# Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$. [closed]

Let $K$ be a field and $f(x)\in K[x]$. Prove that $K[x]/(f(x))$ is a field if and only if $f(x)$ is irreducible in $K[x]$.

How to prove? I really have no idea... Thank you a lot.

• suppose $f(x)$ is reducible $K[x]/(f(x))$ is a field then? – user87543 Jan 8 '14 at 5:56

## 4 Answers

If reducible, then zero divisor exist, so not field.

If irreducible, then all polynomial $g$ can be subjected to extended Euclidean algorithm with $f$ to get multiplicative inverse. The rest of the field axiom are just because of ring quotient.

Hint $$\$$ Recall that $$R/I\,$$ is a field$$\iff I$$ is a maximal ideal. Thus to show that $$\,K[x]/(f)$$ is a field$$\iff f\,$$ is irreducible, it suffices to show, in $$K[x],\,$$ that $$\,(f)\,$$ is maximal$$\iff f$$ is irreducible. Polynomial rings over fields enjoy a (Euclidean) division algorithm, hence every ideal is principal, generated by an element of minimal degree (= gcd of all elements).

But for principal ideals: $$\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$$,  i.e. $$(a)\supset (b)\iff a\mid b,\,$$ thus

$$\qquad\quad\begin{eqnarray} R/(f)\,\text{ is a field} &\iff& (f)\,\text{ is maximal} \\ &\iff&\!\!\ (f)\, \text{ has no proper } \,{\rm\color{#0a0}{container}}\,\ (g)\\ &\iff&\ f\ \ \text{ has no proper}\,\ {\rm\color{#c00}{divisor}}\,\ g\\ &\iff&\ f\ \ \text{ is irreducible}\\ &\iff&\ f\ \text{ is prime,}\ \ \text{by PID} \Rightarrow\text{UFD, so ireducible = prime } \end{eqnarray}$$

Remark $$\$$ PIDs are the UFDs of dimension $$\le 1,\,$$ i.e. where all prime ideals $$\ne 0\,$$ are maximal.

Hints:

(1) Over a field $\;F\;$ a polynomial $\;0\neq p(x)\in F[x]\;$ is irreducible iff the principal ideal $\;\langle\,p(x)\,\rangle\le F[x]\;$ is prime iff it is a maximal ideal.

(2) In a commutative unitary ring $\;R\;$ , an ideal $\;M\le R\;$ is maximal iff the quotient $\;R/M\;$ is a field.

• I guess this is a bit more advanced approach... – user87543 Jan 8 '14 at 6:04
• Indeed so, @PeteL.Clark. Thanks. – DonAntonio Jan 8 '14 at 6:10
• @PraphullaKoushik, do you know any other approach, more or less advanced? – DonAntonio Jan 8 '14 at 6:10
• @DonAntonio : I am trying... hope it come down neatly... – user87543 Jan 8 '14 at 6:11
• I have tried something... please have a look at that! – user87543 Jan 8 '14 at 6:21

For a non zero polynomial $f(x)\in K[x]$

Suppose $K[x]/(f(x))$ is a field and you have factorization:

$f(x)=g(x)h(x)$ with $\text{ Min{deg g(x),deg h(x)} < deg f(x)}$

Can $g(x)$ be in $(f(x))$??

Can $h(x)$ be in $(f(x)$??

Please make use of the fact that $K[x]/(f(x))$ is a field i.e., any element which is not in $(f(x))$ is a unit.

Now, what are all the units in $K[x]$??

The very next step would give you the irreducibility!!

I hope this would help!!