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%begin text



%\vbox{\vskip1.2truein}


\centerline {\bf  {PRIMITIVE HEREDITY IDEALS}}
\centerline {Darren D. Wick}
\rm
\vskip2truecm
%\centerline {Darren D. Wick}
%\centerline {Millsaps College}
%\centerline {Jackson, Mississippi 39210}
%\vskip2truecm
%\midinsert
%\narrower
%\narrower
%\noindent 
\bf {Abstract}. \rm Let $R$ be a left artinian ring. Dlab and Ringel have shown that $R$ is
hereditary
if and only if every chain of idempotent ideals can be refined to an heredity
chain (see [1] Theorem 1). In particular, if $R$ is a basic hereditary ring, then every primitive
ideal
is an heredity ideal. The converse to this is clearly false.  
(See Example 1). We will introduce a class of rings, that includes serial
rings and monomial algebras, for which the converse does hold.


%\endinsert 

\vskip15pt

Throughout this paper, $R$ will be  a  basic left artinian ring (with unity) with a basic set of
primitive idempotents
$\tau =\lbrace e_1, \cdd ,e_n \rbrace$ and with $J$ the Jacobson radical of $R$.
We will refer to the
ideals $Re_i R$ as primitive ideals. If $M$ is an 
$R$-module, we denote the Loewy length of $M$ by $L(M)$, and the composition length of $M$ by
$c(M)$.



Recall that an ideal $I$ of $R$ is {\it heredity} if
$I^2=I$, ${_R}I$ is projective, and $IJI=0$.  The ring $R$
is {\it quasi-hereditary} if there exists a chain of
ideals, 
$$0=I_0\subset I_1 \subset \cdots \subset I_l=R,$$
called an {\it heredity chain}, 
such that $I_k/I_{k-1}$ is an heredity ideal in
$R/I_{k-1}$ for each $k=1, \cdots, l$. (see [2] or [1]).
If $I$ is an idempotent ideal of $R$, then $I=ReR$ for some idempotent $e$ in 
$R$. Moreover, $e$ may be taken to have the form $e=e_1 +\cdd
+e_k$ for some
$k\leq n$ and a suitable ordering of $\tau$ (see [2] Lemma 1.1). 
Furthermore, if $I=R(e_1 +\cdd + e_k)R$ is an heredity ideal, 
then each ideal $Re_i R, \ 
i=1,\cdd ,k$,
is again an heredity ideal of $R$ (see [2] Lemma 1.1).
Thus (see [2] Corollary 1.2) with a suitable ordering of $\tau$, any heredity 
chain can be refined to an heredity chain of the form
$$0\subset Re_1 R\subset R(e_1 +e_2 )R\subset \cdd  \subset R(e_1 +
\cdd + e_{n-1} )R \subset R.$$ 

Dlab and Ringel have shown that $R$ is hereditary if and only if every chain of idempotent
ideals 
can be refined to an heredity chain (see [1] Theorem 1). 
We first consider the trace of projective left modules in the radical of the 
ring $R$. If $Re$ is projective, then $ReJ=\Tr J Re $. Note that $ReJ$ is
the right radical of the idempotent ideal $ReR$. 
\vskip10pt
\ub {Lemma 1}.
Let $ReR$ be a primitive ideal. The following are equivalent.
\item {(a)} $ReR$ is an heredity ideal.
\item {(b)} $ReJ$ is a projective left $R$-module.
\item {(c)} For each $i=1,\cdd ,n$, $\Tr Je_i Re $ is either zero or isomorphic
to a direct 
sum of copies of $Re$.
\vskip10pt
\ub {Proof}. We may assume $e=e_1$. For any module $_R N$, $Re_1 N=\Tr N Re_1 $.
Thus the equivalence of (b) and (c) follows immediately from the direct sum
decomposition $$Re_1 J=\oplus_{i=1}^n Re_1 Je_i =\oplus_{i=1}^n \Tr Je_i
 Re_1 .$$ Since $R$ is basic, $Re_1 Re_i =Re_1 Je_i $ for $i=2,\cdd ,n$. Hence
 we have the decomposition 
$$\eqalign {Re_1 R & =\oplus_{i=1}^n Re_1 Re_i \cr & =Re_1 \oplus 
(\oplus_{i=2}^n Re_1 Je_i) \cr
& =Re_1 \oplus (\oplus_{i=2}^n \Tr Je_i Re_1) .}$$

Assume condition (c). From this last decomposition  we have that $Re_1 R$
is a direct sum of copies of $Re_1$ and is therefore  projective.
If $Re_1 Je_1 \not= 0$, then $1\leq c(Re_1 Je_1 )\leq c(Je_1 )<c(Re_1 )$,
and thus $Re_1 Je_1 $  cannot be isomorphic to a direct sum of copies of
$Re_1$. Hence we must have $Re_1 Je_1 =0$ and $Re_1 R$ is heredity. 


