Definition of field extensions: Can we talk about extensions like $\mathbb{F}_3(5)/\mathbb{F}_3$? First of all, sorry for this absolute beginner question. It has been bothering me a lot and I think I am just getting something conceptually wrong here.
I am learning about extension fields and use the following definition:

An extension field $F(\alpha)/F$ is the smallest field containing $F$ and $\alpha$.

What I am confused about is, what values can be used for $\alpha$ - especially, when working with finite fields of the form $\mathbb{F}_p$.
I have been trying to show that the minimal polynomial of $\sqrt{\alpha}$ over $\mathbb{F}_3(\alpha)$ is $x^2 - \alpha$. First it seemed obvious, since the roots of $x^2 - \alpha$ in $\mathbb{F}_3(\alpha)$ are $\sqrt{\alpha}$ and $2\sqrt{\alpha}$, both of which are not in $\mathbb{F}_3(\alpha)$. On a second thought I got uncertain about my argument, though. What if $\alpha = 1$?
According to above definition $\mathbb{F}_3(1) = \mathbb{F}_3$ and $\mathbb{F}_3(\sqrt{1})/\mathbb{F}_3(1) = \mathbb{F}_3$. Even though we are not really extending the field $\mathbb{F}_3(1)$, we should still be able to talk about the extension field $\mathbb{F}_3(\sqrt{1})/\mathbb{F}_3(1)$. But then both $\sqrt{1} = 1$ and $2\sqrt{1} = 2$ are in $\mathbb{F}_3(1)$ and $x^2 - 1 = (x + 1) (x + 2)$, in which case $x^2 - \alpha$ would not be irreducible.
This confusion made me think generally about what would happen for different values of $\alpha$. If $\alpha = 2$, we would actually extend the field $\mathbb{F}_3(2) = \mathbb{F}_3$, because $2$ has no roots in $\mathbb{F}_3(2)$ and if $\alpha = \sqrt{2}$, then $\mathbb{F}_3(\sqrt{2})/\mathbb{F}_3$ actually extends $\mathbb{F}_3$ and $\mathbb{F}_3(\sqrt{\sqrt{2}})/\mathbb{F}_3(\sqrt{2})$ is also an actual extension of $\mathbb{F}_3(\sqrt{2})$ (i.e. not equal to $\mathbb{F}_3(\sqrt{2})$ itself).
But what about a value like $\alpha = 5$? Would $5$ be treated as $2$ (as if it was an element of $\mathbb{F}_3$)? Or would the elements of $\mathbb{F}_3(5)$ be of the form $a + 5b$, where $a, b \in \mathbb{F}_3$? This would seem a bit strange. On the other hand, why would we interpret $5$ as an element of $\mathbb{F}_3$, when we want to extends the field with something that is explicitly not in the field (like in the case of $\sqrt{2})$.
To summarise, I have the following specific questions:

*

*Is my reasoning that $x^2 - \alpha$ is irreducible in $\mathbb{F}_3(\alpha)$ correct? And if so, what about the case $\alpha = 1$?

*Would the extension $\mathbb{F}_3(5) / \mathbb{F}_3$ be the same as $\mathbb{F}_3(2)/\mathbb{F}_3 = \mathbb{F}_3$ or would the elements of $\mathbb{F}_3(5)$ be of the form $a + 5b$, where $a, b \in \mathbb{F}_3$? In the first case, why would we interpret $5$ as being an element of $\mathbb{F}_3$ as opposed to $\sqrt{2}$?

I know, these are very basic questions, but I really want to get these foundations right and would appreciate some help.
 A: Trying to clear some of the fog.

