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So, I've been reading this book and I've come across two sentences that I find a little confusing.

On pg. 109: The polynomial ring $R[t]$ is generated by the variable $t$ over $R$, and $t$ is transcendental over $R$.

Context: $R[t]$ is the polynomial ring, and, for a fixed $x \in R$, $R[x] = \{ f(x) : f \in R[t] \}$. $x$ is transcendental if $f \mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.

Problem: $t$ doesn't seem to be an element of $R$.

On pg. 117: Let $F$ be a field and $\sigma : F[t] \rightarrow F[t]$ is an automorphism of the polynomial ring such that $\sigma$ restricts to the identity on $F$.

Problem: This seems to act as if $F \subset F[t]$?

Thanks the help.

Roy D.
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    There's a natural embedding $F \hookrightarrow F[t]$ by identifying field elements with constant polynomials. –  Jul 01 '14 at 05:13
  • That would've been my guess, but just wanted to make sure. Thanks! – Roy D. Jul 01 '14 at 05:14
  • Is there something similar for the first question about transcendentals? – Roy D. Jul 01 '14 at 05:15
  • I don't understand the first question. $t$ isn't an element of $R$. It's not supposed to be. Where is the problem? – Gerry Myerson Jul 01 '14 at 05:31
  • The property of something being transcendental applies to elements of $R$, right? – Roy D. Jul 01 '14 at 05:32
  • @RoyD. If $A,B$ are rings and $A$ is a subring of $B$, then it makes sense to ask for an element $b\in B$ whether it is transcendental over $A$. Which is equivalent to the map $A[X]\to B$ induced by $X\mapsto b$ (and the iidentity/inclusion on $A$) is injective. Here $A=R$, $B=R[t]$ and of course $R[X]\to R[t]$ is injective (it doesn't matter what you name your indeterminates) – Hagen von Eitzen Jul 01 '14 at 06:32
  • Sorry, what does your notation $A[X]$ here mean? – Roy D. Jul 01 '14 at 07:08
  • $A[X]$ generally means polynomials in the indeterminate $X$, with coefficients coming from $R$. But Hagen's point is that when we say, for example, that $\pi$ is transcendental, what we really mean is that $\pi$ is transcendental over the rationals (and, of course, $\pi$ isn't in the rationals). It is not true that $\pi$ is transcendental over the reals. – Gerry Myerson Jul 01 '14 at 13:16
  • So, if I'm understanding correctly, this is what's happening: We have a ring $S$ and a subring $R$. We consider $R[t]$, the set of polynomials with coefficients in $R$. However, we can note that this can also be seen as $(R[t])[t]$, the set of polynomials with coefficients that are polynomials in $t$. (These are the same.) – Roy D. Jul 01 '14 at 14:59
  • Thus, we can think of $t \in R[t]$ as an element of the ring generating the polynomials? – Roy D. Jul 01 '14 at 15:01
  • An element of which ring? $t$ is not an element of $R$. – Gerry Myerson Jul 02 '14 at 12:31
  • Sorry, I meant to say the ring of polynomials generated by $R$. I'm interpreting your previous point as: $\pi \in \mathbb{R}, \pi \notin \mathbb{Q}, \mathbb{Q} \subset \mathbb{R}$. So, $t \in R[t]$, and I'm asking if we're thinking of $R \subset R[t]$. – Roy D. Jul 02 '14 at 17:26
  • Yes, we can think of $R$ as being included in $R[t]$. – Gerry Myerson Jul 03 '14 at 12:55

1 Answers1

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From the comments above.

For the first question:

There is a problem with the definition you have of a transcendental element. You say that

[F]or a fixed $x\in R$, $R[x]=\{f(x):f\in R[t]\}$ [and] $x$ is transcendental if $f\mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.

I take it that $R[x]=\{f(x):f\in R[t]\}$ is meant to be a definition of $R[x]$, in which case it is quite restrictive. What should be said is that if $R$ and $S$ are rings with $R \subset S$, then for any fixed $x \in S$ we define $R[x] = \{ f(x) : f \in R[t] \}$. Now, $x \in S$ is said to be transcendental if $f \mapsto f(x)$ is an isomorphism from $R[t]$ to $R[x]$.

The earlier definition defined transcendence (over $R$) only on elements of $R$, which is inappropriate because no element of $R$ is transcendental over $R$.

For the second question:

There is a natural inclusion of $F$ inside $F[t]$, given by identifying $a \in F$ with the constant polynomial $at^0$. In this way we can view $F$ as a subset of $F[t]$.