# Relation between defining polynomials and irreducible components of variety

I've been puzzled about some basic facts in (classical) algebraic geometry, but I cannot seem to find the answer immediately:

1. Let $V=V(f_1,\ldots,f_n)$ be a variety over some field $k$, and let $n > 1$. Suppose that $V$ turned out to be reducible, i.e. it is the union of irreducible varieties $V_1,\ldots,V_m$ for some $m > 1$. Must it be the case that at least one of the $f_i$s are reducible polynomials?

2. Take one of the irreducible components, $V_1$, say. Do the defining equations for $V_1$ have anything to do with the polynomials $f_1,\ldots,f_n$?

3. Suppose varieties $V = V(f_1,f_2)$ and $W = V(f_3)$ shared a common component. Does that mean that $f_3$ shares a factor with either one of $f_1$ or $f_2$? If not, what's a counterexample?

Thanks so much!

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I'm sorry to say that I can't think of good examples offhand. Good things to look at are sources on "primary decomposition", and I think this book will have relevant stuff too. – Dylan Moreland Jun 14 '12 at 4:32
Do you have guesses for any of the items above? Unfortunately I'm quite new to algebraic geometry, so I don't have a good intuition for what the right answers should be :). – Henry Yuen Jun 14 '12 at 4:35
I don't really understands what (2) means. Can you give a clearer explanation of what you mean by "have anything to do with". – caozhu Jun 14 '12 at 4:52
Well, it's purposefully a vague question that I hoped would have small distance to a widely know fact. But, just to give an example of what I was wondering, do we know that, say the defining polynomials of $V_1$ are factors of the $f_1,\ldots,f_n$? – Henry Yuen Jun 14 '12 at 4:55

1) No. The variety $V=V(y,y-x^2+1)\subset \mathbb A^2_k$ is reducible and consists of the two points $(\pm1,0)$ (if $char.k\neq 2$), but the polynomials $y,y-x^2+1$ are irreducible.

2) The question is not very precise. In a sense the answer is "yes", because you can obtain $V_1$ by adding polynomials $g_1,...,g_r$ to the $f_i$'s and write $V_1=V(f_1,...,f_n;g_1,...,g_r)$

3) Yes if $k$ is algebraically closed :
Decompose $f_3$ into irreducibles: $f_3=g_1^{a_1}\ldots g_s^{a_s}$. The irreducible components of $V(f_3)$ are the $V(g_j)$'s. Say $V(g_1)$ is also an irreducible component of $V_1=V(f_1,f_2)$. Then, since $f_1$ vanishes on $V_1$, the polynomial $g_1$ divides $f_1$ (by the Nullstellensatz) and similarly $g_1$ divides $f_2$. So actually $f_3$ shares the same factor $g_1$ with both $f_1$ and $f_2$, which is more than you asked for.

Caveat If $k$ is not algebraically closed, the answer to 3) may be no: for example over $\mathbb R$ the varieties $V=V(x+y,x^4+y^4)$ and $W=V(x^2+y^2)$ are both equal to the irreducible subvariety of the plane consisting of just one point: $V=W=\lbrace (x,y)\rbrace \subset \mathbb A^2(\mathbb R)=\mathbb R^2$ .
However $x^2+y^2$ is irreducible in $\mathbb R[x,y]$ and shares no factor with $x+y$ nor $x^4+y^4$ since $x^2+y^2$ does not divide those polynomials.

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Thanks! The last one was obvious, I should've seen that earlier. What if the second variety was generated by $f_3$ and $f_4$? Then because of the answer to (1), it seems like we can't say anything about the relationship between $f_3$ and $f_1, f_2$. – Henry Yuen Jun 14 '12 at 4:53
Nice work on (3)! I kept thinking that the codimension of $V(f_1, f_2)$ would be $2$. – Dylan Moreland Jun 14 '12 at 4:55
Dear Henry, you are right that one can't say anything in the case you describe: think of $V(x,y)$ and $V(x+y, x-y)$, which both describe the origin in the plane, an irreducible subvariety. No two of the four (irreducible) polynomials have a common non-constant factor. – Georges Elencwajg Jun 14 '12 at 5:03
Wonderful! Thanks for taking the time to answer these elementary questions. I will play with these examples more to gain more intuition. Thanks also Dylan for your example. I wish I could accept both answers. – Henry Yuen Jun 14 '12 at 5:04
@HenryYuen No worries. I'm certainly not in it for the points, and this is the better answer. – Dylan Moreland Jun 14 '12 at 21:17

Here's a silly example to play around with. In $\mathbf A^3$ consider the algebraic set $V(x - yz, x)$. Here the irreducible components are the lines $V(x, y)$ and $V(x, z)$.

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I call the example silly because I could have written $V(yz, x)$ and then it would have seemingly less to do with your (1) and (2). Also, it seems to me that the dimensions in (3) are unlikely to work out. – Dylan Moreland Jun 14 '12 at 4:45
so (1) is false – caozhu Jun 14 '12 at 4:50