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Let $A$ be a non-selfinjective algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\tau(A/AfA)=\Omega^1(D(A))$. We thus have two exact sequences that show that $A/AfA$ and $D(A)$ are closely connected.

The almost split sequence for $A/AfA$: $$0 \rightarrow \tau(A/AfA)=\Omega^1(D(A)) \rightarrow X \rightarrow A/AfA \rightarrow 0$$

The beginning of a minimal projective resoution of $D(A)$: $$0 \rightarrow \Omega^1(D(A)) \rightarrow P \rightarrow D(A) \rightarrow 0.$$

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$?

Let $A$ be a non-selfinjective algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\tau(A/AfA)=\Omega^1(D(A))$. We thus have two exact sequences that show that $A/AfA$ and $D(A)$ are closely connected.

The almost split sequence for $A/AfA$: $$0 \rightarrow \tau(A/AfA)=\Omega^1(D(A)) \rightarrow X \rightarrow A/AfA \rightarrow 0$$

The beginning of a minimal projective resoution of $D(A)$: $$0 \rightarrow \Omega^1(D(A)) \rightarrow P \rightarrow D(A) \rightarrow 0.$$

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^j(A/AfA,A) \neq 0$ for some $j \geq 1$?

Let $A$ be a non-selfinjective algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\Omega^{-1}(\tau(A/AfA))=D(A)$. We thus have two exact sequences that show that $A/AfA$ and $D(A)$ are closely connected.

The almost split sequence for $A/AfA$: $$0 \rightarrow \tau(A/AfA)=\Omega^1(D(A)) \rightarrow X \rightarrow A/AfA \rightarrow 0$$

The beginning of a minimal projective resoution of $D(A)$: $$0 \rightarrow \Omega^1(D(A)) \rightarrow P \rightarrow D(A) \rightarrow 0.$$

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$?

Let $A$ be a non-selfinjective algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\tau(A/AfA)=\Omega^1(D(A))$. We thus have two exact sequences that show that $A/AfA$ and $D(A)$ are closely connected.

The almost split sequence for $A/AfA$: $$0 \rightarrow \tau(A/AfA)=\Omega^1(D(A)) \rightarrow X \rightarrow A/AfA \rightarrow 0$$

The beginning of a minimal projective resoution of $D(A)$: $$0 \rightarrow \Omega^1(D(A)) \rightarrow P \rightarrow D(A) \rightarrow 0.$$

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$?

Let $A$ be an algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\Omega^{-1}(\tau(A/AfA))=D(A)$.

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$?

Let $A$ be a non-selfinjective algebra with dominant dimension at least one and $Af$ a minimal faithful projective-injective left $A$-module (but we use right modules in the following). The Nakayama conjecture states that $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$ (this is equivalent to finite dominant dimension), while the first Tachikawa conjecture states that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$. It is well known that the Nakayama conjecture implies the first Tachikawa conjecture. The other direct seems to be not known.

Now I noted the relation $\Omega^{-1}(\tau(A/AfA))=D(A)$. We thus have two exact sequences that show that $A/AfA$ and $D(A)$ are closely connected.

The almost split sequence for $A/AfA$: $$0 \rightarrow \tau(A/AfA)=\Omega^1(D(A)) \rightarrow X \rightarrow A/AfA \rightarrow 0$$

The beginning of a minimal projective resoution of $D(A)$: $$0 \rightarrow \Omega^1(D(A)) \rightarrow P \rightarrow D(A) \rightarrow 0.$$

Can this somehow (or some other trick) be used to show that $Ext_A^i(D(A),A) \neq 0$ for some $i \geq 1$ implies $Ext_A^i(A/AfA,A) \neq 0$ for some $i \geq 1$?