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The following is from Fundamentals of Mathematics, Volume 1 Foundations of Mathematics: The Real Number System and Algebra, Edited by H. Behnke, F. Bachmann, K. Fladt, W. Süss and H. Kunle. Bold text emphasis added.

[W]e have already seen that in general the function $x\to\sum_{i=0}^{n}a_{i}x^{i}$ does not uniquely determine the $a_{i}.$ But for calculation with such expressions as $\sum_{i=0}^{n}a_{i}x^{i}$ it would be very convenient to be able to assume that the coefficients $a_{i}$ are uniquely determined by the values of the expression. This will unquestionably be the case (for an element $x$ with certain properties) if in $R$ or in a suitable extension of $R$ we can find an element $x$ such that an equation $\sum_{i=0}^{n}a_{i}x^{i}=0$ always implies $a_{0}=a_{1}=\dots=a_{n}=0;$ for then we can recognize, as in [a previous section], that $\sum_{i=0}^{n}a_{i}x^{i}=\sum_{i=0}^{n}b_{i}x^{i}$ implies (comparison of coefficients) the equations $a_{i}=b_{i}\left(i=0,\dots,n\right).$ An element $x$ with this property will be called a transcendent over $R.$ If $R$ is the field of rational numbers, then in agreement with the definition in [the subsequent chapter and section on algebraic numbers], any transcendental number may be chosen as a transcendent over $R$ in the present sense. Since a transcendent $x$ cannot satisfy any algebraic equation $\sum_{i=0}^{n}a_{i}x^{i}=0$ with $a_{n}\ne0,$ it cannot be characterized (i.e., determined) by statements involving only $x,$ elements of $R,$ and equality, addition, and multiplication in $R.$ Thus the transcendents are also called indeterminates.${}^{10}$ But a name of this sort must not be allowed to conceal the fact that a transcendent must be a definite element (of an extension ring of $R$) and that the existence of such elements must in every case be proved. As an indeterminate over the field of rational numbers we may choose any transcendental number such as $\mathrm{e}$ or $\pi$.

${}^{10}$In §§2 and 3 the symbol $x$ will almost always denote an indeterminate; more precisely, $x$ is a variable for which only interterminates can be substituted. On the other hand, in §1 the variable $x$ provided it is not bound may be replaced by any of the elements of the ring.

§1 Entire Rational Functions; §2 Polynomials; §3 Use of Inteterminates as a Method of Proof.

This explains why, as discussed here: In Weyl's 'Classical Groups' is this a proper statement about a polynomial vanishing identically? I balked at Weyl's presentation. But when I posted that, I had not yet realized that the two sources conflict regarding the use of the term indeterminate. Weyl apparently uses it as a synonym for variable. The authors, G. Pickert and W. Rückert of the chapter on polynomials in BBFSK clearly mean something quite different. It is one thing to use terminology in a nonstandard way. It is different to do so while providing good justification for that usage.

Do other authors use the term indeterminate in the specific sense of transcendental presented above?

This is from Weyl's The Classical Groups Their Invariants and Representations:

A formal expression

$$f\left(x\right)=\sum_{i=0}^{n}\alpha_{i}x^{i}$$

involving the “indeterminate” (or variable) $x$, whose coefficients $\alpha_{i}$ are numbers in a field $k$, is called a $\left(k-\right)\text{polynomial}$ of formal degree $n$.