 To see that (a) implies (b) it suffices to
observe that if $Re_1 R$ is heredity, then we have that $\Tr Je_1 Re_1 =Re_1 Je_1 =0$.
\vskip10pt
Dlab and Ringel have shown that the notion of an heredity ideal (and thus the 
notion of a \qh ring) is two-sided. That is, if $I$ is an heredity ideal of $R$, then
$I$ is also projective as a right $R$-module (see [1] statement 7).
 Thus there exists a 
corresponding version of
Lemma 1 for right $R$-modules. In particular, we have the following 
corollary.
\vskip10pt
\goodbreak
 \ub {Corollary 1}. 
For a primitive idempotent $e$ in $R$, $_R ReJ$ is projective if and only if
$JeR_R$ is projective.
\vskip10pt
Note that a primitive ideal $ReR$ which is projective as a left $R$-module is not
an heredity ideal if and only if $eJe\not= 0$. Moreover, if $eJe\not= 0$, then
$1\leq c(ReJe)\leq c(Je)<c(Re)$ so that $ReJe\not= 0$ and $ReJe$ (and
thus $ReJ$) is not 
projective. We thus have the following result.
\vskip10pt
\goodbreak
\ub {Corollary 2}
A primitive ideal $ReR$ is heredity if and only if both $_R ReJ$ and $JeR_R$ are projective
modules.
\vskip10pt
 In particular, if $ReR$ is a primitive heredity ideal, then
 $_R ReJ$ ($JeR_R$) is a direct sum of local left (right) ideals.
 However, for $J$ to be a direct sum of local left ideals, it does not
suffice that each primitive ideal of $R$ be heredity.
 Consider the following example.
\vskip10pt
\ub {Example 1}.
 Let $k$ be a field and consider the incidence algebra of
the poset 

$$\baselineskip=0\baselineskip\vbox{\halign{#\ \ &\hfil #\hfil&\ \ #\cr
&4&\cr
&$\circ$&\cr
&$\bigg/\bigg\backslash $&\cr
2&\raise1pt\hbox{$\circ$\hskip16pt $\circ$}&3\cr
&$\bigg\backslash\bigg/$&\cr
&$\circ$&\cr
&1&\cr
}}$$
%This is the algebra of all $4\times 4$ upper triangular matrices of the 
%form 
%$$\left(\matrix{* &* &* &*\cr
%     0 & * & 0 & *\cr
%     0 & 0 & * & * \cr
%     0 & 0 & 0 & *\cr
%     }\right)$$
%over the field $k$. 

The  indecomposable projective
left $R$-modules have diagrams: 
{\settabs 16\columns
\+&& 1 &&& 2 &&& 3 &&&& 4    \cr
\+&&  &&& 1 &&& 1 &&& 2 && 3\cr
\+&&&&&&&&&&&& 1  \cr} 
Notice that $\gd R =2$ so that $R$ is \qh but not left hereditary.
%For $1\leq i\leq 4$, the primitive ideals $Re_i R$ consist of matrices of the
%form
%\vskip10pt
% $\left(\matrix{* &* &* &*\cr
%    0 & 0 & 0 & 0\cr
%     0 & 0 & 0 & 0 \cr
%     0 & 0 & 0 & 0\cr
%     }\right)$ \hskip5pt 
%$\left(\matrix{0 &* &0 &*\cr
%     0 & * & 0 & *\cr
%     0 & 0 & 0 & 0 \cr
%     0 & 0 & 0 & 0\cr
%     }\right)$ \hskip5pt 
%$\left(\matrix{0 &0 &* &*\cr
%     0 & 0 & 0 & 0\cr
%     0 & 0 & * & * \cr
%     0 & 0 & 0 & 0\cr
%     }\right)$ \hskip5pt 
%$\left(\matrix{0 &0 &0 &*\cr
%     0 & 0 & 0 & *\cr
%     0 & 0 & 0 & * \cr
%     0 & 0 & 0 & *\cr
%     }\right)$
%
%\vskip10pt
%\hskip25pt $i=1$ \hskip55pt  $i=2$ \hskip55pt $i=3$ \hskip55pt  $i=4$
%\noindent and are clearly heredity ideals.
It is clear that the primitive ideals $Re_i R$ ($1\leq i\leq 4$) are heredity ideals.
It is also clear that $J$ 
%consisting of all matrices of the form 
%$$\left(\matrix{0 &* &* &*\cr
%    0 & 0 & 0 & *\cr
%     0 & 0 & 0 & * \cr
%     0 & 0 & 0 & 0\cr
%     }\right)$$
is not a direct sum of local left ideals.
Observe that $R(e_2 +e_3 )R/{Re_2 R}$ is not a projective left 
$R/Re_2 R$-module and that $R(e_2 +e_3 )R/{Re_3 R}$ is not a projective left 
$R/Re_3 R$-module.
Hence the chain of idempotent ideals
$0\subset R(e_2 +e_3 )R\subset R$ 
cannot be refined to an heredity chain.
\vskip10pt
In Example 1, we see that if $e$ is a primitive idempotent, then $ReJ$ is
a direct sum of local left ideals of $R$. However, the trace in $J$ of the 
decomposable projective module
$R(e_2 +e_3 )$ is neither local, nor a direct sum of local left ideals. 



