F[x]/(p(x)) when p(x) irreducible

We have seen $F[x]/(x-a)\cong F$, notice $x-a$ is an irreducible polynomial since it has degree one. In this section, we will show $F[x]/(p(x))$ is always a field contains $F$ if $p(x)$ is irreducible.

Examples

$\mathbb{R}[x]/(x^2+1)$

Let’s consider first $\mathbb{R}[x]/(x^2+1)$, notice $\mathrm{deg}(x^2+1)=2$, so by description of unique representatives in $F[x]/(p(x))$, set theoretically we can write:

$$\mathbb{R}[x]/(x^2+1)=\{a+bx|a,b\in \mathbb{R}\}.$$

Now to determine the ring structure, we need to check the addition and multiplication.

1. Addition: $$(a_1+b_{1}x)+(a_2+b_{2}x)=(a_1+a_2)+(b_1+b_2)x.$$
2. Multiplication: $$(a_1+b_{1}x)(a_2+b_{2}x)=(a_1{a}_2-b_1{b}_2)+(a_1{b}_2+a_2{b}_1)x.$$

In the multiplication statement, we have used $x^2=-1$ to simplify:

$$a_1{a}_2+b_1{b}_2{x}^2=a_1{a}_2+b_1{b}_2(-1)$$

These rules reminds us of complex numbers. In fact, we can establish the following isomorphism:

$$\mathbb{R}[x]/(x^2+1)\cong \mathbb{C}$$

where $a+bx$ corresponds to $a+bi$.

$\mathbb{Q}[x]/(x^2-2)$

In this case, we have $\mathbb{Q}[x]/(x^2-2)$, and we can write:

$$\mathbb{Q}[x]/(x^2-2)=\{a+bx|a,b\in \mathbb{Q}\}.$$

We can also check the addition and multiplication:

1. Addition: $$(a_1+b_{1}x)+(a_2+b_{2}x)=(a_1+a_2)+(b_1+b_2)x.$$
2. Multiplication: $$(a_1+b_{1}x)(a_2+b_{2}x)=(a_1{a}_2+2b_1{b}_2)+(a_1{b}_2+a_2{b}_1)x.$$

Notice that $x^2=2$, so we have used $x^2=2$ to simplify the multiplication.

This ring is a field as well, we have:

$$\mathbb{Q}[x]/(x^2-2)\cong \mathbb{Q}(\sqrt{2})$$

where $a+bx$ corresponds to $a+b\sqrt{2}$.

$\mathbb{Q}[x]/(x^2)$

In this case, we have $\mathbb{Q}[x]/(x^2)$, and we can write:

$$\mathbb{Q}[x]/(x^2)=\{a+bx|a,b\in \mathbb{Q}\}.$$

We can also check the addition and multiplication:

1. Addition:

$$(a_1+b_{1}x)+(a_2+b_{2}x)=(a_1+a_2)+(b_1+b_2)x.$$

2. Multiplication:

$$(a_1+b_{1}x)(a_2+b_{2}x)=a_1{a}_2+(a_1{b}_2+a_2{b}_1)x.$$

Notice that $x^2=0$, so we have used $x^2=0$ to simplify the multiplication.

This ring is not a field, it even fails to be an integral domain. In fact, we have $[x]\cdot [x]=[x^2]=[0]$, so $[x]$ is a non-trivial zero divisor.

Main Theorem

From the previous three examples, we get a feeling when $p(x)$ is irreducible, $F[x]/(p(x))$ is a field, and when $p(x)$ is reducible, $F[x]/(p(x))$ is not even an integral domain.

We will show this is true in general.

Theorem

Let $F$ be a field and $p(x)\in F[x]$ be a non-constant polynomial. Then TFAE:

1. $p(x)$ is irreducible.
2. $F[x]/(p(x))$ is a field.
3. $F[x]/(p(x))$ is an integral domain.

Proof of the theorem

We will show 1. $⟹$ 2. $⟹$ 3. $⟹$ 1.

First we show 1. $⟹$ 2. Suppose $p(x)$ is irreducible, we want to show $F[x]/(p(x))$ is a field. It is sufficient to show every non-zero element $[f(x)]\ne [0_F]$ has a multiplicative inverse.

We can consider $\gcd (f(x),p(x))$. Note $[f(x)]\ne [0_F]$ means $f(x)$ is not divisible by $p(x)$, and $p(x)$ is irreducible, so $\gcd (f(x),p(x))=1$. By Bezout’s theorem, there are $u(x),v(x)\in F[x]$ such that:

$$f(x)u(x)+p(x)v(x)=1.$$

Taking congruence classes, we have:

$$[f(x)][u(x)]+[p(x)][v(x)]=[1]\Rightarrow [f(x)][u(x)]=[1].$$

Thus $[u(x)]$ is the multiplicative inverse of $[f(x)]$, so $F[x]/(p(x))$ is a field.

Second, we show 2. $⟹$ 3. Suppose $F[x]/(p(x))$ is a field, every field is an integral domain, so $F[x]/(p(x))$ is an integral domain.

Finally, we show 3. $⟹$ 1. We can do a contrapositive argument. Suppose $p(x)$ is reducible, then $p(x)=q(x)r(x)$ for some non-constant polynomials $q(x),r(x)\in F[x]$. Then we have:

$$[q(x)][r(x)]\equiv [p(x)]\Rightarrow [q(x)][r(x)]\equiv [0]mod(p(x)).$$

Since $0<\mathrm{deg}(q(x))<\mathrm{deg}(p(x))$ and $0<\mathrm{deg}(r(x))<\mathrm{deg}(p(x))$, we have $[q(x)]\ne [0]$ and $[r(x)]\ne [0]$. Thus $[q(x)]$ and $[r(x)]$ are non-zero divisors, so $F[x]/(p(x))$ is not an integral domain.$■$