First of all: How do I construct quotient rings?
Given $R$ – сommutative ring with 1 and principal ideal
$$\left<x\right> = \{f(x)\cdot x\ |f(x) \in R[x] \}$$
Under what conditions on R it is a prime ideal?
Under what conditions on R it is a maximal ideal?
Prime ideal:
From the definition: Ideal $I$ is prime $\iff$ quotient ring $R/I$ has no zero divisors.
So, as I understood, we are talking about left principal ideals. Recall what is a zero divisor: $a$ is a zero divisor $\iff a \cdot b = 0$ and $b \cdot a = 0$, when $b \neq 0$ and $a \neq 0$. (right?)
To solve my problem I want to have no zero divisors in my quotient ring. Here I got stuck because I have problems with understanding how quotient rings look like $\Rightarrow$ I don't understand how elements in quotient ring look like.
Assume I know the elements in my quotient ring. The next thing I want to do is to pick two non-zero random elements from that quotient ring and see when their product is equal to 0?
It is even harder for me to deal with maximal ideal, because I don't know how to show there is no such ideal $m < I$.
Best Answer
To understand what a quotient ring looks like, I find it useful to think about it as "adding a relation".
Imagine you have some ring such as $\mathbb Z[i]$. This is a commutative ring with 1, and elements look like $a+bi$ where $a,b\in\mathbb Z$, and $i^2 = -1$. We can write this as: $$\mathbb Z[i] = \langle a+b i \mid a,b\in\mathbb Z, i^2 = -1\rangle$$ Now, imagine we want to quotient by some principle ideal (such as $(2+i)$), then we have that: $$\mathbb Z[i]/(2+i) = \langle a+bi\mid a,b\in\mathbb Z,i^2 = -1, 2+i = 0\rangle$$ Essentially, whatever you're quotienting by is now $0$ in the quotient ring. So, we can think about this as a new relation in our ring, where if we quotient by $(a)$ we have that $a = 0$ in the quotient.
In the specific example I've written, as $2+i = 0$, we have that $i = -2$. So, we have that $a+bi = a+b(-2) = a-2b$, so $\mathbb Z[i]/(2+i) \cong\mathbb Z$. This could be verified with an isomorphism theorem, as the homomorphism $\phi:\mathbb Z[i]\to\mathbb Z$ with $\phi(a+bi) = a-2b$ has kernel $(2+i)$, so by the first isomorphism theorem you have that: $$\mathbb Z[i]/\ker\phi\cong\text{im }\phi\cong\mathbb Z$$
For your particular example, we have that: $$R[x]/\langle x\rangle$$ is most clearly attacked with isomorphism theorems. Find some homomorphism with kernel $\langle x\rangle$ (such as $\phi(1) = 1$, and $\phi(x) = 0$). Then, we have that $R[x]/\langle x\rangle\cong\text{im }(\phi)$. Here, the image of $\phi$ is just $R$, as we're essentially "evaluating the polynomial at $0$".
The conclusion to this argument is below. I encourage to try to finish it yourself first (as the difficult part for this was establishing that $R[x]/\langle x\rangle\cong R$).