I have a difficulty when reading "Lectures in Abstract Algebra, II. Linear Algebra" by Nathan Jacobson (GTM #31). The section is excerpted at the end of the question. The sentence I have difficulty is underlined with red. My problem is: the definition of a bilinear form should be given based on a left and a right vector space, i.e., we first have the left vector space $\mathfrak{R}$ then we have the bilinear form $g$. But here the author was using a bilinear form to define a left vector space $\mathfrak{R}$, and the introduction of the bilinear form has already used the $\mathfrak{R}$ as part of its definition which the book is trying to define. Such a circular definition confused me and therefor I ask here. What on earth are $\mathfrak{R}$ and the bilinear form $g$? What is the relation between them? Please try to use the notation and concept before that section in the text. Thank you.
This text is a slightly difficult to follow, but I think it's quite unambiguous. The first paragraph is basically saying "You already know what a bilinear form is. In this chapter, we'll define a similar idea, a product group". The second paragraph actually defines what a product group is. That's the heart. The third paragraph says, "Hey, you already know some product groups! A non-zero bilinear form is a product group iff its value ring $\Delta$ is commutative." The fourth paragraph says, "You can actually construct a product group for any two vector spaces $\mathfrak{R}'$ and $\mathfrak{S}$. We'll show you how."
That fourth paragraph is not defining a product group. It’s showing how to construct a product group, given vector spaces $\mathfrak{R}$, $\mathfrak{R}'$, and $\mathfrak{S}$ and a bilinear form $g:\mathfrak{R}\times\mathfrak{R}'\to\Delta$. The phrase, “$\mathfrak{R}$ dual to $\mathfrak{R}’$”, just means there’s a bilinear form $g(x,y’)$ for $x\in\mathfrak{R}$ and $y’\in\mathfrak{R}’$. By choosing different $\mathfrak{R}$ and $g()$, you'll get different product groups for the same $\mathfrak{R}'$ and $\mathfrak{S}$.
The author said, "We shall give next a way of forming a product group for any pair of vector spaces $\mathfrak{R}'$ and $\mathfrak{S}$". What's perhaps confusing you is that this construction doesn't only depend on $\mathfrak{R}'$ and $\mathfrak{S}.$ It also depends on $\mathfrak{R}$ and $g()$. The author just slips those into the discussion, seemingly out of nowhere. But such a dual space $\mathfrak{R}$ always exists and there's always a corresponding bilinear form $g()$ expressing this duality. (See Wikipedia.) So all is well.