Let's start from the question in the title. When discussing field extensions of the form $k(\alpha)$, where $k$ is a known field, and $\alpha$ is an entity usually not in $k$, we need to exercise some care. The notation $k(\alpha)$ means the smallest field containing both $k$ and $\alpha$, but that simple description is oversimplified without some suitable context. Namely, we are tacitly assuming that there exists at least one field, call it $\Omega$ containing both $k$ and $\alpha$. And then we take the smallest subfield within the umbrella field $\Omega$ that contains both $k$ and $\alpha$. This is actually necessary. After all, for $k(\alpha)$ to be a field we need to know how the arithmetic operations involving $\alpha$ and elements of $k$ work. The definition of the field $\Omega$ contains such details, and we can proceed safely with the usual definition (= take the intersection of all the subfield of $\Omega$ that contain both $k$ and $\alpha$, and then prove that the said intersection is itself a subfield, and obviously minimal among such subfields).
On a first course in extension fields this point does not get much attention. That is largely because we are tacitly working inside $\Omega=\Bbb{C}$, where we can merrily adjoin complex numbers to $\Bbb{Q}$ and friends. But when extending fields like $\Bbb{F}_3$ we don't have a convenient $\Omega$ at our disposal, and need to exercise some care. To make sense of $\Bbb{F}_3(5)$ we first need to interpres $5$ in the language that can extend that of $\Bbb{F}_3$. It is a stretch to assign to $5$ a meaning other than
$$5=1+1+1+1+1.$$
But this gives us the decisive clue. In the field $\Bbb{F}_3$ we have $1+1+1=0$. So, by associativity of addition we must have
$$5=(1+1+1)+(1+1)=0+2=2.$$
Hence we can conclude that $5$ already is an element of $\Bbb{F}_3$, and we don't need to extend it at all!
The fields of the form $\Bbb{F}_3(\alpha)$, where $\alpha$ is a zero of some irreducible polynomial $f(x)\in\Bbb{F}_3[x]$, require yet another way of thinking . The usual process is to define the extension field $K=\Bbb{F}_3[x]/\langle f(x)\rangle$ as a quotient ring of the polynomial ring by the maximal ideal generated by $f(x)$. Then we can declare $\alpha$ to be the coset $x+\langle f(x)\rangle$ that is then a zero of $f(x)$ in $K$. So $K$ can, at least temporarily, assume the role of $\Omega$. At that point the usual argument shows that actually $K=\Bbb{F}_3(\alpha)$.

Another theme in the question was whether $g(x)=x^2-\alpha$ is always irreducible over $\Bbb{F}_3(\alpha)$. Here the answer is it depends. As $g(x)$ is quadratic, it is irreducible over a field $K$ (containing its coefficients) if and only if it does not have a root in $K$. If we select $\alpha$ to be a zero of $x^2+1=x^2-2$, it turns out that then $x^2-\alpha$ has a zero in $\Bbb{F}_3(\alpha)$. Because $\alpha^2+1=0$ I will denote $\alpha=i="\sqrt{-1}"$ to make the calculations easier to follow. We see that $1-i\in\Bbb{F}_3(i)$ and
$$
(1-i)^2=1^2-2i+i^2=-2i=i,
$$
because $-2=1$ and hence also $-2i=i$. So $g(x)=x^2-i$ factors like
$$g(x)=(x-1+i)(x+1-i)\in\Bbb{F}_3(i)[x].$$
On the other hand, if we select $\alpha=1-i\in\Bbb{F}_3(i)=\Bbb{F}_3(\alpha)$ it turns out that $x^2-\alpha$ is irreducible. One way of showing that is to observe that if $\beta$ satisfies $\beta^2=\alpha$, then $\beta^4=i$, and it quickly follows that $\beta$ has multiplicative order sixteen. The multiplicative group of $F_3(\alpha)$ has only $3^2-1=8$ elements, so it cannot contain an element of order sixteen. Therefore $x^2-(1-i)$ is irreducible over $\Bbb{F}_3(1-i)=\Bbb{F}_3(i)$, and $\beta$ generates a quadratic extension of $\Bbb{F}_3(i)$. That is, a field of $3^4=81$ elements.
Compare this with extensions of rationals. The polynomial $x^2-\sqrt2$ is irreducible over $\Bbb{Q}(\sqrt2)$. For its zero, call it $\gamma$, is a root of $x^4-2$ and that polynomial is irreducible over $\Bbb{Q}$ by Eisenstein. On the other hand many elements of $L=\Bbb{Q}(\sqrt2)$ have square roots in $L$. Such as $\alpha=(1+\sqrt2)^2=3+2\sqrt2$. Hence $L(\sqrt{3+2\sqrt2})=L=\Bbb{Q}(3+2\sqrt2)$.
