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Right now I'm trying to find the Hilbert Function , and the corresponding Hilbert Polynomial for the ring $M=k[x,y,z,w]/(x,y) \cap (z,w)$. I just finished reading the first chapter of Eisenbud, so I don't have that much of an advanced toolbox for things like this yet, (I know almost nothing about Gröbner bases). My first attempt was to construct a free resolution, let $ R = k[x,y,z,w]$. We then have the free resolution: $0 \rightarrow R \rightarrow^{d_3} R^4 \rightarrow^{d_2} R^4 \rightarrow^{d_1} R \rightarrow^{d_0} M \rightarrow 0, $where $d_i$ sends the free generators onto the given generators of the next coming module. So, $R^4$ sends the generators onto the generators of the kernel of $d_0$ etc. My answer right now, which I've been getting from this, is that the Hilbert Function should be : $H_m(s) =\binom{3 + s}{3}- 4\binom{1 +s}{ 3} + 3\binom{2 + s}{ 3}.$

However, I have been testing this out with macaulay2, and it doesn't agree with my answer, so I suspect I'm not right. Any help would be nice.

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Solution by Geometry
The projective subscheme $ S=Proj (M) \subset \mathbb P^3_k$ consists of the disjoint union $ S=L\bigsqcup M \;$of two skew lines, so that its Hilbert polynomial is the sum $p_S(x)=p_L(x)+p_M(x)=(x+1)+(x+1)=2x+2$

Solution by Algebra
First prove that $(x,y)\cap(z,w)=(x,y).(z,w)$ Then for $n\geq 2$ you get
$M_n =(\oplus_{i+j=n} kx^iy^j)\oplus (\oplus_{i+j=n} kz^iw^j) $ so that
$dim_k(M_n)=(n+1)+(n+1)=2n+2$ from which again you deduce that $p_S(x)=2x+2$ (and as a bonus you see that the Hilbert function and the Hilbert polynomial coincide for $n\geq 2$.)

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    Dear Dedalus, in the strict logical sense, no, free resolutions are not *necessary* to solve the exercise. But that does not mean that a nice solution cannot be found by using them.2011-08-29