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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?

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  • $\begingroup$ It means the quotient ring of $\mathbb{Q}(x)$ with the ideal generated by $f(x)$. The second statement is not correct for $\mathbb{R}(x)$, the field of fractions of the polynomial ring, but correct for $\mathbb{R}[x]$, the polynomial ring on $\mathbb{R}$. You can check the map sending $ax+b + (x^2+1)$ to $ai+b$ is an isomorphism from the given quotient ring to $\mathbb{C}$ or simply verify that the kernel of the map from $\mathbb{R}[x]$ to $\mathbb{C}$ sending $x$ to $i$ is $(x^2+1)$ and then use the first isomorphism theorem. $\endgroup$ – Levent Jun 23 '16 at 17:58
  • $\begingroup$ Ideal generated by $f(x)$ will have all the elements ${a.f(x)|a \epsilon \Bbb{R}}$.But, I cannot understand properly what a quotient ring means in terms of the elements it has? For e.g does it contain only the polynomials modulo $f(x)$? $\endgroup$ – Mayank Jun 23 '16 at 18:02
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    $\begingroup$ Well the elements of the quotient ring are the equivalence classes but you can simply think as the polynomials in $\mathbb{R}[x]$ modulo $f$. Since $\mathbb{R}[x]$ is an Euclidean domain (i.e. you can use the Euclidean Algorithm to divide one polynomial to another to get a unique quotient and a remainder), you can take a polynomial with degree less than the degree of $f$ to get a representative set for each element of $\mathbb{R}[x]/(f)$ . $\endgroup$ – Levent Jun 23 '16 at 18:05
  • $\begingroup$ The ideal generated by $f$ does not consists of the elements of the form $a\cdot f$, $a\in \mathbb{R}$ but $a\cdot f$ where $a\in \mathbb{R}[x]$. For example when you take $f=x^2+1$, the ideal contains $x\cdot f = x^3+x$. $\endgroup$ – Levent Jun 23 '16 at 18:09
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    $\begingroup$ See this recent thread for more discussion. $\endgroup$ – Jyrki Lahtonen Jun 23 '16 at 18:20
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$\Bbb{Q}[x]/f(x)$ 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: $$\Bbb{R}[x]/\langle x^2+1 \rangle \equiv \Bbb{C}$$

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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.

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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).

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