Are roots of a polynomial $f\in k[x]$ required to be elements of $k$?

Question

This one is a rather basic query. I have only recently started studying fields and in particular finite fields. My question is the following:

Do the roots of a polynomial over a given field have to be inside that particular field ?

For example, consider the polynomial $$f(x) = x^2 + x + 1\in\Bbb{F}_2[x].$$ Then does something like $f(7)$ make sense here ? Or are $f(1)$ and $f(0)$ the only ones which make sense ?

When defining a polynomial over a field it is mentioned that the coefficients have to be in the field over which the polynomial is defined.But I am confused if it also applies to the values which $x$ can take in $f(x)$.I am studying these concepts in the study of rings so usually we deal with operations between different polynomial.

According to me the answer is no because if that were the case $x - 7$ would be a factor of $f(x)$ and $x - 7$ is itself a polynomial which doesn't quite make sense since the field is $\Bbb{F}_2$.

One of my sources of confusion is the concept of extension field. Suppose we consider the extension field $\Bbb{F}_{2^4}$ now can we speak of roots other than $0$ or $1$ for the same polynomial. (I.e can $x$ take up $16$ values instead of $2$ in $f(x)$ )

Thanks.

Answer 1

In $\text{GF}(2)$, $7=1$, so if $f(1)$ makes sense, then so does $f(7)$. Similarly, $x-7$ does make sense as a polynomial over $ \text{GF}(2)$: it is the same polynomial as $x-1$. It is also the same polynomial as $x+1$.

More generally, if $A$ is any algebra over $\text{GF}(2)$, $a \in A$, and $f$ is a polynomial over $\text{GF}(2)$, then $f(a)$ makes sense. You can see this as seeing the calculation implied by $f$ makes sense, or by defining it as the image of $f$ under the homomorphism $\text{GF}(2)[x] \to A$ that sends $r \to r \cdot 1_A$ for $r \in \text{GF}(2)$ and sends $x \to a$. (this uniquely determines a homomorphism by the universal property of polynomial rings)

The roots of $x^2 + x + 1$, incidentally, are the two elements of $\text{GF}(4)$ that are neither $0$ nor $1$, so the roots of a polynomial over a field don't have to lie in that field.

The roots of the polynomial $x^2 + 1$ over $\mathbb{R}$ is a more familiar example of this phenomenon, as its roots are the elements $\pm \mathbf{i} \in \mathbb{C}$. Another familiar example are the roots of the polynomial $x^2 - 2$ over $\mathbb{Q}$.

Answer 2

I think you're asking about the evaluation homomorphism, not about roots. We can evaluate a polynomial by just substituting in some field element everywhere we see $x$; so for a polynomial over $\mathbb{F}_2$, there is no element named $7$, so $f(7)$ doesn't make sense here. If, of course, we regard $\mathbb{F}_2$ as the integers modulo $2$, then $7 = 1 (\mod 2)$, so the evaluation would make sense.

A root of the polynomial is a field element $\alpha$ for which $f(\alpha) = 0$ in the relevant field. I think you may be confusing the evaluation $f(7)$ and the equation $f(x) = 7$.

Answer 3

A field $\mathbb{F}$ having the property that every polynomial $f\in\mathbb{F}[X]$ has a root in $\mathbb{F}$ is called algebraically closed. Not every field is algebraically closed. For any field $\mathbb{F}$, every polynomial $f\in\mathbb{F}[X]$ induces a function $$\mathbb{F}\ \longrightarrow\ \mathbb{F}:\ x\ \longmapsto\ f(x),$$ by evaluation. This function takes values in $\mathbb{F}$, but it does not necessarily take the value $0$, i.e. it does not necessarily have roots in $\mathbb{F}$. A prime example is $X^2+1\in\mathbb{R}[X]$, but $X^2+X+1\in\mathbb{F}_2$ is also an example.

An algebraic extension $\mathbb{K}\supset\mathbb{F}$ is an extension obtained by adjoining a root of some well-chosen $f\in\mathbb{F}[X]$, in which case $\mathbb{K}\cong\mathbb{F}[X]/(f)$. The extension $\mathbb{C}\supset\mathbb{R}$ can be obtained by adjoining $i$, which is a root of $X^2+1$, and hence $\mathbb{C}\cong\mathbb{R}[X]/(X^2+1)$. Similarly $\operatorname{GF}(2^4)\supset\operatorname{GF}(2)$ can be obtained by adjoining a root of $X^4+X+1$, and so $\operatorname{GF}(2^4)\cong\operatorname{GF}(2)[X]/(X^4+X+1)$.