Definition:Transcendental Element of Ring Extension

From ProofWiki
Jump to navigation Jump to search

Definition

Let $\struct {R, +, \circ}$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.

Let $\struct {D, +, \circ}$ be an integral subdomain of $R$.

Let $x \in R$.


Then $x$ is transcendental over $D$ if and only if:

$\displaystyle \forall n \in \Z_{\ge 0}: \sum_{k \mathop = 0}^n a_k \circ x^k = 0_R \implies \forall k: 0 \le k \le n: a_k = 0_R$


That is, $x$ is transcendental over $D$ if and only if the only way to express $0_R$ as a polynomial in $x$ over $D$ is by the null polynomial.


Notation

For such an $x$ transcendental over $D$, it is conventional to use the letter $X$.

Thus a ring of polynomials over $D$ in such a transcendental is therefore usually denoted $D \sqrt X$.


Also see

If $x \in R$ is not transcendental over $D$ then it is algebraic over $D$.

A polynomial in $X$ over $D$ for transcendental $X$ is an instance of a polynomial form.


Historical Note

The term transcendental, in the sense of meaning non-algebraic, was introduced by Gottfried Wilhelm von Leibniz.


Sources