# $\alpha$ algebraic over field $F$ implies $[F(\alpha):F]=\text{degree of minimal polynomial over$\alpha$}$

Let $\alpha$ be some element algebraic over a field $F$. Then $F(\alpha)$ is isomorphic to $F[x]/\langle m_{\alpha,F}\rangle$, where $m_{\alpha,F}$ is the minimal polynomial with root $\alpha$ over $F$. Moreover, $[F(\alpha) : F] = \deg(m_{\alpha,F})$.

The first part of this theorem appears to be clear, but what about the second part (after the word "moreover")? Why does the index of $F(\alpha)$ over $F$ equal the degree of the minimal polynomial in $F[x]$ with root $\alpha$? I think this should be almost obvious, but perhaps I'm lacking some theory.

• The index is defined as the dimension of $F(\alpha)$ as a vector space over $F$. All is remains is to find a basis. Try to write one... – ayberk Apr 11 '17 at 8:37
• @ayberk We have that $\dim_F F(\alpha)=\dim_F F[x]/\langle m_{\alpha,F}\rangle$. But what is the dimension of $F[x]$ over $F$? It is infinite, since a basis is $\{1, x, x^2,\dots\}$. What is $\dim_F F[x]/\langle m_{\alpha,F}\rangle$, however? It seems that it should be infinite as well. But what am I missing? – sequence Apr 11 '17 at 8:45
• Say the degree of the minimal polynomial is $k$. Do you really need the elements $x^l$ where $l\geq k$ in the basis? – ayberk Apr 12 '17 at 5:46
• @ayberk I just don't see how the minimal polynomial is exactly related to $F(\alpha)$. – sequence Apr 12 '17 at 7:55
• Write the minimal polynomial as: $$m_{\alpha,F}(x) = x^k + a_{k-1}x^{k-1} + \ldots + a_0.$$ Then, we have $$x^k = -(a_{k-1}x^{k-1} + \ldots + a_0)$$ hence an $F$ linear combination of $\{1,x,\ldots,x^{k-1}\}$. – ayberk Apr 12 '17 at 8:20

## 3 Answers

Given a field extension $K/F$, the degree of the extension $K/F$ is defined as the dimension of $K$ viewed as a vectorial space over $F$. In other words $$[K:F]=\dim_FK.$$

For your question, I let you to prove (if you haven't already proved it) that if $\alpha$ is algebraic over $F$, then $$F(\alpha)=\{g(\alpha): g(x)\in F[X]\; \text{and}\; g\equiv 0\; \text{or}\; \deg(g)<\deg(m_{\alpha,F})\}.$$

Now let's set $m_{\alpha,F}=p$, then if $p(x)=a_0x^n+\cdots +a_{n-1}x+a_n$ we claim that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a basis for $F(\alpha)$ over $F$. Indeed, every element of $F(\alpha)$ has the form $b_0\alpha^{k}+\cdots +b_{n-1}\alpha+b_n$ for some $g(x)=b_0x^k+\cdots +b_{n-1} x+b_n\in F[x]$ with $k<n$, so it follows easily that $\{1,\alpha,\ldots \alpha^{n-1}\}$ is a generating set of $F(\alpha)$ as a vectorial space over $F$. To prove that the above set is l.i. we write $$c_0+c_1\alpha+\ldots c_{n-1}\alpha^{n-1}=0.$$ Then $q(x)=c_0+c_1x+\ldots c_{n-1}x^{n-1}\in F[x]$ satisfies $q(\alpha)=0$, so either $q\equiv 0$ or $p(x)\mid q(x)$. But the later is impossible because that would mean that $n=\deg(p)\le deg(q)=n-1$, an absurd. Therefore $q\equiv 0$, which implies that $c_0=c_1=\ldots =c_{n-1}=0$.

Hence $\{1,\alpha,\ldots \alpha^{n-1}\}$ is l.i. and thus it's a basis for $F(\alpha)$ over $F$. From this is immediate that $$[F(\alpha):F]=n=\deg(p)=\deg(m_{\alpha,F}).$$

