# Prove $\{a(c):a(x) \in F[x]\}$ is an integral domain isomorphic to $F[x]$

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, $$S_c\cong F[x]$$.

• $F$ might be a field, but $F[x]$ never is.
– user239203
Apr 4 '20 at 20:32
• 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. Apr 4 '20 at 20:51

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.