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In the answer of this question:Question about the category of projective $A$-modules of an Artin algebra $A$

There is a place I don't know: Here $A$ is an artin algebra, if there is a projective module $P$ non-injective, then there is a monomorphism $i:R \rightarrow S$ and a map $f: R \rightarrow P$, there is no map $g:S \rightarrow P$ suc that $f=gi$, then the answer of this question gets a non-split short exact sequence of $A$-modules of the form : $0 \rightarrow R \rightarrow S \rightarrow S/R \rightarrow 0$, then since it is not split, $Ext^1 _A(S/R,R) \neq0$. Then how to get the projective dimesion of $S/R=1$? (In my mind, $pd M=1$ iff $Ext^1(M,-) \neq0$ and $Ext^2(M,-)=0$)

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    please post all information on R and S from the previous question to make the context clear. – 2017-01-06
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    @Mare I think there are no additional conditions on $R$ and $S$, I have post how to get the non-split short exact sequence. – 2017-01-06
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    The short exact sequence $0 \rightarrow R \rightarrow S \rightarrow S/R \rightarrow 0$ is a projective resolution of $S/R$ of length 1, so its projective dimension is at most 1. – 2017-01-08
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    @HughThomas Why $R$ and $S$ are projective? – 2017-01-09
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    They're projective because they're in the category of projective $A$-modules -- how you would check if $P$ is injective in the category of projective $A$-modules is by starting with a mono in that category. – 2017-01-09
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    @HughThomas Oh, I misunderstand the original meaning. I thought he means $P$ is injective in the category $mod \Lambda$. Get it, thank you. – 2017-01-09

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