Question about maximal ideals? I'm reading Freligh and introduction to Abstract algebra and I'm getting confused. The set generated by $\langle x^2 + 1\rangle$ is a maximal ideal in $R[x]$. First, I don't understand it. $\langle x^2 + 1\rangle$ isn't the only irreducible element in $R[x]$, but yet it's the largest ideal. Why? 
I want a better intuition of maximal ideals, but I can't find it in textbooks, the internet, or anything. Someone help me out. 
Thanks
 A: (Below i am assuming that $R$ is the field of real numbers.) The point you are missing is that the set of ideals of a ring is in general a partially ordered set under inclusion. In the case of $R[x]$ you are right that each irreducible element generates a maximal ideal. But these maximal ideals are not comparable in terms of inclusions. Take for example the ideal generated by $x$ and that generated by $x+1$. In fact, there are as many maximal ideals in $R[x]$ as irreducible elements. It may help if you revisit the basic theory of partially ordered sets: maximal elements need not be unique, yet they are maximal. To conclude, in general there is no such thing as the maximal ideal but simply a maximal ideal. Rings that have a unique maximal ideal are called local rings, e.g. $R[x]/(x^2)$.
A: Froom the fact there is no greater (proper) ideal, which is the definition of a maximal ideal, you cannot  infer it is greater than all other ideals, because, given two ideals, one is not necessarily contained in the other.
Rings with only one maximal ideal are precisely known as local rings (or quasi-local rings, depending on terminology) and are quite important in algebraic geometry and number theory.
A: Hint $ $ For principal ideals: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides},\,$ i.e.  $\,(a)\supseteq (b)\iff a\mid b.\,$ Thus in a PID
$$\begin{align} (p) \text{ is maximal}\iff &
\ (p)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (a)\\
\iff&\ \ \,p\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ a\\
\iff&\,\ \ p\ \text{ is irreducible}\\
\end{align}\qquad$$
A: An ideal I in R , any commutative integral domain is maximal if and only R/I is a field. The proof is not hard to establish once you have  the correspondence theorem. Now in your case, $x^{2}+1$ is irreducible in $\mathbf{R}[x]$. So it is maximal. If I is an indeal containing ($x^{2}+1$), then I=$(f(x))$ (since $\mathbf{R}[x]$ is PID) and $f(X)$ divides $x^{2}+1$. So $f(x)=1$ or $f(x)=x^{2}+1$. Now, the definition of maximal I does not mean that any proper ideal must be contained in I, but that I is not contained is any other proper ideal, other than itsel.
