MATH 415 Lecture 18

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Embedded Quotient Fields

Continuing from previous lecture, after proving that $F$ is a field, we want to show that integral domain $D$ is isomorphic to the field $F$.

Construct $i:D\to F$ such that $i(a)=\left[\left(a,1\right)\right]$. We claim $i$ is an isomorphism between $D$ and $i(D)$. Indeed, $i$ is

1. homomorphism
   • $i(a)+i(b)=[(a+b,1)]=i(a+b)$
   • $i(a)\cdot i(b)=[(a\cdot b,1)]=i(a\cdot b)$
2. one-to-one (if $i(a)=i(b)$, then $(a,1)\sim (b,1)$ implies $a=b$)
3. onto $i(D)$ (tautology by definition of image; trivial)

Thus we justify the following theorem:

Theorem 21.5

Any integral domain $D$ can be enlarged to (or embedded in) a field $F$ such that every element of $F$ can be expressed as a quotient of two elemens of $D$.

Such $F$ is called a field of quotients of $D$.

Note: This theorem explains existence. There is another theorem in the book that explains uniqueness.

Rings of Polynomials

We begin with a ring $R$.

Let $x$ be an indeterminate or unknown variable.

We denote the ring of polynomials in one indeterminate with coefficients in $R$ as $R[x]$

In general, a polynomial is an infinite formal sum of form:

$p(x)=\sum _{i=0}^{\infty }a_{i}\,x^{i}=a_{0}+a_{1}\,x+\dots +a_{n}\,x^{n}+\dots$

Where $a_{i}\in R$ and $a_{i}=0$ for all but a finite number of values of $i$.

If $a_{n}\neq 0$, then we call $n$ the degree of $p(x)$.

• A polynomial of the form $p(x)=a_{0}$ (degree 0) is called a constant polynomial
• $p(x)=a_{0}+a_{1}\,x$ (degree 1) is called linear
• $p(x)=a_{0}+a_{1}\,x+a_{2}\,x^{2}$ (degree 2) is called quadratic
• etc.

We define addition and multiplication as follows:

Let $f,g\in R[x]$ with

${\begin{aligned}f(x)&=\sum _{i=0}^{\infty }a_{i}\,x^{i}\\g(x)&=\sum _{i=0}^{\infty }b_{i}\,x^{i}\end{aligned}}$

Then

${\begin{aligned}(f+g)(x)&=\sum _{i=0}^{\infty }(a_{i}+b_{i})\,x^{i}(f\cdot g)(x)&=\sum _{i=0}^{\infty }\left(\sum _{j=0}^{i}a_{j}\cdot b_{i-j}\right)\,x^{i}\end{aligned}}$

Note that $a\,b\neq b\,a$ in general.

But if $R$ is commutative, then $R[x]$ is also commutative.

Similarly, $R[x]$ has unity if and only if $R$ has unity, and unity in $R[x]$ will be constant polynomial $p(x)=1$.

$R$ is isomorphic to subring $R_{1}\subset R[x]$, where $R_{1}$ is the set of constant polynomials.

Example

$\mathbb {Z} _{2}$ is a (finite) field, but $\mathbb {Z} _{2}[x]$ is an infinite field.

$(x+1)^{2}=x^{2}+2x+1$, but $2x\equiv 0{\pmod {2}}$, so $(x+1)^{2}=x^{2}+1$ in $\mathbb {Z} _{2}[x]$.

Multiple Indetermintes

We write $R[x,y]$ to represent all polynomials in two indeterminates with coefficients in $R$ and are expressed as follows:

$p(x,y)=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }a_{ij}\,x^{i}\,y^{j}$

In this case, $x\,y=y\,x$.

Evaluation Homomorphisms

Let $E$ and $F$ be fields with $F\leq E$ ($F$ is a subfield of $E$).

Theorem 22.4

If $F\leq E$, $\alpha \in E$, then the map $\phi _{\alpha }:F[x]\to E$ defined by

$\phi _{\alpha }(a_{0}+a_{1}\,x+\dots +a_{n}\,x^{n})=a_{0}+a_{1}\,\alpha +\dots +a_{n}\,\alpha ^{n}\,\!$

for $a_{0}+a_{1}\,x+\dots +a_{n}\,x^{n}\in F[x]$ is a homomorphism of $F[x]$ into $E$. Furthermore, $\phi _{\alpha }(x)=\alpha$, and $\phi _{\alpha }$ maps $F$ isomorphically by the identity map; that is, $\phi _{\alpha }(a)=a$ for $a\in F$.

The homomorphism $\phi _{\alpha }$ is an evaulation at $\alpha$.

To take a page from Haskell, an evaluation homomorphism has the following type signature:

eval :: (E a, E b) => (a -> b) -> a -> b

E {$E } ; phialpha [ below of = E] {$\phi_\alpha(x) = \alpha \in \phi_\alpha(F[x])$} ; Fx [ left of = phialpha ] {$p \in F[x]$} ; F [ below of = Fx ] {$a \in F$} ; phia [ below of = phialpha] {$\phi_\alpha(a) = a \in F$} ;

\draw [->] (Fx) (phialpha) {$\phi_\alpha$} ; \draw [--] (Fx) (F) ; \draw [->] (F) (phia) {identity map} ; \draw [--] (E) (phialpha) ; \draw [--] (phialpha) (phia) ;