Does $\operatorname{Ext}_{\Lambda}^i (D\Lambda,X)=0$ for any $i\ne 0$ imply that $X$ is injective?

Question

Let $K$ be a field, let $\Lambda$ be a finite-dimensional $K$-algebra with global dimension $\le n$, let $D=\operatorname{Hom}_K(-,K)$.

Assume that $X\in \bmod \Lambda$ satisfies $\operatorname{Ext}_{\Lambda}^i (D\Lambda,X)=0$ for any $i\ne 0$, is $X$ injective?

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As we know, $X$ is injective iff. $\operatorname{Ext}_{\Lambda}^i (M,X)=0$ for all $i\ne 0$ and all $M\in \bmod \Lambda$. What is the relation between $M$ and $D\Lambda$? Thank you!

Answer 1

Yes.

Let $X$ be noninjective, and let $$0\to X\to I^0\to I^1\to\dots\to I^m\to0$$ be a minimal injective resolution of $X$.

Then this represents a nonzero element of $\operatorname{Ext}^m_\Lambda(I^m,X)$.