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Let $P_1$, $P_2$, $\ldots$ , $P_t$ be $t$ (possibly non homogeneous) polynomials in $n$ variables over the field of complex numbers. If the variety $V(P_1, P_2, \ldots, P_t)$ defined by these polynomials is non-empty, then is it true that the dimension of $V$ is at least $n-t$ ?

This is true if each $P_i$ is homogeneous, but what happens in the non-homogeneous case, assuming that the variety is non-empty? I would really appreciate any pointers.

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This is true. Consider the following lemma:

Let $X$ be an irreducible variety, $f\in k[X]$ a nonzero nonconstant function, and $Y$ an irreducible component of $V(f)$. $Y$ has codimension 1 in $X$.

Proof: We may restate as follows. Let $A$ be an an affine $k$-algebra which is a domain with $f\in A$ a nonzero nonunit. Let $P$ be a minimal prime ideal containin $f$- then $P$ must have height one. The height is at most one, by Krull's principal ideal theorem, and at least one, as $X$ irreducible.

I may then apply this lemma $t$ times to the chain $V(0)\supset V(P_1)\supset V(P_1,P_2)\supset\cdots\supset V(P_1,\cdots,P_t)$. This proves the claim. (The reason it must be at most instead of exactly is that a function $P_i$ may fail to be a nonzero nonunit on some irreducible component of $V(P_1,\cdots,P_j)$ where $i<j$. In this case, the final line of the lemma where it is shown that the height is at least one fails.)

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  • $\begingroup$ Thanks for your answer. Is there a place to look up the proof for the non-homogeneous case? I could find a proof for when Pi's are homogeneous using the definition of dimension of a variety as the degree of the Hilbert polynomial. It is not clear to me if that proof extends to the non-homogeneous case. $\endgroup$
    – tintin t
    Jan 7 '17 at 1:04
  • $\begingroup$ Actually I am a bit confused by the following example. In the 3 dimensional space over complex numbers, V(xz, yz) has dimension 2, but V(xz, yz, z-1) has dimension 0. So the codimension seems to have increased by 2 when we add z-1. $\endgroup$
    – tintin t
    Jan 7 '17 at 2:14
  • $\begingroup$ Good point- the failure here is that $V(xz,yz)$ isn't equidimensional. It's the $z$-axis, which is 1-dimensional, plus the $xy$-plane, which is two dimensional. The plane $z=1$ only intersects the $z$-axis, though. The correct fix is that each irreducible component of $V(P_1,\cdots,P_t)$ has codimension at most $t$. I'll edit the answer to make this clearer in the morning. $\endgroup$
    – KReiser
    Jan 7 '17 at 9:40
  • $\begingroup$ Or now, I guess. $\endgroup$
    – KReiser
    Jan 7 '17 at 9:48
  • $\begingroup$ Thanks for the clarification. $\endgroup$
    – tintin t
    Jan 7 '17 at 14:17

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