Nonzero nilpotent elements in $\Bbb C\otimes\Bbb Q[x]/(f)$?

Question

I have to find if this affirmation is true:

Let $f\in \mathbb{Q}[x]$ such that $\gcd(f,f')=1$, then in $\mathbb{C}\otimes_{\mathbb{Q}} \mathbb{Q}[x]/(f)$ there are no nonzero nilpotent elements.

My idea is: $\gcd(f,f')=1\implies $ f has not multiply irreducible factors in $\mathbb{Q}[x]$, but I don't know how to prove that $\mathbb{C}\otimes_{\mathbb{Q}} \mathbb{Q}[x]/(f)$ is isomorphic with $\mathbb{C}[x]/(f)$ and then that f has not multiply irreducible factors even in $\mathbb{C}[x]$. Can anyone help me? Thank you in advance.

Answer 1

$\mathbb{C}\otimes_{\mathbb{Q}} \mathbb{Q}[x] \cong \mathbb{C}[x]$ is due to the following lemma:

Let $R, S $ be commutative rings.

If the $R$-module $M$ is free with basis $\lbrace m_1, \ldots m_k \rbrace$ and there is a homomorphism of commutative rings $\phi : R \to S$ , then $M \otimes_R S$ is a free $S$-module with basis $$\lbrace m_1 \otimes 1, \ldots m_k \otimes 1 \rbrace$$

This implies $$R[x] \otimes_R S \cong S[x]$$

See for example the notes of Keith Conrad about tensor product

Answer 2

You want to prove $\Bbb C\otimes_{\Bbb Q}\Bbb Q[x]/(f)$ is isomorphic to $\Bbb C[x]/(f)$. First, think of maps between the two, then prove they are mutually inverse. What are the most obvious maps to pick? Remember, $\otimes$ is supposed to function like multiplication, and you can multiply complex numbers with rationals.

Next on the agenda, factor $f$ and use the Chinese Remainder Theorem to decompose $\Bbb C[x]/(f)$.

The summands you get should each look like $\Bbb C[x]/(x-z)^e$, which is isomorphic to $\Bbb C[\varepsilon](\varepsilon^e)$, which has a nilpotent iff what? A direct sum of rings has a nilpotent iff one of its summands do; can you prove this simple fact?

Finally, show $\gcd(f,f')=1$ iff $f$, $f'$ share no roots iff $f$ is separable (no repeated roots).

There is a good motivation for involving tensor products with separability. The usual equivalence $L/K$ separable $\Leftrightarrow L\otimes_K\overline{K}$ reduced extends the notion of separability to infinite field extensions.