Eisenbud Unmixedness Example I am struggling with the following example in Chapter 18 of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry , in which the author uses the unmixedness theorem to show that a genus 1 degree 4 irreducible curve $C$ in $\mathbb{P}^3_k$ has a homogeneous ideal generated by two quadrics in $S = k[x_0,...,x_n]$.
He shows first that there exist linearly independent irreducible quadrics $Q_1$, $Q_2$ such that $(Q_1,Q_2) \subset I$, where $I$ is the homogeneous ideal of $C$. Next he uses irreducibility to assert that $Q_1, Q_2$ form a regular sequence in $S$. I am okay with this part. 
Now he says that the curve cut out by $Q_1, Q_2$ has degree 4 by Bezout's theorem. I believe I understand this part. The degree of the scheme-theoretic intersection is equal to the product of the degrees (I believe this is what Bezout's theorem says, although I can't find it phrased in these terms), which is 2*2 since we deal with two quadric hypersurfaces. This does not depend on the fact that $Q_1, Q_2$ are a regular sequence. Please correct me if I'm mistaken anywhere here.
Now, here is the part I do not understand. "Since $(Q_1,Q_2)$ is contained in $I$, we must have $(Q_1,Q_2) = I \cap J$ where codim$J > 2$." Where does this decomposition come from?
Finally, he concludes that $J=\emptyset$ since by the Unmixedness Theorem every associated prime of $(Q_1,Q_2)$ has codimension 2. This part again I believe I understand.
Can anyone explain the middle part to me? I do not get it at all. Disclaimer: I am not interested in the actual assertion, only the technique of his proof, so while an alternate approach may be edifying, it's not what I'm looking for. 
 A: I'm not sure exactly how Eisenbud intends to argue, because I don't know what
he's assuming known at this point.  But here is some kind of explanation, which perhaps you can adapt to what you know:
The inclusion $(Q_1,Q_2) \subset I$ shows that $V(Q_1,Q_2) \supset V(I) = C$.
Now $V(Q_1,Q_2)$ is a curve of degree $4$ (by Bezout), as is $C$ (by assumption) and so this inclusion has to be
an equality at generic points of $C$.  (Essentially by definition,
``degree'' of a curve measures the behaviour of a curve (or whatever dimensional
objects we are applying it to, but in this particular case it is curves) at its generic points.)
Thus the only way that $V(Q_1,Q_2)$ and $V(I)$ can differ is by the former having some "extra" scheme-structure at some finite number of closed points.
These points are what are being cut out by $J$.  (They are what are called
embedded points of $V(Q_1,Q_2)$, and ring-theoretically correspond to associated primes that are not minimal.)
Unmixedness then shows that these embedded points don't exists.
Technically, the decomposition $(Q_1,Q_2) = I \cap J$ is probably the primary decomposition (although I've never been very sure about the algebraic terminology for these sorts of considerations), and $J$ can't have any codimension $2$ contributions because
these would be extra curve components of $V(Q_1,Q_2)$ (above and beyond $C$),
and we've already seen for degree reasons that these don't exist.
A: Recasting Matt E's answer in terms I feel more comfortable with: 
Let $L=(Q_1,Q_2)$. By the unmixedness theorem, the associated primes of $L$ are precisely the minimal primes over $L$ and all such have height 2. Since $I$ is such a prime, the minimal primary decomposition of $L$ looks like $q \cap q_1 \cap ... \cap q_l$, where $q$ is $I$-primary. Since each prime in the decomposition has codimension equal to that of $L$, we have deg $S/L$ = deg $S/q$ + deg $S/q_1$ + ...+ deg $S/q_l$ (i.e. the degree is the sum of the degrees of the components). 
I believe I can prove, using an exercise in Ravi Vakil's notes (18.6F), that if $q$ is $I$ primary, deg $S/q$ $\geq$ deg$S/I$ with equality if and only if $q=I$ (in other words, putting nilpotent structure on a component of a curve always strictly increases the degree). Since we have
4 = deg $S/L$ $\geq$ deg $S/q$ $\geq$ deg $S/I$ = 4,
all the inequalities must be equalities, so that $q=I$ and $q_1$,...,$q_l$ are not really there at all. That is, $L=I$. 
