Note: the precise formulation of the problem is at the end of the post. $ \def \y {\boldsymbol y} \def \z {\boldsymbol z} \def \a {\boldsymbol a} \def \Z {\mathbb Z} $
Consider a finite set of variables consisting of $ x $, $ y _ 1 $, $ \dots $ and $ y _ k $. $ \y $ will be an abbreviation for the sequence $ y _ 1 , \dots , y _ k $ of variables. We're given finitely many polynomial $ p _ 1 ( x , \y ) , \dots , p _ l ( x , \y ) \in \Z [ x , \y ] $. Denote their GCD by $ q ( x , \y ) $, and for any $ \a \in \Z ^ k $, denote by $ q _ \a ( x ) $ the GCD of $ p _ 1 ( x , \a ) $, $ \dots $ and $ p _ l ( x , \a ) $, where each $ p _ i ( x , \a ) $ is taken as a polynomial in $ \Z [ x ] $.
Here is a rather loose version of my question:
Assume that $ q ( x , \y ) = 1 $, and for any $ \a \in \Z ^ k $, $ q _ \a ( x ) = b x + c $ for some $ b , c \in \Z $ with $ \gcd ( b , c ) = 1 $. What can we say about the set of the common roots of $ p _ 1 ( x , \y ) $, $ \dots $ and $ p _ l ( x , \y ) $ in $ \Z ^ { k + 1 } $?
Already for $ k = 1 $ and $ l = 2 $, there are different situations in different examples. Of course, one example is when there are no common roots: take $ p _ 1 ( x , y ) $ to be the constant $ 1 $ polynomial. There are examples where there is a unique common root: take $ p _ 1 ( x , y ) = x $ and $ p _ 2 ( x , y ) = y $. There are also more interesting examples where several common roots exist: take $ p _ 1 ( x , y ) = x + y $ and $ p _ 2 ( x , y ) = x + y ^ 3 $. In any case, the condition $ q ( x , y ) = 1 $ leaves no room for infinitely many common roots, as one can see from this post.
For $ k = 2 $, the finiteness of the set of the common roots is lost: take $ p _ 1 ( x , y _ 1 , y _ 2 ) = x - y _ 1 + y _ 2 $ and $ p _ 2 ( x , y _ 1 , y _ 2 ) = x - y _ 1 ^ 2 + y _ 2 $. I can imagine that for larger $ k $ and $ l $, more complicated scenarios can arise.
But the last example I've given here is somehow suggesting that the set of common roots is "not so much infinite", in this sense: take $ z = x + y _ 2 $, and note that $ \tilde p _ 1 ( z , y _ 1 ) = z - y _ 1 $ and $ \tilde p _ 2 ( z , y _ 1 ) = z - y _ 1 ^ 2 $ have only two common roots in $ \Z ^ 2 $, namely $ ( 0 , 0 ) $ and $ ( 1 , 1 ) $. This is the main motivation behind asking my main question. I'm looking for those properties of the set of common roots that are of some sort of "finiteness" nature, while I'm not sure there is always such a property. Maybe one can give an example where the set of common roots is "infinite" in every apparent way?
What makes me hopeful to find a suitable "finite" property for the set of common roots is the strong condition on the $ q _ \a ( x ) $: fixing the values of the variables $ \y $, we can find at most one value of $ x $ to give a common root of the $ p _ i ( x , \y ) $. I don't think I've exploited the full potential of this assumption. What I've done is only trying to find examples satisfying this property, with the set of common roots "as infinite as possible".
EDIT:
I think I've come up with a suitable candidate for formulating the problem. Before presenting it, let's first consider a couple of further examples that motivated the formulation. The first one is not exactly an example, as it does not exactly satisfy the condition on $ q _ \a ( x ) $: for some $ \a $ we have $ q _ \a ( x ) = b x + c $ but $ \gcd ( b , c ) \ne 1 $. Yet it gives some insight on how to formulate the problem.
