# Question about maximal ideals in a Polynomial ring

I'm reading Freiligh and he has an example in a book, here it is:

Let $F = R$ and let $f(x) = x^2 + 1$. Which is well known to have no zeros in $R$ and thus is irreducible over $R$ by a theorem previously stated.

Then $<x^2+1>$ is a maximal ideal in $R[x]$. So $R[x]/<x^2 + 1>$ is a field.

This quotient ring is somehow isomorphic with the Complex numbers. I guess I'm having a hard time understanding why that is the case.

Being as simple as possible, can someone explain to me why that is the case. I don't know really advanced math.

• For a proof see here, and here. – Dietrich Burde May 12 '15 at 17:39
• It's not so much a proof I would like to see as it's an understanding about what's happening in a more intuitive level. IF thats possible. I hear that proofs in abstract algebra are not really intuitive based. I guess I'm wondering why this quotient ring is bigger than the Real numbers, shouldnt it be smaller? – user121615 May 12 '15 at 17:47
• It is not smaller, because already $\mathbb{R}[x]/(x)\simeq \mathbb{R}$. Such insights (to understand what a quotient really is) will help you more than fuzzy ideas on the intuitive level. – Dietrich Burde May 12 '15 at 18:07
• Okay, how can I understand what a quotient really is. From my text book, this construction should be the quotient ring, which is a set defined with the operations of Coset multiplication and coset addition. An example would be the Integers. Let's take $Z \3Z$ This defines the cosets of all multiples of 3. So one set will have {...0,3,6...} another {...1,4,7..} and the last {...2,5,8...}. This is the general Idea right? – user121615 May 12 '15 at 18:17
• Does this construction work differently when we are talking about polynomials? – user121615 May 12 '15 at 18:18

$\mathbb{R}[x] \rightarrow \mathbb{C}$, $f(x) \rightarrow f(i)$.
• But why is the ring constructed by $R[x]\<x^2 + 1>$ isomorphic to the complex numbers? Shouldn't the ring be smaller? I don't get it – user121615 May 12 '15 at 17:55