This is a bit of a weird idea, so any advice on how to make it clearer would be welcome. Please leave a comment.
Suppose we're using mathematics to study electronic circuits. We're probably interested not only in unitless quantities (like the ratio associated with a voltage divider) but also units-possessing quantities like resistances, currents etc.
Now suppose $r$ and $r'$ denote resistances. Then $rr'$ is no longer a resistance, but rather, a resistance squared. Therefore:
Resistances do not form a field.
That's because a field is closed under multiplication, but the product of two resistances is no longer a resistance. The same can be said of currents, voltages, etc. In general, variables that possess units do not belong to fields. So, lets make up a word for what they do belong to: lets say that such variables belong to unifields. So for example, if $R$ denotes a resistance, we'll say it belongs to the unifield of all resistances.
Question. Is there a technical, correct name for these so called "unifields"?
Where can I learn more about?
Discussion. We'll make the notion of a unifield more precise in just moment. First, let us observe that we want addition keep us within a unifield (for example, the sum of two resistances is still a resistance) but multiplication to move us between unifields (for example, the product of two resistances is not a resistance, but a resistance squared). Focusing on the case of electronics, we're at least going to need:
- A unifield of all resistances $\mathcal{R}$
- A unifield of all currents $\mathcal{I}$
- All unifields of the form $\mathcal{I}^a\mathcal{R}^b$, where $a$ and $b$ denote integers. For example, $\mathcal{I}^2$ would denote the set of all currents squared, and $\mathcal{I}\mathcal{R}$ would denote the set of all voltages.
Formally, we can define this "system of unfields" as the free Abelian group $\mathbf{A}$ generated by $\{\mathcal{I},\mathcal{R}\}$, with the law of composition denoted multiplicatively. We also assert that every unifield $X \in \mathbf{A}$ is equipped with
- An additive identity $0_X \in X$
- A law of addition $+_X : X^2 \rightarrow X$, and
- A negation operator $-_X : X \rightarrow X$
such that $(X,+_X,0_X,-_X)$ is an additive Abelian group. Further, to every pair of unifields $X,Y \in \mathbf{A},$ we associate a multiplication function $\cdot_{(X,Y)} : X \times Y \rightarrow XY$, where $X \times Y$ is the set-theoretic Cartesian product and $XY$ is interpreted by recalling that $\mathbf{A}$ is multiplicative Abelian group. We require:
["Associativity"] For any three unifields $X,Y$ and $Z$ and all $x \in X,y \in Y$ and $z \in Z$, we have $(x \cdot_{(X,Y)} y) \cdot_{(Y,Z)} z = x \cdot_{(X,Y)} (y \cdot_{(Y,Z)} z).$ Suppressing the subscripts, this just says $(xy)z=x(yz).$
["Commutativity"] For any two unifields $X$ and $Y$ and all $x \in X,y \in Y$ we have $x \cdot_{(X,Y)} y = y \cdot_{(Y,X)} x.$ Suppressing the subscripts, this just says $xy=yx.$
["Distributivity"] For any two unifields $X$ and $Y$ and all $x,x' \in X,y \in Y$ we have $(x +_X x')\cdot_{(X,Y)} y = (x \cdot_{(X,Y)} y) +_X (x' \cdot_{(X,Y)} y).$ Suppressing the subscripts, this just says $(x+x')y = xy+x'y.$
In addition, lets assert that the multiplicative identity of $\mathbf{A}$ (which we'll denote $F \in \mathbf{A}$) has a multiplicative identity (which we'll denote $1 \in F$). We'll also assert that every $X \in \mathbf{A}$ is associated with a reciprocation (partial) function $*^{-1} : X \rightarrow X^{-1}$ that is defined on $X \setminus \{0_X\}$.
We demand:
- ["Inverses"] For any unifield $X$ and any $x \in X$ not equal to $0_X$, we have that $x^{-1} \cdot_{(X^{-1},X)} x = 1.$
It follows that $F$ is a field. We might call $\mathbf{A}$ "the system of unifields generated by $\{\mathcal{I},\mathcal{R}\}.$"
To make it all work, we'd also have to equip each $X \in \mathbf{A}$ with an order relation $\leq_X$ and demand that a few more axioms hold, thereby yielding a "system of complete ordered unifields."