• Can you please clarify the following? 1. In $F(\alpha)=\{g(\alpha): g(x)\in F[X]\; \text{and}\; g\equiv 0\; \text{or}\; \deg(g)<\deg(m_{\alpha,F})\}$, how can $\deg(g)<\deg(m_{\alpha,F})$? If $m_{\alpha,F}$ is the minimal polynomial for $\alpha$ over $F$, then why can't there be polynomials of bigger degree? Also, why isn't it true that the degree of the minimal polynomial is always $1$, i.e. $x-\alpha$? 2. Why does every element of $F(\alpha)$ has the form $b_0\alpha^{k}+\cdots +b_{n-1}\alpha+b_n$ with $k\le n-1$? – sequence Apr 11 '17 at 23:06
• Take an arbitrary polynomial $h (x)$ and divide it by $m_{\alpha, F}$ applying the euclidean division. Then you'll find that $g (x)$ is the remainder so is either the zero polynomial or its degree is less than... The minimal polynomial has its coefficients in $F$ whereas $\alpha$ is an element of $K$ so it's not neccesarily true that $x-\alpha \in F [x]$. For your last question I've edited my answer. – Xam Apr 12 '17 at 1:12
• Why does the set of $F(\alpha)$ have to contain the remainder $g(x)$ after the Eucledian division of $f(x)$ by $p(x)$? I don't see the connection. It's still not clear how the minimal polynomial for $\alpha$ is related to $F(\alpha)$. That is, if $F[x]$ contains polynomials of all degrees over $F$, then why should there be only finitely many elements in the basis for $F(\alpha)$? Looks like I'm quite confused. @Xam – sequence Apr 12 '17 at 8:35
• When $\alpha$ is algebraic $F [\alpha]=F (\alpha)$ and this is proved in user1952009's answer. An arbitrary element of $F [\alpha]$ has the form $h(\alpha)$ for some $h(x)\in F [x]$. By the euclidean division $h(x)=p (x) q(x)+g(x)$, so evaluating at $\alpha$ give us $h(\alpha)=p (\alpha) q(\alpha)+g(\alpha)$, but $p(\alpha)=0$ so $h(\alpha)=g(\alpha)$. The connection between minimal polynomial of $\alpha$ and $F(\alpha)$ is given by the isomorphism stated in user195's answer. For your last question note that the polynomials used in $F(\alpha)$ have degree $< \deg (p)$. – Xam Apr 12 '17 at 13:35

The minimal polynomial gives you a way of expressing $\alpha^n$ (n the degree of the minimal polynomial) as a combination of the other powers of $\alpha$ and the coefficients of the minimal polynomial. Therefore the dimension of $F(\alpha)$ over $F$ is not infinite.

• Sorry, I don't get what "as a combination of the other powers of α and the coefficients of the minimal polynomial" means. – sequence Apr 12 '17 at 7:57

It seems the other answer and comment are about $F[\alpha]$ being a $F$-vector space of dimension $deg(p)$ ($F[\alpha]$ is the smallest ring containing $F$ and $\alpha$, i.e. $F[\alpha] = \{ \sum_{n =0}^d c_n \alpha^n, c_n \in F\}$).

For showing that $F[\alpha] = F(\alpha)$ you need to prove that $\varphi : F[x]/(p(x)) \to F[\alpha],\ \ \varphi(f(x)) = f(\alpha)$ is an isomorphism of rings ($\varphi$ is clearly surjective, and it is injective by definition of the minimal polynomial)

Hence $F[\alpha]$ is a field, so $F[\alpha] = F(\alpha)$ and $[F(\alpha):F]= deg(p)$

• Can you please clarify why $\phi$ is injective by the definition of the minimal polynomial? Also, why does it so follow that if $F[\alpha]=F(\alpha)$ then $[F(\alpha):F]=deg(p)$? – sequence Apr 11 '17 at 23:15
• @sequence : Did you read any course about field extensions ? $[F[\alpha]:F]= deg(p)$ is proven in the other answers. $F[x]/(p(x))$ is by definition a $F$-vector space of dimension $deg(p)$. And $\varphi$ is injective because otherwise for some $f \in F[x], deg(f) < deg(p), f \ne 0$ : $\varphi(f(x)) = f(\alpha) = 0$ contradicting that $p$ is the minimal polynomial of $\alpha$. – reuns Apr 11 '17 at 23:40
• Why is $F[x]/(p(x))$ of dimension $\deg(p)$? Consider $f(x) + (p(x))$, where $f(x)$ is of degree 1000, and $p(x)$ is of degree 1. Then $f(x)+(p(x))$ is of degree 1000. @user1952009 – sequence Apr 12 '17 at 8:01
• @sequence You can define $\mathbb{C}=\mathbb{R}(i)$ as $\mathbb{R}[x]/(x^2+1)$. Do you see why $[\mathbb{C}:\mathbb{R}] = 2$ ? – reuns Apr 12 '17 at 13:54
• I can see why $[\mathbb{C}:\mathbb{R}]=2$ from $\mathbb{R}(i)$, since it is a vector space over the basis $\{1,i\}$. But I don't know how to see this from $\mathbb{R}[x]/(x^2+1)$. There must be a theorem which says that the index of a field extension equals the degree of the minimal polynomial over the primitive element, but I don't know how to see this intrinsically. @user1952009 – sequence Apr 12 '17 at 17:27