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Let $I$ be an ideal in $S=K[x_1,\dots,x_n]$. Can we compute $\operatorname{depth}(I\cap K[x_1,\dots,x_r])$ with $r \leq n$? Is there any relation between depth $I$ and $\operatorname{depth}(I\cap K[x_1,\dots,x_r])$?

And what can we tell about $\operatorname{depth}(M \cap N)$ when $M,N$ are $S$-modules? Does it have any bounds?

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What do you mean by "depth"? That is usually associated to an ideal where the elements are taken from. Or do you mean the grade? For an ideal that is actually $\mathrm{depth}_I(S)$. Is that what you mean? –  Sándor Kovács Jun 14 '11 at 12:48
    
en.wikipedia.org/wiki/Depth_%28ring_theory%29 this is the definition for depth –  Andrei Jun 14 '11 at 14:01
    
and yes I mean depth$_S I$, at least that's the notation I know. –  Andrei Jun 14 '11 at 14:06
    
I think that if we note $J=(x_{r+1},...,x_n)$ then $ I \cap K[x_1,...,x_r] = \frac{I+J}{J}$ and then we get depth $(I \cap K[x_1,...,x_r])$ from depth lemma. In general for $M \cap N$ I don't see anything. –  Andrei Jun 14 '11 at 20:30
    
A relation between $\text{depth} I$ and $\text{depth} I \cap K[x_1,\dots,x_r]$ can hardly exist. If $r=n/2$, say, then $I=(x_1,\dots,x_r)$ and $I' = (x_{r+1}, \dots, x_n)$ are two ideals, both containing a regular sequence of length $r$. In the first case the entire regular sequence survives, in the second case nothing of it survives. –  Thomas Oct 12 '11 at 5:54
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