We will use the following characterization
of tree subsets due to Burgess, Fuller, Green and Zacharia (see [3] Proposition 1.2).
\vskip10pt
\goodbreak
\ub {Proposition 1}. (Burgess, Fuller, Green and Zacharia)
Let $m=\LL R $.
Let $X$ be a subset of $R\setminus \lbrace 0\rbrace$ such that $X=\cup_{i=1}^n
 e_i X$ and if $x,y\in X$ with $x\not= y$ then $Rx\not= Ry$.
 Then $X$ is a tree subset for $R$ if and only if 
$X$ can be written $X=Y_0 \cup \cdd \cup Y_{m-1}$ so that
$R=\oplus_{y\in Y_0} Ry$; and for each $l, \ 1\leq l\leq m-1$, and 
$x\in Y_{l-1}$, there are subsets $Y_{lx}\subseteq Y_l$ so that
$Y_l =\cup_{x\in Y_{l-1}} Y_{lx}$ and $Jx=\oplus_{y\in Y_{lx}} Ry$.
Moreover, under these conditions, $J^l =\oplus_{y\in Y_{l}} Ry$ for
$l=1,\cdd ,m-1$.
\vskip10pt
\goodbreak
\ub {Theorem 1}.
The following are equivalent:
\item {(a)} $J^k$ is a direct sum of local left ideals for all $k=1,\cdd ,L(R)-1$.
\item {(b)} $ReJ^k$ is a direct sum of local left ideals for all $k=1,\cdd ,L(R)-1$ and for all
idempotents $e\in R$.
\item {(c)} There exists a tree subset for the regular module ${_R}R$.
\vskip10pt
\ub {Proof}. 

(a $\Leftrightarrow$ c) See [3] Corollary 1.3.

(b $\Rightarrow$ a) By hypothesis, $R\cdot 1\cdot J^k =J^k$ is
a direct sum of local left ideals for each $k=1,\cdd ,\LL R -1$.

(c $ \Rightarrow$ b) Suppose there exists a tree subset 
  $X=Y_0 \cup \cdd \cup Y_{L(R) -1}$
 for $_R R$. Let $e$ be an idempotent
in $R$ and let $x\in X$. 
We claim that $ReRx$ is either zero or a direct
sum of local left ideals. To prove this, we induct on $l=\LL Rx $.

$l=1:$ In this case $Rx$ is semisimple so that $ReRx$ is either zero or 
semisimple.

$l>1:$ We first note that $x\notin Y_{L(R) -1}$. 
Since $x=e_i x$ for some basic idempotent $e_i$,
$Rx$ is a local left ideal of $R$.
Thus if $ReRx=Rx$ we are done. Suppose $ReRx\not= Rx$. Then $ReRx\subset Jx$ and
therefore $ReRx=ReJx$. But $Jx=\oplus_{y\in Y_{(l+1)x}} Ry$ by Proposition 1
 For $y\in Y_{(l+1)x}$, $Ry
\subset Jx$ so that $\LL Ry <\LL Rx $. Thus by induction $ReRy$ is either zero
or a direct sum of local left ideals.
Thus $ReRx=ReJx=\oplus_{y\in Y_{(l+1)x}} ReRy$ is either zero or
 a direct sum of local left ideals.

