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This question originates from Pinter's Abstract Algebra, Chapter 27, Exercise G1.

Let $F$ be a field, and let $c$ be transcendental over $F$. Prove {$a(c):a(x) \in F[x]$} is an integral domain isomorphic to $F[x]$.

[Edited]

In general, if $A$ is an integral domain, then $A[x]$ is an integral domain.

Let $S_c =$ {$a(c):a(x) \in F[x]$}. $S_c$ is an integral domain iff $S_c$ is a commutative ring with unity that has no divisor of zero.

  1. $S_c$ is obviously commutative and closed under addition and multiplication by the properties inherited from $F[x]$, which is an integral domain. Note $F[x]$ is an integral domain for $F$ is a field.
  2. $a(x)=1\in F[x]\implies a(c)=1\in S_c$.
  3. Suppose $a_1(c)a_2(c) = 0$ for nonzero polynomials $a_1(x), a_2(x)\in F[x]$. Let $b(x)=a_1(x)a_2(x)$. Then $b(c) = 0$, contradicting that $c$ is transcendental. Hence $a_1(c)a_2(c) = 0$ implies either $a_1(x)$ or $a_2(x)$ is zero; that is, $S_c$ has no divisor of zero.

This completes the proof that $S_c$ is an integral domain.

Let $\sigma_c: F[x]\rightarrow S_c$ such that $\sigma_c(a(x)) = a(c)$. Note $\sigma_c$ is a ring homomorphism from $F[x]$ onto $S_c$. Furthermore, $a(c)= b(c)\implies a(x)= b(x)$, for otherwise suppose $a(x)\ne b(x)$, then $a(x)-b(x)\ne 0\implies a(c)-b(c) \ne 0$, as $c$ is transcendental, contradicting $a(c)=b(c)$. Hence $\sigma_c$ is bijective and therefore an isomorphism; that is, $S_c\cong F[x]$.

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    $\begingroup$ $F$ might be a field, but $F[x]$ never is. $\endgroup$
    – user239203
    Apr 4 '20 at 20:32
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    $\begingroup$ Shorter: The map $F[x]\to \{\,a(x):a(x)\in F[x]\,\}$ induced by $x\mapsto c$ is obviously an epimorphism of rings. As $c$ is transcendental, its kernel is trivial,hence it is an isomorphism. $\endgroup$ Apr 4 '20 at 20:51
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Your claim that that $\sigma_c$ is injective is rather unconvincing. Also, as suggested in the comments, you can skip the entire proof that $S_c$ is an integral domain; once you prove that $S_c\cong F[x]$ it follows that $S_c$ is an integral domain because $F[x]$ is. Overall your proof focuses on the wrong points; you gloss over key facts without any argument, and elaborate on completely irrelevant facts.

To prove that $S_c\cong F[x]$ indeed the most efficient approach is to note that the ring homomorphism $$\sigma_c:\ F[x]\ \longrightarrow\ S_c:\ a\ \longmapsto\ a(c),$$ is surjective by definition of $S_c$, and injective because $a(c)=0$ implies $a=0$ because $c$ is transcendental. Then $S_c$ is an integral domain because $F[x]$ is.

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  • $\begingroup$ Agree. I've fixed the injectivity argument in the OP anyway. Thank you for the answer. $\endgroup$
    – hchar
    Apr 5 '20 at 18:20

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