# 4 – Algebraic elements

(This is part of my series Field theory for high-schoolers.)

In this post we study important results and terminology about the roots of polynomials in real coefficients only.

3.1 Algebraic elements

Let $F$ be a field and $E$ be an extension of $F$. An element $\alpha$ of $E$ is said to be algebraic over $F$ if there exists some non-zero polynomial with coefficients in $F$ of which $\alpha$ is a root. If there exists no such polynomial, $\alpha$ is said to be transcendental over $F$.

Note that in general, when one speaks of an “algebraic number”, it means an element of $\mathbf{C}$ that is algebraic over $\mathbf{Q}$. Since here we are only concerned with the extension $\mathbf{C}/\mathbf{R}$, we will not speak of algebraic numbers in the proper sense (but of course some elements that are algebraic over $\mathbf{R}$ are also algebraic over $\mathbf{Q}$).

Examples:

• $i$ is algebraic over $\mathbf{R}$, since it is a root of the polynomial $x^2+1$ with coefficients in $\mathbf{R}$. (It is also algebraic over $\mathbf{Q}$.)
• $\sqrt{2}$ is algebraic over $\mathbf{R}$ since it is a root of the polynomial $x-\sqrt{2}$ with coefficients in $\mathbf{R}$. In general, an element of a field $F$ is of course always algebraic over $F$. ($\sqrt{2}$ is also algebraic over $\mathbf{Q}$, since it is a root of the polynomial $x^2-2$ with coefficients in $\mathbf{Q}$.)
• $\pi$ is algebraic over $\mathbf{R}$, since it is a root of the polynomial $x-\pi$ with real coefficients. (It is known (but not easy to prove!) that $\pi$ is transcendental over $\mathbf{Q}$.)

3.2 Degree of an algebraic element

Let $F$ be a field and $E$ be an extension of $F$. We have seen that an element $\alpha$ of $E$ is said to be algebraic over $F$ if there is some polynomial $f(x) \in F[x]$ such that $f(\alpha) = 0$. It can be shown that if $\alpha$ is algebraic over $F$, then there exists exactly one polynomial $f(x)$ with coefficients in $F$ such that $f(x)$ is monic, is irreducible over $F$, and has $\alpha$ as a root. This polynomial is called the minimal polynomial of $\alpha$ (over $F$).

Examples:

• We know that $f(x) = x^2+1 \in \mathbf{R}[x]$ is monic and irreducible, and that $f(i) = 0$. Thus $f(x)$ is the minimal polynomial of $i$ (over $\mathbf{R}$).
• We know that $g(x) = x^2-1 \in \mathbf{R}[x]$ is monic and that $g(1) = 0$, but $g(x)$ is not the minimal polynomial of $1$, because it is not irreducible. Of course, for any element $\alpha$ of $F$, the minimal polynomial of $\alpha$ over $F$ is $x-\alpha$.
• Likewise, we know that $h(x) = x^4-1 \in \mathbf{R}[x]$ is monic and that $h(i) = 0$, but $h(x)$ is not the minimal polynomial of $i$ because $h(x)$ is not irreducible (and we have seen that the minimal polynomial of $i$ is $x^2+1$).
• $k(x) = 2x^2+2 \in \mathbf{R}[x]$ is irreducible, and $k(i) = 0$, but $k(x)$ is not the minimal polynomial of $i$ because it is not monic (and we have seen that the minimal polynomial of $i$ is $x^2+1$).

For an element $\alpha$ of $E$ algebraic over $F$, the degree of the minimal polynomial of $\alpha$ over $F$ is called the degree of $\alpha$ over $F$.

Examples:

• We have seen that the minimal polynomial of $i$ over $\mathbf{R}$ is $x^2+1$, which is of degree $2$, and so the degree of $i$ over $\mathbf{R}$ is $2$.
• We have of course that the degree of $\alpha$ over $F$ is $1$ if and only if $\alpha$ is an element of $F$.

Finally, it can be shown that if $g(x)$ is a polynomial with coefficients in $F$ such that $g(\alpha) = 0$ for some $\alpha \in E$, if we let $f(x)$ be the minimal polynomial of $\alpha$ over $F$, then $g(x) = f(x)h(x)$ for some polynomial $h(x)$ with coefficients in $F$. In other words, the minimal polynomial of $\alpha$ over $F$ divides any polynomial with coefficients in $F$ which has $\alpha$ as a root. This implies that the degree of $g(x)$ is at least as large as the degree of $f(x)$, and so that $f(x)$ is of minimal degree among all the polynomials with coefficients in $F$ having $\alpha$ as a root (hence the term “minimal”).

Examples:

• Let $g(x) = 2x^4+5x^2+3 \in \mathbf{R}[x]$. We have $g(i) = 0$ and we know that the minimal polynomial of $i$ over $\mathbf{R}$ is $x^2+1$. This implies that there exists some polynomial $h(x)$ in real coefficients such that $g(x) = (x^2+1)h(x)$. Indeed, we have $h(x) = 2x^2+3$.
• We have seen that the degree of $i$ over $\mathbf{R}$ is $2$. This means that every polynomial $g(x)$ with coefficients in $\mathbf{R}$ such that $g(i) = 0$ is of degree at least $2$. This makes sense, since we know that any polynomial of degree $1$ in real coefficients has exactly one root, and that this root is real. Thus it can’t be $i$.

Next: Simple extensions

Questions:

• What is the minimal polynomial of $2i$ over $\mathbf{R}$?
• What is the minimal polynomial of $i+\sqrt{2}$ over $\mathbf{R}$?
• Is there some $\alpha \in \mathbf{C}$ such that the degree of $\alpha$ over $\mathbf{R}$ is $3$?
• What is the minimal polynomial of $i+\sqrt{2}$ over $\mathbf{Q}$?
• What is the minimal polynomial of $\sqrt{2}+\sqrt{3}$ over $\mathbf{Q}$?

## 2 thoughts on “4 – Algebraic elements”

1. Teja

In the examples about minimal polynomials, the example must be h(x)=x^4 – 1 ; i^4 + 1 = 2 \ne 0.

1. Firas Post author

I was testing you. ;)

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