Let $S$ be an algebra essentially of finite type over a field $k$. Then ${\mathrm{Hom}}_k^{}(S,k)$ is an injective $S$--module, and the Matlis structure theorem (Matlis, E.: Pacific J. Math. 8, 511–528 1958) tells us that it can be written as a direct sum of indecomposable injectives. We compute the multiplicities of these injectives. Let $\mathfrak{p}$ be a prime ideal in $S$, and let $I(\mathfrak{p})$ be the injective hull of $S/\mathfrak{p}$. If the residue field $k(\mathfrak{p})$ is algebraic over $k$ then the multiplicity of $I(\mathfrak{p})$ is $\mu (\mathfrak{p})=1$. If the transcendence degree of $k(\mathfrak{p})$ over $k$ is $\ge 1$ then $\mu (\mathfrak{p})\ge {|\mathrm{\#}k|}^{{\mathrm{\aleph }}_0}$, that is the multiplicity is no less than the cardinality of the field $k$ raised to the power ${\mathrm{\aleph }}_0$. If $S$ is finitely generated over $k$ then equality holds, that is $\mu (\mathfrak{p})={|\mathrm{\#}k|}^{{\mathrm{\aleph }}_0}$. For $k(\mathfrak{p})$ of transcendence degree $\le 1$ the result is not surprising, but for $k(\mathfrak{p})$ of transcendence degree $\ge 2$ it is not clear that $\mu (\mathfrak{p})\ne 0$. We prove the result by induction on the transcendence degree, and the key is that we produce an injective map, from a space whose dimension we know by induction and into the space whose dimension we want to estimate. The interest in the result comes from the fact that the size of $\mu (\mathfrak{p})$ measures the failure of a natural map $\psi (f):f^{\times }⟶f^{!}$ to be an isomorphism. Here $f^{\times }$ and $f^{!}$ are the twisted inverse image functors of Grothendieck duality.