Prove that $L=K[x]/\langle m(x)\rangle$ is an algebraic extension of $K$

Question

Let $m(x)$ be an irreducible polynomial of degree $n$. How can I prove that $$L=\frac{K[x]}{\langle m(x)\rangle}$$ is:

1. A vector space of dimension $n$,

2. An algebraic extension of $K$?

Edit: I'm going to post what I've come up with so far. Please point out any errors within my proof.

Let $\varphi:K[x]\rightarrow A,\varphi(p(x))=p(a),a\in A$

We can, by the 1st Isomorphism Theorem, say that

$$\frac{K[x]}{\langle m(x)\rangle}\simeq K[a]$$ because $ker(\varphi)=\langle m(x)\rangle$.

Due to the Division Algorithm, we can say that $$p(x)=q(x)m(x)+r(x)$$ which means that every $p(a)$ can be written as $$p(a)=r_0+r_1a+...+r_{n-1}a^{n-1}$$

As such, if we take $B=\{1,a,a^2,...,a^{n-1}\}$ we can see that $B$ generates $K[a]$ (which is isomorphic to $\frac{K[x]}{\langle m(x)\rangle}$) over $K$, which has dimension $n$.

Answer 1

Hint:

• Show that as $m (x)$ is irreducible then through the homomorphism $\varphi : K[x] \to \Omega$ defined as $\varphi(f)(x) = f(a)$ as $a \in \Omega$ is such that $m (a) = 0$ you have that $$\frac{K[x]}{\langle m(x) \rangle} \simeq K[a]$$

(Use the First Isomorphism Theorem)

• Show that for any $f(x) \in K[x]$, $f(a)$ can be written uniquely as

$$f(a) = \alpha_0 + \alpha_1 a + \ldots + a_n a^n$$

where $\alpha_i \in K$.

(Use the Division Algorithm)

• Notice that if $[K[a] : K] = n < \infty$ then $1, a, \dots, a^n$ are L.D. And also that $$\{1, a, \ldots, a^{n-1}\}$$

is a basis of $K[a]$ over $K$.

Note: By definition $K[a] = \{f(a) ; f(x) \in K[x]\}$. And is possible to show that it is a integral domain such that $$K \subseteq K[a] \subseteq L$$ where $L \supseteq K$ is an extension over $K$.