hence the PID vanishes iff the above colimit stables at $0$.

(3 → 4). Note that flat module with a projective cover is projective.
(4 → 5). Clear.
(5 → 6). We construct countably preserted flat module $F$ for any descending chain of principal left ideals. Such chain stabilises as $F$ is projective.
(6 → 1). $\frac{R}{J(R)}$ satisfies descending chain condition of principal left ideals, hence semi-simple. We show $J(R)$ is transfinite nilpotent.
For any decsending principal left ideal $\{(r_n \cdots r_0)\}_{n ∈ ℕ}$ with each $r_∙ ∈ J$, there is $r_{n+1}$ s.t. $(r_n \cdots r_0) = (r_{n+1} \cdots r_0)$. We take $a$ s.t. $r_n \cdots r_0 = ar_{n+1} r_n \cdots r_0$. Note that $(1- ar_{n+1})$ is a unit, we see $r_n \cdots r_0 = 0$.

(1 → 2). Over a right perfect ring, any flat right modules are projective. Over a left coherent ring, any product of flat right modules is flat. Hence, any product of projective left modules is projective.
(2 → 3). Clear.
(3 → 1). The product of free right modules is flat iff the ring is left coherent.