# Understanding principal ideals and quotient rings

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 $m < I$.

• In a somewhat "loose" sense, what "modding by an ideal $I$" does, is to effectively set everything in $I$ to zero. What polynomials in $R[x]$ survive, when all the "$x$" parts are all $0$? – David Wheeler Dec 11 '16 at 23:35
• Ones that have a constant? – False Promise Dec 11 '16 at 23:36
• Quotient rings are just a generalisation of $\mathbf Z/n\mathbf Z$. – Bernard Dec 11 '16 at 23:49
• You can consider $R/I=\{r+I:r\in R\}.$ Note that (1) Elements in $R/I$ are sets (Or equivalent class). (2) Each element in $R/I$ can be written as $r+I$ but not unique. (3) $r_1+I=r_2+I$ if and only if $r_1-r_2\in I.$ (4) Operations: $(r_1+I)+(r_2+I)=(r_1+r_2)+I$ and $(r_1+I)(r_2+I)=(r_1r_2)+I$ – Worawit Tepsan Dec 12 '16 at 0:18

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 $R[x]/\langle x\rangle\cong R$).

So, $\langle x\rangle$ is prime if $R$ is a domain, and $\langle x\rangle$ is maximal if $R$ is a field.

• For example $\varphi: R[x] \mapsto c$. Any element of R[x] can be expressed in form of $p = p_1x^n + \dotso + p_n$ and $c = p_n$? – False Promise Dec 12 '16 at 0:30
• With that notation, you generally say something like "Let $p(x)\in R[x]$. Let $\phi: p(x)\mapsto p(0) = p_n\in R$". That homomorphism does work though, and using First Isomorphism Theorem with it will let you find what $R[x]/\ker\phi$ is isomorphic to (which should give you your answer). – Mark Dec 12 '16 at 0:34