# GCD in multivariate polynomial ring and quotient ring

1. GCD in multivariate polynomial ring


I would like to prove the following but couldn't figure out how to.
Let $$d$$ and $$h_1, h_2, \cdots, h_k$$ be multivariate polynomials with $$l$$ number of indeterminates, i.e. $$d, h_1, h_2, \cdots, h_k \in \mathbb{C}[x_1, x_2, \cdots, x_l]$$. If
$$\tag 1 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k$$holds for arbitrary $$a, b_1, b_2, \cdots, b_k \in \mathbb{C}[x_1, x_2, \cdots, x_l]$$, then $$d$$ is a (greatest) common divisor (GCD) of $$h_1, h_2, \cdots, h_k$$.
I think I can prove that the converse is true, i.e. if $$d$$ is a GCD of $$h_1, h_2, \cdots, h_k$$, then $$(1)$$ holds, but I have no clue how to proceed with the original claim. Does Hilbert's Nullstellensatz somehow come into play in proving the above claim?
Also, if $$(1)$$ holds, can I say that the principal ideal generated by $$d$$ equals the ideal generated by $$h_1, h_2, \cdots, h_k$$, i.e. $$ = $$? Can this be true even if the multivariate polynomial ring is not a principal ideal domain?

2. GCD in multivariate polynomial quotient ring


Now I have an ideal $$I$$ in $$\mathbb{C}[x_1, x_2, \cdots, x_l]$$. Let $$d$$ and $$h_1, h_2, \cdots, h_k$$ be multivariate polynomials in the quotient ring $$R=\mathbb{C}[x_1, x_2, \cdots, x_l]/I$$. If
$$\tag 2 a \odot d = b_1 \odot h_1 + b_2 \odot h_2 + \cdots + b_k \odot h_k$$holds for arbitrary $$a, b_1, b_2, \cdots, b_k \in R$$ , where $$\odot$$ denotes the polynomial multiplication operation in the quotient ring $$R$$, can I call $$d$$ a (greatest) common divisor of $$h_1, h_2, \cdots, h_k$$ in $$R$$? How would I compute $$d$$ given $$h_1, h_2, \cdots, h_k$$ in this case?
The following is my initial thoughts. Since I know how to compute the Groebner basis $$G=\{g_1, g_2, \cdots, g_m\}$$ of $$I$$, I can lift $$(2)$$ to $$\mathbb{C}[x_1, x_2, \cdots, x_l]$$ and have $$\tag 3 a \cdot d + r_1 = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + r_2$$where $$r_1$$ and $$r_2$$ are unique polynomials in $$$$. Collecting the terms, I get $$\tag 4 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + r_3$$where $$r_3 = r_2-r_1$$. Since $$r_3 \in $$, I can write the above into $$\tag 5 a \cdot d = b_1 \cdot h_1 + b_2 \cdot h_2 + \cdots + b_k \cdot h_k + c_1 \cdot g_1 + c_2 \cdot g_2 + \cdots + c_m \cdot g_m$$ for some $$c_i$$.
Now $$(5)$$ looks awfully like $$(1)$$ so I want to say $$d$$ is a GCD of $$\{h_1, h_2, \cdots, h_k, g_1, g_2, \cdots, g_m\}$$.
Am I on the right track?
I have only taken linear algebra and no other algebra course but I have access to most textbooks. A gentle nudge in the right direction and pointers to relevant materials will be greatly appreciated.

• for the Groebner basis (GB), I'm assuming that I have fixed a monomial order and computed an unique reduced GB. Jun 30, 2019 at 18:06
• Does $(1)$ means something different than said ideal equality? If so please be more precise about its denotation. If the ideal equality holds then clearly $d$ is the gcd (same as classical Bezout case in $\Bbb Z).\,$ But not every ideal is principal in such multivariate rings, e.g. $\gcd(x_1,x_2) = 1$ but $\,(x_1,x_2)\neq (1)\,$ (else eval at $\,x_1 = 0 = x_2\,$ to deduce $0 = 1)\ \$ Jun 30, 2019 at 18:25
• @BillDubuque Thanks for your comment. By (1) I meant, for every $a \in \mathbb{C}[x_1,x_2,\cdots,x_l]$, we can find a set of $b_1, b_2, \cdots, b_k \in \mathbb{C}[x_1,x_2,\cdots,x_l]$ and vice versa. Is this called ideal equality? More specifically, can I say that $h_1, h_2, \cdots, h_k$ generate the principal ideal generated by $d$? And I don't know how to deduce that $d$ is the GCD from $(1)$. From your counterexample, what if the GCD is something other than the unit $1$? In that case, the principal ideal generated by the GCD always equals the ideal generated by the polynomials involved? Jun 30, 2019 at 20:53
• @BillDubuque Actually, if $gcd(f_1,f_2)=1$ for some polynomials $f_1$ and $f_2$, then aren't they called relatively prime and considered to not have a GCD? Jul 1, 2019 at 5:30
• See here for the definition of GCD & LCM in general rings. Jul 1, 2019 at 12:41

First, $$d$$ is a common divisor of $$h_1,\ldots,h_k$$.
Second, if $$s$$ is a common divisor of $$h_1,\ldots, h_k$$, then $$s$$ divides $$d$$.