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:


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*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?

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

*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!
 A: 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.
A: 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 $V(x, z)$.
