## Section8.4Algebraic Elements

An element $\alpha$ in an extension field $E$ over $F$ is algebraic over $F$ if $f(\alpha)=0$ for some nonzero polynomial $f(x) \in F[x]\text{.}$ An element in $E$ that is not algebraic over $F$ is transcendental over $F\text{.}$ An extension field $E$ of a field $F$ is an algebraic extension of $F$ if every element in $E$ is algebraic over $F\text{.}$ If $E$ is a field extension of $F$ and $\alpha_1, \ldots, \alpha_n$ are contained in $E\text{,}$ we denote the smallest field containing $F$ and $\alpha_1, \ldots, \alpha_n$ by $F( \alpha_1, \ldots, \alpha_n)\text{.}$ If $E = F( \alpha )$ for some $\alpha \in E\text{,}$ then $E$ is a simple extension of $F\text{.}$

###### Example8.13.

Both $\sqrt{2}$ and $i$ are algebraic over ${\mathbb Q}$ since they are zeros of the polynomials $x^2 -2$ and $x^2 + 1\text{,}$ respectively. Clearly $\pi$ and $e$ are algebraic over the real numbers; however, it is a nontrivial fact that they are transcendental over ${\mathbb Q}\text{.}$ Numbers in ${\mathbb R}$ that are algebraic over ${\mathbb Q}$ are in fact quite rare. Almost all real numbers are transcendental over ${\mathbb Q}\text{.}$ 2 (In many cases we do not know whether or not a particular number is transcendental; for example, it is still not known whether $\pi + e$ is transcendental or algebraic.)

The probability that a real number chosen at random from the interval $[0, 1]$ will be transcendental over the rational numbers is one.

A complex number that is algebraic over ${\mathbb Q}$ is an algebraic number. A transcendental number is an element of ${\mathbb C}$ that is transcendental over ${\mathbb Q}\text{.}$

###### Example8.14.

We will show that $\sqrt{2 + \sqrt{3} }$ is algebraic over ${\mathbb Q}\text{.}$ If $\alpha = \sqrt{2 + \sqrt{3} }\text{,}$ then $\alpha^2 = 2 + \sqrt{3}\text{.}$ Hence, $\alpha^2 - 2 = \sqrt{3}$ and $( \alpha^2 - 2)^2 = 3\text{.}$ Since $\alpha^4 - 4 \alpha^2 + 1 = 0\text{,}$ it must be true that $\alpha$ is a zero of the polynomial $x^4 - 4 x^2 + 1 \in {\mathbb Q}[x]\text{.}$

There are lots of extension fields that contain elements transcendental over the base field. For example, we could consider $\Q(\pi)\text{.}$ It turns out this will be isomorphic to $\Q(x)\text{,}$ the field of all rational functions with rational coefficients. So in some sense, this is not that interesting: no matter what transcendental element you add, you get the same extension field. We have a more interesting situation in the case of algebraic extensions.

###### Proof.

Let $\phi_{\alpha} : F[x] \rightarrow E$ be the evaluation homomorphism. The kernel of $\phi_{\alpha}$ is a principal ideal generated by some $p(x) \in F[x]$ with $\deg p(x) \geq 1\text{.}$ We know that such a polynomial exists, since $F[x]$ is a principal ideal domain and $\alpha$ is algebraic. The ideal $\langle p(x) \rangle$ consists exactly of those elements of $F[x]$ having $\alpha$ as a zero. If $f( \alpha ) = 0$ and $f(x)$ is not the zero polynomial, then $f(x) \in \langle p(x) \rangle$ and $p(x)$ divides $f(x)\text{.}$ So $p(x)$ is a polynomial of minimal degree having $\alpha$ as a zero. Any other polynomial of the same degree having $\alpha$ as a zero must have the form $\beta p( x)$ for some $\beta \in F\text{.}$

Suppose now that $p(x) = r(x) s(x)$ is a factorization of $p(x)$ into polynomials of lower degree. Since $p( \alpha ) = 0\text{,}$ $r( \alpha ) s( \alpha ) = 0\text{;}$ consequently, either $r( \alpha )=0$ or $s( \alpha ) = 0\text{,}$ which contradicts the fact that $p$ is of minimal degree. Therefore, $p(x)$ must be irreducible.

