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This is a curiosity question that came out of teaching abstract algebra.

Let $F$ be a field, and $n>1$ an integer.

Let $F^{n \leq n}$ be the $F$-algebra of all upper-triangular $n\times n$-matrices $\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ 0 & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n,n} \end{pmatrix}$ with entries in $F$.

Let $J$ be the subset of $F^{n \leq n}$ that consists of all such matrices that satisfy $a_{i,j} = 0$ for all $\left(i,j\right) \neq \left(1,n\right)$. In other words, $J$ is the set of all $n\times n$-matrices whose only nonzero entry (if any) is in the northeasternmost corner. It is easy to see that $J$ is an ideal of $F^{n \leq n}$. Thus, a quotient $F$-algebra $F^{n \leq n} / J$ is defined (and its elements can be thought of as upper-triangular matrices whose northeasternmost entry is undetermined).

Question. What is the smallest $m$ such that there is an injective $F$-algebra homomorphism from $F^{n\leq n} / J$ to the matrix ring $F^{m\times m}$? In other words, what is the smallest dimension of a faithful representation of the $F$-algebra $F^{n \times n} / J$?

My suspicion is that it is $2n-2$. Indeed, it is certainly $\leq 2n-2$, since there is an injective $F$-algebra homomorphism $F^{n\leq n} / J \to F^{\left(2n-2\right)\leq \left(2n-2\right)}$ that sends the residue class of a matrix $\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n} \\ 0 & a_{2,2} & \cdots & a_{2,n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n,n} \end{pmatrix}$ to $\begin{pmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1,n-1} & 0 & 0 & \cdots & 0 \\ 0 & a_{2,2} & \cdots & a_{2,n-1} & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{n-1,n-1} & 0 & 0 & \cdots & 0 \\ 0 & 0 & \cdots & 0 & a_{2,2} & a_{2,3} & \cdots & a_{2,n} \\ 0 & 0 & \cdots & 0 & 0 & a_{3,3} & \cdots & a_{3,n} \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & 0 & 0 & \cdots & a_{n,n} \end{pmatrix}$.

For $n = 3$, this is an embedding $F^{3 \leq 3} / J \to F^{4 \leq 4}$, and I think I have convinced myself by a long and messy argument that no embedding $F^{3 \leq 3} / J \to F^{3 \leq 3}$ exists, but this doesn't rule out an embedding $F^{3 \leq 3} / J \to F^{3 \times 3}$ into arbitrary (rather than triangular) matrices.

The whole thing originated in my attempts to illustrate the variety (common-sense meaning) of quotient rings. Specifically, I was trying to show that not every quotient ring is just a subring in disguise. The easiest example for this is $\mathbb{Z} / n\mathbb{Z}$, but I was looking for something less obvious.

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    $\begingroup$ This algebra is split basic so every representation is equivalent to an upper triangular one since the simples are one dimensional $\endgroup$ Commented Feb 3, 2023 at 2:24
  • $\begingroup$ @BenjaminSteinberg: Nice! (I don't know what "split basic" means, but I can confirm that all simple $F^{n\leq n}/J$-modules are $1$-dimensional, since the same is true for $F^{n\leq n}$, and therefore any module has a complete flag of submodules.) So the $m = 2n-2$ conjecture is confirmed for $n=3$. $\endgroup$ Commented Feb 3, 2023 at 2:35
  • $\begingroup$ This is the truncated path algebra of the straightline path quiver with n-vertices where you force the path of length n to be zero. I calculated with Mazorchuk the minimum degree faithful representation of the corresponding truncated path semigroup but this should be lower dimensional than the algebra. Ill have to look what we had $\endgroup$ Commented Feb 3, 2023 at 2:35
  • $\begingroup$ Yes, I thought about including the quiver-with-relations viewpoint in the question (particularly as it means we're looking at $3$-term complexes in the case $n=3$), but I wasn't sure it would be of much use. $\endgroup$ Commented Feb 3, 2023 at 2:36
  • $\begingroup$ Split basic means all representations are 1 dimensional. More precisely split means F is a splitting field and basic means the wedderburn components of the semisimple quotient are division rings $\endgroup$ Commented Feb 3, 2023 at 2:37

3 Answers 3

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Here's an elementary proof that $2n-2$ is a lower bound.

Suppose that $$V_1\xrightarrow{\alpha_1}V_2\xrightarrow{\alpha_2}\dots\xrightarrow{\alpha_{n-2}}V_{n-1}\xrightarrow{\alpha_{n-1}}V_n$$ is a representation of the linearly ordered $A_n$ quiver $Q$ that is a representation of $FQ/I$, where $I$ is the one-dimensional ideal of the path algebra $FQ$ generated by the longest path, so that $\alpha_{n-1}\dots\alpha_1=0$, and that it is faithful as a representation of $FQ/I$.

