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Let $X$ be a variety, $V$ is a hypersurface in $X$, and $Y$ is a closed subvariety in $X$ which intersects with $V$ transverally. Suppose the corresponding closed embeddings are:$i:V \to X,\quad j: Y \to X$ . Then we have the following exact sequence:

$0 \to L(-V) \to O_X \to i_*O_V \to 0$

One can tensor it with $j_*O_Y$ to get an exact sequence. But I don't know how does transversality play the role to make it left exact.

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    @MattE Thank you for point out this to me, which I take for granted when I wrote it down. Frankly speaking, I don't know. But since this question comes from Hartshorne Chapt 5, Lemma 1.3, I use the definition of transversal intersection of curves given in the beginning of the same section, and it does not assume smoothness. And I also found a definition of effective divisors meeting transversally in Liu Qing's book(Def.1.6 p378). It would be very helpful if you can point out to me the definition in the smooth case, because there is case I considering which is smooth but not divisors.2012-09-16

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There is a weaker condition then being transversal (which is usually defined under smoothness condition as in the comment of Matt E): namely $V$ and $Y$ cut properly if $V$ doesn't contain any irreducible component of $Y$. If $Y$ is irreducible (or pure), this just means that $Y\cap V$ has dimension $\dim Y-1$.

Now suppose $Y$ is integral and cuts $V$ properly, then tensoring your exact sequence by $O_Y$ is exact. Indeed, the question is local on $X$. Let $x\in Y$, let $f$ be a generator of $L(-V)_x$ (local equation of $V$ at $x$). We have to show that $ fO_{X,x} \otimes_{O_{X,x}} O_{Y,x} \to O_{Y,x}$ is injective. Write for simplicity $A=O_{X,x}$, $O_{Y,x}=A/I$. Then the above map is $ fA/fI \to A/I.$ Its kernel if $(fA\cap I)/fI$. Let $g=fa\in fA\cap I$. Then in $A/I$, we have $\bar{f}\bar{a}=0$. As $A/I=O_{Y,x}$ is an integral domain and $\bar{f}\ne 0$ because $V$ doesn't contain $Y$, we find $\bar{a}=0$, hence $g\in fI$ and $fA\cap I=fI$. Thus the injectivity.

In general (without integral hypothesis on $Y$), we see by this proof that the exactness at left is equivalent to the image of $f$ in $O_{Y,x}$ is a regular element. This condition is little stronger than $V, Y$ cut properly (we could probably say something like them are secant).

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    Thank you SOOOOOO much!!! I cannot expect any answer clearer than that!2012-09-15