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I’m doing an exercise that guides me through the steps of a weird construction of the quotient field of a commutative, unitary integral domain $R$. I’m supposed to consider the polynomial ring $R[X_a|_{a\in R \setminus \{0\}}]$ in the variables $X_a$ and the ideal generated by the polynomials $aX_a-1$ for $a \in R \setminus \{0\}$, called $\mathfrak{m}$.

Then we of course notice that $X_a$ plays the role of $\frac{1}{a}$ in the factor ring $R[X_a|_{a\in R \setminus \{0\}}]/ \mathfrak{m}$ and so on... At some point the exercise wants me to show that each $f \in R[X_a|_{a\in R \setminus \{0\}}]/\mathfrak{m} $ can be written as $f=bX_{a_1}^{\nu_1}…X_{a_r}^{\nu_r}, \ b \in R$. Somehow I can’t see why this is the case, a general $f$ is a finite sum of coefficients times a product of these $X_{a_i}$, why should only one term in that sum remain? We only know that there we can rewrite terms like: $cX_{a_1}^{\nu_1}…X_c^{\nu_j}…X_{a_r}$ that occur in that sum as $X_{a_1}^{\nu_1}…X_c^{\nu_j-1}…X_{a_r}$, but how does this help me?

Henry T.
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    This is a very abstract version of being able to add two fractions to get another fraction. It may help to think about how that process works. – TomKern Nov 23 '22 at 06:13
  • Ahhaa, that’s a very nice observation, many thanks!! – Henry T. Nov 23 '22 at 06:17
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    As hinted, if you view it in terms of fractions, you are essentially rewriting the sum using a common denominator. For full details of this method of construction see the links in this answer in the linked dupe (which includes simplifications and generalizations). – Bill Dubuque Nov 23 '22 at 08:40
  • Though this method of construction might appear a bit strange at first glance, algebraically it is the most natural(universal) way to proceed. The more common pair-based construction can then be derived by transporting the ring structure to the normal form pairs ("fractions"), as explained in the link. – Bill Dubuque Nov 23 '22 at 08:50
  • So the trick in $X_b+X_c$ is to use that $aX_a=1$ to get that $X_b+X_c=cX_cX_b+bX_bX_c=(b+c)X_bX_c$. It works the same way with more powers of the $X_a$. – reuns Nov 23 '22 at 09:47
  • @reuns Yes, it's essentially scaling by a common denominator as I said above, e.g.

    $$\overbrace{\dfrac{1}b+\dfrac{1}c}^{\textstyle X_b + X_c} = \overbrace{\dfrac{c+b}{bc}}^{\textstyle(c+b)X_bX_c!!!!!!!!!!!!!}\qquad$$

    which follows by multiplying $X_b+X_c,$ by $1 = X_bX_c:! bc,,$ where $,bc,$ is a common denominator.

     – Bill Dubuque Nov 24 '22 at 15:00

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