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

Again, the reader is probably familiar with the process of factoring polynomials in real or complex coefficients, and with the concept of irreducible polynomials. We are really doing high school algebra, but with more general formulation and terminology.

**2.1 Factoring polynomials**

Since polynomials can be multiplied, it is natural to wonder whether a given polynomial can be written as the product of two others. Of course, any polynomial can be written as the product of a constant polynomial and some other polynomial, for example $x^2+2x+1 = 2\left(\frac{1}{2}x^2+x+\frac{1}{2}\right)$. This case is not important here.

Instead, we wonder whether a given polynomial can be expressed as the product of two *non-constant* polynomials. We note that since the degree of a product of two polynomials is the sum of their degrees, if $f(x)$ has degree $n$ and if $f(x) = g(x)h(x)$ with $g(x)$ and $h(x)$ both non-constant, both $g(x)$ and $h(x)$ will also have degree less than $n$ (since a non-constant polynomial has degree at least $1$).

The reader probably knows that a polynomial of degree $2$ in real coefficients can be written as the product of two polynomials of degree $1$ in real coefficients if and only if it has a root in $\mathbf{R}$. For example, the polynomial $2x^2-4$ has $\sqrt{2}$ as a root in $\mathbf{R}$, and thus it can be written as $(x-\sqrt{2})g(x)$, where $g(x)$ is a polynomial of degree $1$ with real coefficients.

More generally, a polynomial $f(x)$ of degree $n$ with coefficients in a field $F$ can be factored into $(x-\alpha)g(x)$ with $\alpha \in F$ and $g(x)$ of degree $n-1$ with coefficients in $F$ if and only if $\alpha$ is a root of $f(x)$ in $F$.

*Examples:*

- The polynomial $f(x) = x^3-1$ of degree $3$ with coefficients in $\mathbf{R}$ has $1$ as a root in $\mathbf{R}$ and thus can be written as $(x-1)g(x)$, where $g(x)$ is a polynomial of degree $2$ with coefficients in $\mathbf{R}$.
- The polynomial $f(x) = x^2+1$ of degree $2$ with coefficients in $\mathbf{R}$ has no root in $\mathbf{R}$, and thus cannot be written as a product of two polynomials of degree $1$ with coefficients in $\mathbf{R}$.
- However, it has $i$ as a root in $\mathbf{C}$, and so it can be written as $(x-i)g(x)$, with $g(x)$ a polynomial of degree $1$ with coefficients in $\mathbf{C}$.
- As was said in the last post, $\mathbf{C}$ is algebraically closed, and thus any non-constant polynomial with coefficients in $\mathbf{C}$ has a root in $\mathbf{C}$. Thus any polynomial $f(x)$ of degree $n$ with coefficients in $\mathbf{C}$ can be written as $(x-\alpha)g(x)$, with $g(x)$ a polynomial of degree $n-1$ with coefficients in $\mathbf{C}$.

The above implies two important results:

- Given a polynomial $f(x)$, if we can write it as $f(x) = g(x)h(x)$, then $\alpha$ is a root of $f(x)$ if and only if it is a root of $g(x)$ or $h(x)$. This makes sense, since then we have $f(\alpha) = g(\alpha)h(\alpha)$, and thus if $f(\alpha) = 0$, it must be that $g(\alpha) = 0$ or $h(\alpha) = 0$.
- A polynomial of degree $n$ can have at most $n$ roots. This also makes sense since for each root, we can factor it away, and be left with a polynomial of degree $n-1$, thus we cannot repeat the process more than $n$ times.

**2.2 Irreducible polynomials**

A polynomial $f(x)$ with coefficients in a field $F$ that can be written as the product of two non-constant polynomials also with coefficients in $F$ is said to be *reducible over $F$* (or *in $F$*). A polynomial that is not reducible over $F$ is said to be *irreducible over $F$*.

It is easy to see that any polynomial $f(x)$ with coefficients in a field $F$ that has a root $\alpha$ in $F$ is reducible over $F$, since it can then be written as $(x-\alpha)g(x)$. We wonder whether the converse is true: if $f(x)$ has no roots in $F$, does that imply that it is irreducible over $F$?

The answer is no. For a counterexample, consider the polynomial $f(x) = x^2+1$ with coefficients in $\mathbf{R}$. We know that it has no roots in $\mathbf{R}$. Now consider the polynomial $g(x) = (x^2+1)^2 = x^4+2x^2+1$. It is certainly reducible, since $g(x) = f(x)f(x)$, but it can have no root in $\mathbf{R}$ since a root of $g(x)$ would be a root of $f(x)$, but $f(x)$ has no roots in $\mathbf{R}$.

The answer is yes, however, if $f(x)$ is of degree $2$ or $3$. To see this, we recall that any polynomial $f(x) = ax+b$ of degree $1$ with coefficients in $\mathbf{R}$ has a root in $\mathbf{R}$ (namely, $-b/a$). This remains true in any field, since if the field properties are satisfied we can always let $\alpha = -b/a$ regardless of what those operations actually mean, and $\alpha$ will be a root of $f(x)$ in $F$.

Now let $f(x)$ be a reducible polynomial of degree $2$ or $3$. Then it can be written as $f(x) = g(x)h(x)$, with one of $g(x)$ and $h(x)$ being of degree $1$ (the other is of degree $1$ or $2$, depending on whether $f(x)$ is of degree $2$ or $3$). Say $g(x)$ is of degree $1$, then it surely has a root, and thus $f(x)$ has a root as well. Thus if a polynomial of degree $2$ or $3$ has no root, then it is impossible for it to be reducible.

*Examples:*

- The polynomial $f(x) = x^2+1$ is irreducible over $\mathbf{R}$ since it is of degree $2$ and has no root in $\mathbf{R}$. It is surely reducible over $\mathbf{C}$, since $\mathbf{C}$ is algebraically closed (and indeed it is known that $f(x) = (x-i)(x+i)$).
- As above, the polynomial $g(x) = x^4+2x^2+1$ of degree $4$ has no roots in $\mathbf{R}$, but is not irreducible over it.
- The polynomial $h(x) = x^2-1$ is reducible over $\mathbf{R}$ since it has a root in $\mathbf{R}$, and we have of course $h(x) = (x+1)(x-1)$.

**Next:** Kronecker’s theorem

**Questions:**

- $\mathbf{Q}$ is a field. Is the polynomial $f(x) = x^2-2$ with coefficients in $\mathbf{Q}$ reducible over $\mathbf{Q}$?
- $\mathbf{F}_5$ is a field. Is the polynomial $f(x) = x^2+1$ with coefficients in $\mathbf{F}_5$ reducible over $\mathbf{F}_5$?
- Find an example of a degree-6 polynomial with coefficients in $\mathbf{R}$ that has no root in $\mathbf{R}$.