My book says the following:

Let $K\subseteq L$, so $L$ is an extension field of $K$ and $\alpha\in > L$ algebric over $K$, with $p(x) = irr(\alpha, K)$ (the irreducible polynomial) such that $\partial(p(x)) = n$. Then:


$$\frac{K[x]}{\langle p(x) \rangle} \simeq K[\alpha]$$

and $K[\alpha] = K(\alpha)$ and every element of $K(\alpha)$ is written in a unique way as $r(\alpha)$ where $r(x)\in K[x], r\equiv 0$ or $\partial (r(x)) < \partial(p(x))$

Proof of $i$:

Let $\phi: K[x]\to K[\alpha]$ given by $f(x)\to f(\alpha)$, then $\phi$ is a surjective homomorphism such that $\ker \phi = \langle p(x) \rangle$.

Indeed, $\ker \phi = \{f(x) \in K[x] / \phi(f(x)) = 0\}$

Note that $\phi(f(x)) = 0 \iff f(\alpha) = 0 \implies p(x) | f(x) \implies f(x) = t(x)p(x)$ for a $t(x)\in K[x]$, that is, $f(x)\in \langle p(x) \rangle$, so

$$\ker \phi \subseteq \langle p(x) \rangle$$

If $g(x)\in \langle p(x) \rangle$, then $g(\alpha) = 0 \implies g(x) \in \ker \phi$ so

$$\langle p(x) \rangle \subseteq \ker \phi \\ \ker \phi = \langle p(x) \rangle$$

So by the first isomorphism theorem,

$$\frac{K[x]}{\langle p(x) \rangle} \simeq K[\alpha]$$

The greatest problem I'm having to understand this proof is to understand what is $K[\alpha], K(\alpha), f(\alpha)$. For me, $K[x]$ are the polynomials with coefficients in $K$, like $3x^2 + 2x + 1$ for when $K= \mathbb{Q}$ for example. But what is $K[\alpha]$? Polynomials over $\alpha$ instead of $x$? What's the difference? Also, $f(x)$ should be a polynomial from $K[x]$, right? But isn't $f(\alpha)$ a number? We're mapping lots of functions $f(x)$ to a single number? Also, shouldn't this number be $0$ since $\alpha$ is algebric over $K$?


$K$ is a field (otherwise it gets more complicated)

  • $K[\alpha]$ is the smallest ring containing $K$ and $\alpha$, ie. the ring of $K$-linear combinations of powers of $\alpha$.

    Iff $\alpha$ is algebraic over $K$ then $\{\alpha^m\}_{m \ge 0}$ is not a $K$-linearly independent family and only finitely many powers of $\alpha$ are needed to generate $K[\alpha]$ as a $K$-vector space. This is encoded by the minimal polynomial.

    Thus $K[\alpha] \cong K[x]/(p(x))$ almost by definition of the minimal polynomial (aka the first $K$-linear relation occurring between the powers of $\alpha$, which by minimality encodes all the $K$-linear relations satisfied by the powers of $\alpha$).

  • $K(\alpha)$ is the smallest field containing $K$ and $\alpha$, ie. the fraction field of $K[\alpha]$.

    Iff $p$ is irreducible over $K$ then $K[x]/(p(x))$ is an integral domain, and it is a field by the $\gcd$ algorithm and the Bezout identity.

    Or, because any element $\beta \in K[\alpha]$ is algebraic over $K$ (since $K[\alpha]$ is a finite dimensional $K$-vector space) and inversible in $K[\beta]\subset K[\alpha]$ (since $\sum_{m=0}^d b_m \beta^m = 0 \implies \beta^{-1} = \frac{-1}{b_0} \sum_{m=0}^{d-1} b_{m+1} \beta^{m}$).

  • $\alpha$ is a number, a root of an irreducible polynomial, a matrix $\in K^{n \times n}$, a linear map, abstract-algebra is made to unify all this. Note that $\mathbb{C} = \mathbb{R}[i] \cong \mathbb{R}[{\scriptstyle\begin{pmatrix} 0 & 1 \\ - 1 & 0 \end{pmatrix}}]\cong \mathbb{R}[x]/(x^2+1)$ and $\mathbb{Q}[x] = \mathbb{Q}[\pi]$.

  • $\begingroup$ what is $(p(x))$? $\endgroup$ – Paprika Sep 11 '17 at 2:41
  • $\begingroup$ yes, sorry, but I forgot to mention that I don't understand exactly what $\langle p(x) \rangle$ means $\endgroup$ – Paprika Sep 11 '17 at 2:50
  • $\begingroup$ @Paprika Work on $\mathbb{C} = \mathbb{R}[x]/(x^2+1)$, which really means "$x$ behaves like $i$ algebraically". Concretely $(x^2+1)$ is the ideal of $\mathbb{R}[x]$ generated by $x^2+1$, which imposes in the quotient ring the linear relation $x^2+1= 0$ $\endgroup$ – reuns Sep 11 '17 at 2:54
  • $\begingroup$ What about $f(\alpha)$? $\endgroup$ – Guerlando OCs Sep 11 '17 at 5:20
  • $\begingroup$ @GuerlandoOCs what with $f(\alpha)$ ? $\endgroup$ – reuns Sep 11 '17 at 5:32

Let $f \in f[X]$ be a monic irreducible polynomial and $E$ an extension of $F$ so that $E=F[\alpha]$ and $f(\alpha)=0$.

Then the evaluation map, $g(X) \mapsto g(\alpha):F(X) \to E$ has kernel exactly $f$ by irreducibility.

But then $E$ is isomorphic to $f[X]/(f)$, so all elements are of the form $$a_0+a_1 \alpha+\dots+a_{m-1} \alpha^{m-1}$$ with $a_i \in F$ and $m=\partial (f)$.

On the other hand, note that $f[X]/(f)$ is a field implies that $F[\alpha]$ is as well, so $f[\alpha]=f(\alpha)$.


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