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  "body": "\n## Definition\n\nLet $\\mathcal{A}$ and $\\mathcal{B}$ be [[abelian categories|Y9CZ3MTNF1]], and assume that every object of $\\mathcal{A}$ has an [[injective resolution|ZW3W6FX4PW]].\nLet $F \\colon \\mathcal{A} \\to \\mathcal{B}$ be an [[additive functor|KAN07YC9T2]] that is left [[exact|S1G61TDW0K]].\nFor an object $X$ of $\\mathcal{A}$, choose an injective resolution $X \\to I^\\bullet$.\nFor $n \\ge 0$, the $n$-th *right derived functor* of $F$ at $X$ is the [[cohomology|2J4JJJYP22]]\n$$R^n F(X) = H^n(F(I^\\bullet)).$$\nDually, if every object has a [[projective resolution|HK2AA1YSQ2]] $P_\\bullet \\to X$ and $F$ is additive and right exact, the *left derived functors* are $L_n F(X) = H_n(F(P_\\bullet))$.\n\n## Remarks\n\nUp to a canonical isomorphism, $R^n F(X)$ does not depend on the resolution, since any two injective resolutions are chain homotopy equivalent; the same argument makes $R^n F$ a functor.\nLeft exactness gives $R^0 F \\cong F$, and a short exact sequence in $\\mathcal{A}$ gives a long exact sequence of the $R^n F$.\nExample: let $F = \\Hom_\\Z(\\Z/2\\Z, -)$ on abelian groups and $X = \\Z$, with the injective resolution $0 \\to \\Z \\to \\Q \\to \\Q/\\Z \\to 0$.\nThen $F(\\Q) = 0$ and $F(\\Q/\\Z) = \\set{0, 1/2} \\cong \\Z/2\\Z$, so $R^0 F(\\Z) = 0$ and $R^1 F(\\Z) \\cong \\Z/2\\Z$, which is $\\operatorname{Ext}^1_\\Z(\\Z/2\\Z, \\Z)$.\nNon-example: the complex $0 \\to \\Z \\xrightarrow{\\id} \\Z \\to 0$ is exact, but $\\Z$ is not injective, and using it in place of $I^\\bullet$ would give $0$ in degree $1$ instead of $\\Z/2\\Z$.\nIn general $R^n \\Hom_R(M, -) = \\operatorname{Ext}^n_R(M, -)$ and $L_n(M \\otimes_R -) = \\operatorname{Tor}_n^R(M, -)$.\n",
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