Let $E$ be an extension field of $F$ and $\alpha \in E$ be algebraic over $F\text{.}$ The unique monic polynomial $p(x)$ of the last theorem is called the minimal polynomial for $\alpha$ over $F\text{.}$ The degree of $p(x)$ is the degree of $\alpha$ over $F$.

###### Example8.16.

Let $f(x) = x^2 - 2$ and $g(x) = x^4 - 4 x^2 + 1\text{.}$ These polynomials are the minimal polynomials of $\sqrt{2}$ and $\sqrt{2 + \sqrt{3} }\text{,}$ respectively.

###### Proof.

Let $\phi_{\alpha} : F[x] \rightarrow E$ be the evaluation homomorphism. The kernel of this map is $\langle p(x) \rangle\text{,}$ where $p(x)$ is the minimal polynomial of $\alpha\text{.}$ By the Fundamental Homomorphism Theorem for rings, the image of $\phi_{\alpha}$ in $E$ is isomorphic to $F( \alpha )$ since it contains both $F$ and $\alpha\text{.}$

Recall that $F(\alpha)$ is simply the smallest field that contains $F$ and $\alpha\text{.}$ What do elements in this set look like? Since $F(\alpha)$ is a field, it is closed under addition and multiplication, so $a + b \alpha + c \alpha^2 + \cdots$ are elements we can get in $F(\alpha)\text{.}$ Do we need to consider arbitrarily large powers of $\alpha\text{?}$ No, and this is why it is helpful to recognize $F(\alpha)$ as a quotient ring.

###### Proof.

Since $\phi_{\alpha} ( F[x] ) \cong F( \alpha )\text{,}$ every element in $E = F( \alpha )$ must be of the form $\phi_{\alpha} ( f(x) ) = f( \alpha )\text{,}$ where $f(\alpha)$ is a polynomial in $\alpha$ with coefficients in $F\text{.}$ Let

\begin{equation*} p(x) = x^n + a_{n - 1} x^{n - 1} + \cdots + a_0 \end{equation*}

be the minimal polynomial of $\alpha\text{.}$ Then $p( \alpha ) = 0\text{;}$ hence,

\begin{equation*} {\alpha}^n = - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0\text{.} \end{equation*}

Similarly,

\begin{align*} {\alpha}^{n + 1} & = {\alpha} {\alpha}^n\\ & = - a_{n - 1} {\alpha}^n - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\\ & = - a_{n - 1}( - a_{n - 1} {\alpha}^{n - 1} - \cdots - a_0) - a_{n - 2} {\alpha}^{n - 1} - \cdots - a_0 {\alpha}\text{.} \end{align*}

Continuing in this manner, we can express every monomial ${\alpha}^m\text{,}$ $m \geq n\text{,}$ as a linear combination of powers of ${\alpha}$ that are less than $n\text{.}$ Hence, any $\beta \in F( \alpha )$ can be written as

\begin{equation*} \beta = b_0 + b_1 \alpha + \cdots + b_{n - 1} \alpha^{n - 1}\text{.} \end{equation*}

To show uniqueness, suppose that

\begin{equation*} \beta = b_0 + b_1 \alpha + \cdots + b_{n-1} \alpha^{n-1} = c_0 + c_1 \alpha + \cdots + c_{n - 1} \alpha^{n - 1} \end{equation*}

for $b_i$ and $c_i$ in $F\text{.}$ Then

\begin{equation*} g(x) = (b_0 - c_0) + (b_1 - c_1) x + \cdots + (b_{n - 1} - c_{n - 1})x^{n - 1} \end{equation*}

is in $F[x]$ and $g( \alpha ) = 0\text{.}$ Since the degree of $g(x)$ is less than the degree of $p( x )\text{,}$ the irreducible polynomial of $\alpha\text{,}$ $g(x)$ must be the zero polynomial. Consequently,

\begin{equation*} b_0 - c_0 = b_1 - c_1 = \cdots = b_{n - 1} - c_{n - 1} = 0\text{,} \end{equation*}

or $b_i = c_i$ for $i = 0, 1, \ldots, n-1\text{.}$ Therefore, we have shown uniqueness.

