*(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 $\mathbf{Q}$?

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

FirasPost authorI was testing you. ;)