Let $1\leq k\leq \LL R -1$. Then by Proposition 1, $J^k =\oplus_{x\in Y_{k}}
 Rx$. Thus $ReJ^k =\oplus_{x\in Y_{k}} ReRx$ is either zero or a direct sum
of local left ideals. 
\vskip10pt
Now, the existence of a tree subset  for the regular module $_R R$ is a defining 
property of left monomial rings (see [3] Definition 2.2).
We thus have 
the following examples of rings satisfying the conditions of Theorem 1 (see [3] p. 6).
\vskip10pt
\goodbreak
\ub {Example 2}.
The following rings satisfy the conditions of Theorem 1.
\item {(a)} Left monomial rings.
\item {(b)} Left serial rings.
\item {(c)} Monomial algebras.
\item {(d)} Left hereditary left artinian rings.
\item {(e)} Left artinian rings with $J^2 =0$.
\vskip10pt
Recall that a ring $R$ is \it l-hereditary \rm if for every two indecomposable
projective $R$-modules $P$ and $Q$, every non-zero 
homomorphism $h:P \rightarrow Q$ is monic (see [4]). We note that
 l-hereditary rings are quasi-hereditary  
(see [2] Proposition 1.3).
% This fact relies on the existence of a
%simple projective module which generates an heredity ideal in an l-hereditary ring.
%However, it is easy to see that not every indecomposable projective module $Re_i$
%in an l-hereditary ring $R$ generates a projective trace ideal $Re_i R$. 
As was shown by Burgess and Fuller, if $ReR$ is a primitive heredity ideal, then
every non-zero homomorphism $h:Re \longrightarrow R$ is monic ([2] Lemma 1.7). Hence if
every 
primitive ideal is heredity, $R$ is l-hereditary.
Conversely, suppose that $R$ is l-hereditary and $ReR$ a primitive ideal.
Then any non-zero 
homomorphism $h:Re\longrightarrow Re$ is monic. Thus $eJe \subset ReJe =
\Tr Je Re =0$. However, it is easy to see that not every primitive ideal
in an l-hereditary ring is an heredity ideal. Consider the following example.
\vskip10pt
\ub {Example 3}. 
Let $\Gamma$ be the digraph \vskip10pt
\vbox{\settabs 7\columns
\+&&&  \hskip8.8pt $\buildrel a \over \longrightarrow$ &  
 $\buildrel c \over \longrightarrow$  \cr \vskip-10pt
\+&&& 1 & \hskip-15pt 2 & \hskip-25pt 3  \cr \vskip-10pt
\+&&&   \hskip8.8pt $\buildrel b \over \longrightarrow$ 
& $\buildrel d \over \longrightarrow$   \cr
\+&&&&&   \cr}
Let $k$ be a field and let $R=k\Gamma /I$ where $I=(cb-da)$.
Then the indecomposable projective left $R$-modules have diagrams
\goodbreak
\vskip12pt
\vbox{{\settabs 16\columns
\+&&& 1 & &&&&& 2 &&&& 3   \cr
\+&& 2 && 2  &&& & 3&& 3 &&\cr
\+&3 && 3 && 3 &&&&    \cr
}}
\vskip10pt
Notice that $_R Re_2 R=Re_2 \oplus Je_1$, and $Je_1 =Re_2 Je_1 =\Tr Je_1 Re_2 $. Since 
$c(Je_1 )=5$ and $c(Re_2 )=3$, we see that $Je_1$ (and hence $_R Re_2 R$) is
not projective.
Thus $R$ is not a left hereditary ring.
We claim that $R$ is l-hereditary.
To see this, it will suffice to show that any non-zero homomorphism
$h:Re_2 \rightarrow Je_1 $ is monic. But any such homomorphism $h$ is given by
right multiplication by some $x=e_2 xe_1 \in e_2 Je_1 =\langle a,b\rangle$.
Thus $x=\alpha a +\beta b$ for some $\alpha ,\beta \in k$.

Now, $Re_2 =\langle e_2 ,c,d\rangle$. Suppose $\lambda_1 e_2 +\lambda_2 c+\lambda_3 d
\in \rm Ker\it (h)$, with $\lambda_1 ,\lambda_2 ,\lambda_3 \in k$.
Then $0=h(\lambda_1 e_2 +\lambda_2 c +\lambda_3 d)=
\lambda_1 \alpha a + \lambda_1 \beta b +
\lambda_2 \alpha ca +(\lambda_3 \alpha +\lambda_2 \beta )da +
\lambda_3 \beta db$.
If either $\alpha \not= 0$ or $\beta \not= 0$, we see that $\lambda_i =0$ for
$i=1,2,3$.
Thus $h$ is monic and $R$ is l-hereditary.
\vskip10pt
The rings in Examples 1 and 3 are l-hereditary rings which are not left hereditary.
We will show that under conditions weaker than those of Theorem 1,
 the notions of l-hereditary and left hereditary are the same.
\vskip10pt
\goodbreak
\ub {Theorem 2}.
Suppose that 
$J$ is a direct sum of local left ideals.
Then the following are equivalent.
\item {(a)} $R$ is left hereditary.
\item {(b)} $R$ is l-hereditary.
\item {(c)} Every primitive ideal of $R$ is an heredity ideal.
\vskip10pt
\ub {Proof}.

