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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. $<d> = <h_1, h_2, \cdots, h_k>$? 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 $<G>$. 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 $r_3 \in <G>$, 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.

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  • $\begingroup$ for the Groebner basis (GB), I'm assuming that I have fixed a monomial order and computed an unique reduced GB. $\endgroup$ – D.Park Jun 30 at 18:06
  • $\begingroup$ 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)\ \ $ $\endgroup$ – Bill Dubuque Jun 30 at 18:25
  • $\begingroup$ @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? $\endgroup$ – D.Park Jun 30 at 20:53
  • $\begingroup$ @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? $\endgroup$ – D.Park Jul 1 at 5:30
  • $\begingroup$ See here for the definition of GCD & LCM in general rings. $\endgroup$ – Bill Dubuque Jul 1 at 12:41
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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 $d$.

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  • $\begingroup$ Thanks for your comment, but what you wrote would be the definition of a GCD, not an answer to my question. Perhaps you can tell me why the 'First' comment can be concluded given (1)? And, if that can be answered, can the statement be extended to the case of (2) as well? $\endgroup$ – D.Park Jun 30 at 18:03

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