Number of homomorphisms from $Q[x]/\langle f(x)\rangle$ to $\mathbb{C}$

Question

How many homomorphisms are there from $Q[x]/\langle f(x)\rangle$ to $\mathbb{C}$ that take $1$ to $1$ for an arbitrary polynomial $f(x)\in Q[x]$?

I took some examples and tried to figure out the connection between the number of homomorphisms, $n$ and $f(x)$. For say $x^2+1$ we have $n=2$ i.e., $f(x)+(x^2+1)\mapsto f(i)$ and $f(x)+(x^2+1)\mapsto f(-i)$. Similarly for $x^3-1$ we have $f(x)+(x^3-1)\mapsto f(1)$, $f(x)+(x^3-1)\mapsto f(\omega)$ and $f(x)+(x^3-1)\mapsto f(\omega^2)$. I suspect that $n$ is the number of distinct roots of $f(x)$ in $\mathbb{C}$. Can anyone please help me through?

I found the question randomly on the internet. If this is not a valid question I will be very happy to get corrected.

Answer 1

There is a bijection $$\{\alpha\in\mathbb C:f(\alpha)=0\}\simeq\hom(\mathbb Q[x]/f(x),\mathbb C).$$ Given a $\alpha\in\mathbb C$, the corresponding homomorphism $\varphi_\alpha\colon\mathbb Q[x]/f(x)\to\mathbb C$ sends $p(x)\pmod{f(x)}$ to $p(\alpha)$. Conversely, a homomorphism $\varphi$ corresponds to $\varphi(x\pmod{f(x)})$.

Please check that these maps are well-defined, and that they are inverses of each other.

Answer 2

Elaborating the answer of Kenta- Let $\varphi:\mathbb{Q}[x]/{\langle f(x)\rangle}\to \mathbb{C}$ is a ring homomorphism taking $1$ to $1$ and $\varphi(\overline{x})=\alpha$. Then for any $a\in \mathbb{Q}$ we will have $\varphi(a)=a$. We will also have $\varphi({\overline{x}}^i)=\alpha ^i$ for any $i\geq 0$. Now for any $p(x)=\sum_{i=0}^{n}a_ix^i \in \mathbb{Q}[x]-$ \begin{align*} \varphi\left(\overline{p(x)}\right) &=\varphi\left(\sum_{i=0}^{n}a_ix^i(mod\ f(x))\right)\\ &=\sum_{i=0}^{n}\varphi\left(a_ix^i(mod\ f(x))\right)\\ &=\sum_{i=0}^{n}\varphi (a_i)\cdot \varphi\left(x^i(mod\ f(x))\right)\\ &=\sum_{i=0}^{n} a_i\cdot{\alpha}^i\\ &=p(\alpha). \end{align*} Thus, $\varphi\left(f(x)(mod\ f(x)\right)=0$ implies $\varphi\left(f(x)(mod\ f(x)\right)=f(\alpha)=0$. Hence, $\alpha$ is a root of the polynomial $f(x)$ and the homomorphism, $\varphi$ corresponds to some homomorphism of the first type.