By faithfulness, $\alpha_{n-2}\dots\alpha_1(v_1)\neq0$ for some $v_1\in V_1$. For $1<i<n$ let $v_i=\alpha_{i-1}\dots\alpha_1(v_1)\in V_i$.

Also by faithfulness, $\alpha_{n-1}\dots\alpha_2(w_2)\neq0$ for some $w_2\in V_2$. For $2<i\leq n$ let $w_i=\alpha_{i-1}\dots\alpha_2(w_2)\in V_i$.

For $1<i<n$, $v_i$ and $w_i$ are linearly independent (consider their images in $V_{n-1}$ and $V_n$), so $v_1,\dots,v_{n-1},w_2,\dots,w_n$ span a ($2n-2$)-dimensional subrepresentation (which is isomorphic to the representation described in the OP).

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  • $\begingroup$ Very nice! The quiver viewpoint really clears the fog. $\endgroup$ Commented Feb 4, 2023 at 5:24
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Take $Q=Q_n$ to be the linear oriented line quiver with n vertices and $KQ$ its path algebra. Take any admissible ideal $I$ and let $A=KQ/I$ (such algebras are called Nakayama algebras and are in nice bijection with Dyck paths.) The smallest faithful module for this algebra is the direct sum of all indecomposable projective-injective modules, which correspond bijectively to the peaks of the corresponding Dyck path. If $I=rad^l KQ$ ($l=n-1$ is your case), then the projective-injective indecomposables are for example simply the ones with dimension $l$. There are $n-l+1$ such modules and thus the minimal dimension is $l (n-l+1)$, which is 2(n-1) in your case.

The argument is this: A module $M$ is faithful if there exists an exact sequence $0 \rightarrow A \rightarrow M^n$ for some $n \geq 1$. Thus every faithful module has all indecomposable projective-injective modules as a direct summand. On the other hand, the direct sum of all indecomposable projective-injective modules of a Nakayama algebra are already faithful (this is equivalent to a property called dominant dimension at least 1 or being a QF-3 algebra, which is a well known property of Nakayama (=serial) algebra, see for example the book by Anderson and Fuller). So there is a unique smallest faithful module and it is the "canonical" one.

In general the dimension of the minimal faithful module of a linear Nakayama algebra corresponding to a Dyck path will be the sum of (the height +1) of the peaks in the corresponding Dyck path.

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  • $\begingroup$ It seems your answer for the algebra is the same as the answer for the semigroup. That's curious since usually it is easier to get a faithful representation of a semigroup than its algebra. $\endgroup$ Commented Feb 3, 2023 at 11:12
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The answer should be $2n-2$, and in the case that $F$ is an uncountable field of characteristic $0$, this is a consequence (assuming I understood correctly the notation) of the result in 4.2 of Forsberg, Love, Effective representations of path semigroups, Semigroup Forum 92 (2016), no. 2, 449–459. . Added. Examining the paper it seems to not require any assumption on the field for the lower bound on the minimum faithful degree and since you have an upper bound this should work

This paper is about minimum faithful degree of truncated path semigroups. You take a finite quiver $Q$ and fix $N>0$. Then the truncated path semigroup $P_N(Q)$ is the finite semigroup consisting of $0$ and all paths in $Q$ of length less than $N$ (including the empty path) with product concatenation with the understanding that paths of length $\geq N$ and undefined concatenations are $0$. This contracted semigroup algebra of $P_N(Q)$ is the truncated path algebra $FQ/J^N$ where $J$ is the arrow ideal.

Forsberg compute the minimum degree of a faithful representation of $P_N(Q)$ over a field $F$ of characteristic $0$ which is not algebraic over $\mathbb Q$. (I suspect this might be overkill if $Q$ is acyclic.) This is a lower bound for the minimum degree of a faithful representation of $FQ/J^N$ of course.

If you take the $A_n$-quvier $Q$ with $n$ vertices $1,\ldots, n$ with a directed edge from $i$ to $i+1$, then the truncated path algebra $FQ/J^{n-1}$ is your algebra. Now 4.2 of Forsberg implies (if I understood the notation) that 2n-2 is the minimum degree of a faithful representation of the path semigroup and hence your algebra has no faithful representation of smaller degree over $F$ (assuming $F$ is as above).

More generally, he gets if you take $Q$ the $A_n$-quiver, oriented as above, and if $N\leq n$, then the minimum faithful degree for the truncated path semigroup $P_N(Q)$ is $N(n+1-N)$ and maybe this also works for the corresponding algebra.

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