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 $K[x]$.

Question

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.

Answer 1

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 of the ideal).

Recall that for principal ideals we have: $\ \rm\color{#0a0}{contains} = \color{#c00}{divides}$, $ $ i.e. $\,(a)\supset (b)\!\iff\! a\mid b,\,$ so

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

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

Answer 2

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.

Answer 3

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!!

Answer 4

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.