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  "body": "\n## Definition\n\nLet $R$ be a [[ring|0B1261B5ED]] and let $M$ and $N$ be [[$R$-modules|0NZ8JA25WK]].\nChoose a [[projective resolution|HK2AA1YSQ2]] $\\cdots \\to P_1 \\xrightarrow{d_1} P_0 \\to M \\to 0$ of $M$.\nApplying $\\Hom_R(-, N)$ to $P_\\bullet$ gives a cochain complex\n$$0 \\to \\Hom_R(P_0, N) \\xrightarrow{d_1^*} \\Hom_R(P_1, N) \\xrightarrow{d_2^*} \\Hom_R(P_2, N) \\to \\cdots,$$\nwhere $d_i^*(f) = f \\circ d_i$.\nFor $n \\ge 0$, the $n$-th *Ext group* is its cohomology\n$$\\operatorname{Ext}^n_R(M, N) = \\ker d_{n+1}^* / \\im d_n^*,$$\nwith $d_0^* = 0$.\n\n## Remarks\n\nUp to a canonical isomorphism, $\\operatorname{Ext}^n_R(M, N)$ does not depend on the choice of the resolution.\nOne has $\\operatorname{Ext}^0_R(M, N) \\cong \\Hom_R(M, N)$, and the same groups are obtained from an injective resolution of $N$.\nIf $R$ is commutative, the Ext groups are $R$-modules.\nExample: from $0 \\to \\Z \\xrightarrow{\\cdot n} \\Z \\to \\Z/n\\Z \\to 0$ with $n \\ge 1$, the complex is $\\Z \\xrightarrow{\\cdot n} \\Z$, so $\\operatorname{Ext}^0_\\Z(\\Z/n\\Z, \\Z) = 0$ and $\\operatorname{Ext}^1_\\Z(\\Z/n\\Z, \\Z) \\cong \\Z/n\\Z$.\nNon-example: the sequence $0 \\to \\Z/n\\Z \\xrightarrow{\\id} \\Z/n\\Z \\to 0$ is exact but $\\Z/n\\Z$ is not projective over $\\Z$ for $n \\ge 2$; using it would give $\\operatorname{Ext}^1 = 0$, which is wrong.\nA module $M$ is projective if and only if $\\operatorname{Ext}^1_R(M, N) = 0$ for all $R$-modules $N$.\n",
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