Let $S = k[x_1,x_2,x_3]$ be a polynomial ring of dimension $3$ over an infinite field, and let $I$ be a homogeneous ideal of height $3$. Since $S$ has no zero divisors, the Krull dimension of $I$ is $3$. Since $\dim S/I =0$, $I$ contains a power of the irrelevant ideal and so its depth is $3$ as well. Then by Grothendieck's vanishing theorem on local cohomology (e.g. B&H, Thm. 3.5.7) we must have that $H_{\mathfrak{m}}^3(I) \neq 0$ and $H_{\mathfrak{m}}^i(I)=0, \, \forall i< 3$. Now, we have an exact sequence \begin{align} H_{\mathfrak{m}}^i(S) \rightarrow H_{\mathfrak{m}}^i(S/I) \rightarrow H_{\mathfrak{m}}^{i+1}(I) \rightarrow H_{\mathfrak{m}}^{i+1}(S). \, \, \, (\dagger) \end{align} Again by the vanishing theorem, we must have that $H_{\mathfrak{m}}^i(S)=0, \, \forall i<3$ and $H_{\mathfrak{m}}^0(S/I) \neq 0$. Substituting $i = 0$ in $(\dagger)$ we get $0 \neq H_{\mathfrak{m}}^0(S/I) \cong H_{\mathfrak{m}}^1(I)$.
This is a contradiction on the earlier conclusion that $H_{\mathfrak{m}}^i(I)=0, \, \forall i< 3$. Now i suspect that this earlier conclusion is wrong, but i can't see why, since I am viewing $I$ as an $S$-module and am simply applying the vanishing theorem.
PS: There is a tendency in commutative algebra literature to implicitly refer to $S/I$ when one is talking about attributes of $I$ (not always, but sometimes). So perhaps i applied erroneously the definition of dimension of $I$. What I do find confusing is that when we have a theorem holding for any $S$-module $M$, then if we substitute $I$, the resulting statement should be true. From this standpoint, what is your opinion about my little "paradox" presented above?
Why do you think $\operatorname{depth}_SI=3$? I'd say that it is $1$.