Steven Thomas Hatton
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  • A possible connection between the two usages: if $\alpha\in \mathbb{C}$ is transcendental then $\mathbb{Q}(\alpha)$ (a subfield of $\mathbb{C}$) is isomorphic to $\mathbb{Q}(x)$ (the field of formal rational functions with coefficients in $\mathbb{Q}$) in which $x$ plays the role of a formal variable. – Daniel Schepler Jan 18 '19 at 00:36
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    A variable is something that takes different values; an indeterminate is not. One possible source of confusion lies in confusing polynomials with polynomial functions (the former have indeterminates, the latter have variables, but they “look” the same). Neither is the same as “transcendental”, which is a term of art to refer to certain real or complex numbers (that are not roots of any polynomial with rational coefficients). – Arturo Magidin Jan 18 '19 at 00:43
  • @ArturoMagidin Are you arguing that both sources are incorrect? I added a direct quote from Weyl's book to show why I conclude he is using indeterminate as a synonym for variable. How else would you explain the quoted texts? – Steven Thomas Hatton Jan 18 '19 at 01:12
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    I have never seen transcendental numbers called “indeterminate”, as your quote states, before this particular quote. Strikes me as perverse, given the already rife confusion between “indeterminates” and “variables” that exists and persists. But note that you are talking about terminology. Terminology is neither correct nor incorrect; it can be common, uncommon, good, bad (prone to confusion), standard, non-standard, etc. But saying that a term is “correct” or “incorrect” is like saying that someone’s name is correct or incorrect. – Arturo Magidin Jan 18 '19 at 02:19
  • @ArturoMagidin It is likely the above quoted usage has precedence, and speaks to the original definitions. – Steven Thomas Hatton Jan 18 '19 at 03:04
  • @MorganRodgers I suggest you consider the source before jumping to conclusions https://mitpress.mit.edu/contributors/h-behnke The book failed to achieve their goal, not because they are inaccurate or wanting in content. They failed because they were written to the German mathematician's standard of "easy". – Steven Thomas Hatton Jan 18 '19 at 09:12
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    @StevenHatton: I don’t know what you mean by “precedence”. If you mean, it was prior to others, then historical meanings don’t carry a lot of weight in determining common usage (or even good or bad usage). We don’t use “fluxions” anymore, for example. In group theory (my area), “metabelian” used to mean what is now called “nilpotent of class 2”, while “metabelian” has a different meaning (extension of abelian by abelian). We don’t defer to the older terminology, because the newer one fits better inside a more general theory. – Arturo Magidin Jan 18 '19 at 12:35
  • Final comments: I note that you are conflating “transcendent” with “transcendental number”, which your quote differentiates. And that your comment about the reason for their “failure” is what in usenet parlance was termed IYHO, not a statement of fact as you present it. – Arturo Magidin Jan 18 '19 at 12:39
  • @ArturoMagidin that is spelt USENET. – Steven Thomas Hatton Jan 18 '19 at 13:12

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We have the following theorem:

Let $ R $ and $ S $ be (commutative) rings, $ \eta $ a homomorphism of $ R $ into $ S $, $ u $ an element of $ S $. Let $ R[x] $ be the ring of polynomials over $ R $ in the indeterminate $ x $. Then $ \eta $ has one and only one extension to a homorphism $ \eta_u $ of $ R[x] $ into $ S $ mapping $ x $ into $ u $. [Basic Algebra, Jacobson]

Sketch of proof:

Map the coefficients of $ f(x)\in R[x] $ to the image of $ \eta $ and map $ x $ to $ u $. It is plain to check that this is a ring homomorphism. For the uniqueness part, it suffices to note that $ R[x] $ is generated by $ R $ and $ x $, then we are done.

From here we have the corollary:

$ R[u]\cong R[x]/I $ where $ x $ is an indeterminate and $ I $ is an ideal in $ R[x] $ such that $ I\cap R=0 $.

Therefore, $$ u\ \text{is transcendental}\longleftrightarrow I=0\longleftrightarrow R[u]\cong R[x] .$$ Note that when I write $ R[x]\cong R[u] $ above, I mean the homomorphism is identity on $ R $.

Bach
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  • How does that address my question 'Do other authors use the term indeterminate in the specific sense of transcendental presented above?' – Steven Thomas Hatton Jan 18 '19 at 02:57
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    @StevenHatton As you see, in that case, they are isomorphic as rings. If you don't like one of the terms, substitute one for another as long as you are sure only ring structures are involved. The major difference between transcendental and indeterminate is that we can have an indeterminate from nowhere, however, we can only say an element is transcendental if there does exist such an element in the overring, which has already been specified. Once we have such a transcendental element, then we are safe to replace one with the interminate. – Bach Jan 18 '19 at 03:07
  • I've added the remainder of the quoted paragraph and a footnote which appears to address what you are suggesting. Those authors appear to be saying that the only allowable indeterminates are transcendent elements of an extension field of $R$. They are also saying that an indeterminate is a specific element, and not a symbol which may be replaced by a value. – Steven Thomas Hatton Jan 18 '19 at 04:35
  • @StevenHatton: “They are also saying that an indeterminate is a specific element, and not a symbol which may be replaced by a value.” Isn’t that what I said? “A variable is something that takes different values; an indeterminate is not.” – Arturo Magidin Jan 18 '19 at 12:40
  • @ArturoMagidin In regarding the "variability" of indeterminates I was referring to Weyl, who appears to be treating the as synonymous with variables. The other source clearly identifies trancendents with indeterminates. – Steven Thomas Hatton Jan 18 '19 at 13:01