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.
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 $L(\sqrt{3+2\sqrt2})=L=\Bbb{Q}(3+2\sqrt2)$.