(a $\Rightarrow$ b) This is clear.

(b $\Rightarrow$ c) Let $R$ be l-hereditary and 
let $I=Re_i R$ be a primitive ideal of $R$. 
As we have seen, if $R$ is l-hereditary then $eJe=0$
for every primitive idempotent $e$. Thus it will suffice to
show that $_R I$ is projective.
For $\ds {j\not= i}$, we have $\ds {Re_i Re_j =} \Tr Re_j Re_i =Re_i Je_j$. Thus
$$ Re_i R=\bigoplus_{j=1}^n Re_i Re_j =Re_i \oplus \bigoplus_{j\not= i}
 Re_i Je_j.$$ 
Now, each non-zero $Re_i Je_j$ is a direct sum of local left ideals of the form 
Im$(\rho_x )$, where the homomorphism $\rho_x :Re_i \rightarrow Je_j$ is given by right
multiplication
by $x$. By assumption, $\rho_x$ is monic and thus $Re_i R$ is projective. 

(c $\Rightarrow$ a)
 Assume that every primitive ideal is heredity. Recall that $R$ is left hereditary if
and only if
  $Je_k$ is projective for each $k$, $\ds {1\leq k\leq n}$ (see [5] ex. 18.10).

Let $1\leq k\leq n$.
Since $J=\oplus_{i=1}^n Je_i$ is a direct sum of local left ideals, 
$\ds Je_k =\oplus_{j=1}^l L_j$ for some collection
$\ds {L_1,\cdd ,L_l}$ of local left ideals. Let $1\leq i\leq l$ and consider
the local left ideal $L_i$.
 Assume
$L_i$ has projective cover $Re_t$. Notice that $\ds {t\not= k}$ since 
$\ds {Re_k Je_k =0}$. Since $Re_t R$ is heredity,  $\ds {Re_t Re_k =Re_t Je_k}$ is
a direct sum of copies of $Re_t$. But   $\ds {Re_t Je_k =\oplus_{j=1}^l Re_t L_j}$
and hence $Re_t L_i =\Tr L_i Re_t =L_i$ is isomorphic (by Krull-Schmidt)
to a copy of $Re_t$. Thus $L_i$ is projective and $R$ is left hereditary. 
\vskip10pt
Since the radical of a left hereditary ring is necessarily a direct sum of local left ideals, 
we also have the following
result.
\vskip10pt
\goodbreak
\ub {Corollary 3}.
Suppose $R$ is l-hereditary. Then $R$ is left hereditary if and only if $J$ is a direct sum of local
left ideals.
\vskip10pt
Suppose $J$ is a direct sum of local left ideals. We  then have the following dichotomy of 
the left global dimensions of such
 rings $R$ with the property that each primitive ideal is a projective
left $R$-module.
\vskip10pt
\goodbreak
\ub {Corollary 4}.
 If $J$ is direct sum of local left ideals and each primitive ideal is projective as a left
$R$-module,  
 then either $R$ is left hereditary or 
$\gd R =\infty$.
\vskip10pt
\ub {Proof}.
If $R$ is not left hereditary, then by Theorem 2 there exists
an  $\ds {i,\ 1\leq i\leq n}$, such that $Re_i R$ is not an heredity ideal. Hence
 $\ds {e_i Je_i \not= 0}$ and by Corollary 1.5 of [6], we have that $\gd R =\infty$. 
\vskip10pt
As noted above, if $ReR$ is an heredity ideal, then both $_R ReR$ and $ReR_R$ are projective.
Thus there exists a right-hand version of Theorem 2 and we have
the following result.
\vskip10pt
\goodbreak
\ub {Corollary 5}.
Suppose $_R J$ is a direct sum of local left ideals, and $J_R$ is a 
direct sum of local right ideals. Then $R$ is hereditary if and only if 
every primitive ideal of $R$ is an heredity ideal.



\bigskip
\vskip.5truein
\centerline {\ub {\it References}}
\vskip10pt
\item {1.} \rm  V. Dlab and C.M. Ringel, \it Quasi-hereditary algebras, 
\rm Illinois J. Math., 
33 
(1989), 280-291.
\vskip10pt
\item {2.} \rm  \rm  W.D. Burgess and K.R. Fuller, \it On quasihereditary rings, 
\rm Proc. Amer. 
Math. Soc., 106 (1989), 321-328. 
\vskip10pt 
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