- Let $ k = 3 $, $ l = 2 $, $ p _ 1 ( x , \y ) = x y _ 1 + y _ 2 $ and $ p _ 2 ( x , \y ) = x y _ 3 + y _ 2 ^ 3 $. It's easy to verify that the common roots of $ p _ 1 $ and $ p _ 2 $ in $ \Z ^ 4 $ are exactly the quadruples $ ( u , v , - u v , - u ^ 2 v ^ 3 ) $, where $ u $ and $ v $ are arbitrary integers. The triples $ \a = ( a _ 1 , a _ 2 , a _ 3 ) $ for which there exists a common root of $ p _ 1 ( x , \a ) $ and $ p _ 2 ( x , \a ) $ are exactly those for which there are $ u , v \in \Z $ with $ \a = ( v , - u v , - u ^ 2 v ^ 3 ) $. For such $ \a \in \Z ^ 3 $, we have $ q _ \a ( x ) = a _ 1 x + a _ 2 $; for other $ \a \in \Z ^ 3 $, $ q _ \a ( x ) = 1 $.
- Let $ k = 2 $, $ l = 2 $, $ p _ 1 ( x , \y ) = ( x - y _ 1 ) ^ 2 - 1 $ and $ p _ 2 ( x , \y ) = ( x - y _ 2 ) ^ 2 - 4 $. The common roots are exactly the triples of the form $ ( u , u \pm 1 , u \pm 2 ) $.
From the first example, we can see that whenever $ \a $ is such that the $ p _ i ( x , \a ) $ have a common root, such root is given as a rational function of $ \a $; namely, the common roots of the $ p _ i ( x , \y ) $ are given by the roots of $ x - f ( \y ) $, where $ f ( \y ) = - \frac { y _ 2 } { y _ 1 } $ (excluding the cases where $ y _ 1 $ takes the value $ 0 $, of course). Furthermore, the necessary and sufficient condition for $ \a $ to give rise to a common root is the existence of certain integers on which the components of $ \a $ have a polynomial dependence.
From the second example, we can see that the rational function producing the value of $ x $ from the values of the other variables may not be unique. One can, of course, imagine that other examples exist with a combination of the properties we discussed for these two examples.
As suggested by playing around with different examples I could come up with, I conjectured that in any case, the set of common roots of the $ p _ i $ share the properties of the above examples. So, here is a more definitive version of my question:
Assume that $ q ( x , \y ) = 1 $, and for any $ \a \in \Z ^ k $, $ q _ \a ( x ) = b x + c $ for some $ b , c \in \Z $ with $ \gcd ( b , c ) = 1 $. Then can we conclude the following?
- There are $ r _ 1 ( \y ) , \dots , r _ m ( \y ) , s _ 1 ( \y ) , \dots , s _ m ( \y ) \in \Z [ \y ] $ such that any common root of all the $ p _ i ( x , \y ) $ is a root of one of the polynomials $ r _ j ( \y ) x + s _ j ( \y ) $.
- In case the previous item holds, for every $ 1 \le j \le m $, consider the following set: those $ \a \in \Z ^ k $ for which there is $ d \in \Z $ with $ r _ j ( \a ) d + s _ j ( \a ) = 0 $ and $ p _ i ( d , \a ) = 0 $ for all $ 1 \le i \le l $. There are $ n $ variables $ \z $, and polynomials $ t _ { j \ 1 } ( \z ) , \dots , t _ { j \ k } ( \z ) \in \Z [ \z ] $ such that the mentioned set coincides with $ \left \{ \bigl ( t _ { j \ 1 } ( \z ) , \dots , t _ { j \ k } ( \z ) \bigr ) \bigm | \z \in \Z ^ n \right \} $.
Item 1 is itself of the finite nature that I was looking for. Comparing to the condition we had on the $ q _ { \a } ( x ) $, it seems plausible to me. Item 2 is a bonus one; I found it to be true in all the examples I came up with, yet it could be wrong in general. For example, one could find examples for which some of the $ t _ { j \ h } ( \z ) $ are not in $ \Z [ \z ] $, but in $ \mathbb Q [ \z ] $; or one might find a case where the assertion is wrong altogether. this is of less importance to me, than the validity of item 1. Finally, in case item 1 is not valid, I'm still curious to find other "finiteness" properties to replace it.