What is the meaning of $\Bbb{Q}[x]/f(x)$?

Question

I am very confused with the meaning of $\Bbb{Q}(x)/f(x)$. Does it mean the set of all polynomials modulo $f(x)$? If it does then how can we say that $\Bbb{R}[x]/(x^2+1)$ is isomorphic to set of complex numbers?

Answer 1

$\Bbb{Q}[x]/f(x)$ refers to the quotient ring between $\Bbb{Q}[x]$ and the ideal generated by $f(x)$. This quotient ring is important in studying field extensions of $\Bbb{Q}$ and in Galois theory.

Now, to prove $\Bbb{R}[x]/\langle x^2+1 \rangle$ is isomorphic to $\Bbb{C}$, we simply use the First Isomorphism Theorem.

Let's say we map every element in $\Bbb{R}[x]$ to $\Bbb{C}$ using the evaluation homomorphism $\phi_i$, which evaluates a polynomial at $x=i$ to convert it into a complex number. We need to find the kernel of $\phi_i$, where: $$\phi_i(f(x))=f(i)=0$$ Now, if $f(i)=0$, this means that the minimal polynomial of $i$ divides $f$. The minimal polynomial of $i$ is $x^2+1$, so $x^2+1 \mid f$ for all $f$ in the kernel of $\phi_i$. Therefore, the kernel is simply $\langle x^2+1 \rangle$ and by the First Isomorphism Theorem, we have: $$\Bbb{R}[x]/\langle x^2+1 \rangle \equiv \Bbb{C}$$

Answer 2

Yes that's what it means. Take the isomorphism $\phi: \mathbb{R}[x]/(x^2+1)\rightarrow \mathbb{C}$ by $\phi: (a+bx) \mapsto (a+bi)$. I'll leave it to you to show that it is an isomorphism.

Answer 3

You are indeed correct, this quotient ring simply means the set of polynomials over $\mathbb{Q}$ put into equivalence classes according to their remainders modulo $f(x)$. This is in general a ring but if $f(x)$ is irreducible then you get a field.

Another way to think of this is that you have decided to take the free variable $x$ and declare that it now satisfies $f(x) = 0$ (i.e. that $x$ is an abstract root of $f(x)$). Thinking in this way it is now intuitive why $\mathbb{R}[x]/(x^2+1)$ is isomorphic to $\mathbb{C}$...you have forced $x$ to behave like the complex number $i$ (or even $-i$, but this doesn't matter).