I have a conjecture:

Let $ R $ be a commutative ring and $ a_1,\cdots,a_n $ be $ n $ distinct regular elements in $ R $ and $ m \geq n $ a positive integer and $ I $ the ideal of $ R[x_1,\cdots,x_m] $ generated by $ a_1x_1-1,\cdots,a_nx_n-1 $, then $ I \cap R = \{ 0 \} $.

Is this true? How to prove it?

Given a commutative ring $ R $ and a multiplicatively closed subset $ S $ of $ R $, there are two ways to consturct $ S^{-1}R $:

1. define an equivalence relation $ \sim $ on $ R\times S $ and then take $ S^{-1}R := (R\times S)/\sim $.

2. Let $ R[S] $ be the free commutative algebra over $ R $ generated by $ S $ and $ i:S\to R[S] $ the canonical embedding and $ I $ the ideal of $ R[S] $ generated by $ i(s)s-1 $ for each $ s $ in $ S $ and then take $ S^{-1}R := R[S]/I $.

Suppose $ S $ dosen't contain zero divisors, if one works in the first way, it's easy to prove that the canonical map $ j:R\to R^{-1}S $ is injective. But if one works in the second way, then it's equivalent to state that $ I \cap R = \{ 0 \}$. Can we directly prove this without considering $ (R\times S)/\sim $ ?

I have a conjecture:

Let $ R $ be a commutative ring and $ a_1,\cdots,a_n $ be $ n $ distinct regular elements in $ R $ and $ m \geq n $ a positive integer, then for each $ f_1,\cdots,f_n $ in $ R[x_1,\cdots,x_m] $, there's no nonzero $ r $ in $ R $ such that $$ (a_1x_1-1)f_1+\cdots+(a_nx_n-1)f_n = r .$$

Is this true? How to prove it?

I have a conjecture:

Let $ R $ be a commutative ring and $ a_1,\cdots,a_n $ be $ n $ distinct regular elements in $ R $ and $ m \geq n $ a positive integer and $ I $ the ideal of $ R[x_1,\cdots,x_m] $ generated by $ a_1x_1-1,\cdots,a_nx_n-1 $, then $ I \cap R = \{ 0 \} $.

Is this true? How to prove it?

I have a conjecture:

Let $ R $ be a commutative ring and $ a_1,\cdots,a_n $ be $ n $ distinct regular elements in $ R $ and $ m \geq n $ a positive integer, then for each $ f_1,\cdots,f_n $ in $ R[x_1,\cdots,x_m] $, there's no nonzero $ r $ in $ R $ such that $$ (a_1x_1-1)f_1+\cdots+(a_nx_n-1)f_n = r .$$

It's this true? How to prove it?

I have a conjecture:

Let $ R $ be a commutative ring and $ a_1,\cdots,a_n $ be $ n $ distinct regular elements in $ R $ and $ m \geq n $ a positive integer, then for each $ f_1,\cdots,f_n $ in $ R[x_1,\cdots,x_m] $, there's no nonzero $ r $ in $ R $ such that $$ (a_1x_1-1)f_1+\cdots+(a_nx_n-1)f_n = r .$$

Is this true? How to prove it?