###### Example8.19.

Since $x^2 + 1$ is irreducible over ${\mathbb R}\text{,}$ $\langle x^2 + 1 \rangle$ is a maximal ideal in ${\mathbb R}[x]\text{.}$ So $E = {\mathbb R}[x]/\langle x^2 + 1 \rangle$ is a field extension of ${\mathbb R}$ that contains a root of $x^2 + 1\text{.}$ Let $\alpha = x + \langle x^2 + 1 \rangle\text{.}$ We can identify $E$ with the complex numbers. By Proposition 8.17, $E$ is isomorphic to ${\mathbb R}( \alpha ) = \{ a + b \alpha : a, b \in {\mathbb R} \}\text{.}$ We know that $\alpha^2 = -1$ in $E\text{,}$ since

\begin{align*} \alpha^2 + 1 & = (x + \langle x^2 + 1 \rangle)^2 + (1 + \langle x^2 + 1 \rangle)\\ & = (x^2 + 1) + \langle x^2 + 1 \rangle\\ & = 0\text{.} \end{align*}

Hence, we have an isomorphism of ${\mathbb R}( \alpha )$ with ${\mathbb C}$ defined by the map that takes $a + b \alpha$ to $a + bi\text{.}$

##### Summary.

At this point, it might seem like we are running around in circles, as it often seems whenever we get close to the Fundamental Homomorphism Theorem. Before moving on, let's try to sort out the different things the theorems and propositions are doing in this section.

First, if we are start with a given polynomial $p(x)$ over some field $F\text{,}$ we know there is some extension field $E$ above $F$ in which $p(x)$ has a root $\alpha$ (and therefore factors). This is Theorem 8.10.

On the other hand, we can start a extension field $E$ of a base field $F$ that contains some element $\alpha$ algebraic over $F\text{.}$ By definition, $\alpha$ is the root to some polynomial with coefficients from $F\text{,}$ and by Theorem 8.15 there is a minimum polynomial $p(x)$ for $\alpha\text{.}$ By Proposition 8.17, the field extension $F(\alpha)$ (which is a subfield of $E\text{,}$ if not $E$ exactly) is isomorphic to the quotient ring $F[x]/\langle p(x) \rangle\text{.}$

So if you have a minimal polynomial $p(x)\text{,}$ you can find an extension field that contains a root $\alpha\text{,}$ namely $F[x]/\langle p(x)\rangle\text{.}$ Or if you have an extension field with a root $\alpha\text{,}$ you can find a minimal polynomial $p(x)$ and the extension field that you would get by starting with $p(x)$ will be precisely the smallest field containing $F$ and $\alpha\text{.}$

### ExercisesCollected Homework

###### 1.

Consider the number $\alpha = \sqrt{3}+\sqrt{5}$ and the field $\Q(\alpha)\text{,}$ a simple extension of $\Q\text{.}$

###### (a)

Prove that $\alpha$ is an algebraic number.

###### (b)

Is $\Q(\alpha)$ an extension field of $\Q(\sqrt{3})\text{?}$ Is it an extension field of $\Q(\sqrt{5})\text{?}$ Explain.

Hint

First decide whether $\Q(\alpha)$ contains the element $\sqrt{15}\text{.}$ If it does, note that it must also contain $\sqrt{15}(\sqrt{3}+\sqrt{5}) - 3(\sqrt{3}+\sqrt{5})\text{.}$ Why is this helpful?

###### (c)

If we take $\Q(\sqrt{3})$ and adjoin $\sqrt{5}\text{,}$ do we get a bigger field? We can write $\Q(\sqrt{3})(\sqrt{5})$ as $\Q(\sqrt{3},\sqrt{5})\text{.}$

###### (d)

Find the minimal polynomials for $\sqrt{3}$ over $\Q$ and of $\alpha = \sqrt{3}+\sqrt{5}$ over $\Q(\sqrt{3})\text{.}$ How do the degrees of these relate to the degree of the polynomial you found for $\alpha$ in part (a)?