Does it matter whether we consider field extensions as true supersets of a field?

Question

I am trying to wrap my head around field extensions, and am using Pinter's abstract algebra. He presents a basic theorem on field extensions, namely that given any field $F$, and any polynomial $a(x)$ over $F$, we can always find a field $E$ containing $F$, and some element $c \in E$ s.t. $c$ is a root of $a(x)$ w.r.t. the operations in $E$.

As presented, the theorem seems to imply that $F$ is a 'genuine' subset of $E$. Also prior to stating and proving this theorem, we developed some theory where we assume a priori that we already have some extension $E$, with $F \subset E$.

The reason I am wondering, is that the proof we are presented with doesn't quite show that given any $F$ we have an extension $E$ with $F \subset E$. Rather, we show that actually, $F \cong H$, where $H$ is actually the subfield of $F[x]/(p(x))$ containing all the constant polynomials, and $p(x)$ is an irreducible factor of $a(x)$ in $F[x]$. So we actually prove that $F$ is isomorphic to another field $H$, and that this field $H$ has an extension, $F[x]/(p(x))$.

I assumed that since $F$ and $H$ are isomorphic, then defining a field $E_2$ such that $F \subset E_2$ and $E_2 \cong F[x]/(p(x))$ would be very easy. However I ran into problems actually coming up with this set and isomorphism. I took $E_2 = (E_1 \setminus H) \cup F$, and defined addition and multiplication on $E_2$ in a 'natural' way, but there were so many cases to consider that I struggled to actually prove it was even a field, specifically, with proving additive associativity in the most general case.

This leaves me with two questions: (i) can we simply consider $F$ and $H$ to be the same? That is, that $F[x]/(p(x))$ really is just an extension of $F$? (ii) Is there an easy way to get a 'true' extension of $F$ easily from the extension of $H$?

Answer 1

This is a very good question. Look, $F$ and $H$ are not the same by the rules of set theory. However, they are so naturally isomorphic that mathematicians usually just say they are the same thing. The only difference is that we call the elements $\alpha+(p(x))$ instead of $\alpha$. This is the only difference, this change of names respects all the operations of $F$. So in some sense, yes, they are the same object.

That being said, I like when things are as formal as possible. Such extensions are used a lot, and I don't always feel very comfortable knowing that it is not a "real" extension. But there is an easy way to fix it.

Note that the cardinality of $F[x]/(p)$ is $\leq\max\{\aleph_0, |F|\}$. So take any larger cardinal $\kappa$ and consider the set $X:=F\cup\kappa$. We have the natural isomorphism $H\to F$, and also can define any injective function $(F[x]/(p))\setminus H\to X\setminus F$. So together we obtain an injective function $f:F[x]/(p)\to X$ which satisfies $f(\alpha+(p))=\alpha$ for all $\alpha\in F$.

Let $L=Im(f)$, this is a set which really contains $F$. And now we can naturally give it a field structure by:

$f(\alpha)+f(\beta)=f(\alpha+\beta)$

$f(\alpha)f(\beta)=f(\alpha\beta)$

You can easily check that this turns $L$ into a field, and it extends the usual operations we had on $F$ before. And $f$ is now an isomorphism of fields. So now we have a real extension of $F$, and the polynomial $p$ indeed has a root in it. (it is the image of the "root" from $F[x]/(p)$)

Answer 2

(i) NO. (ii) NO. (iii) Yes.

$\;\;\;\;\;$ By definition, a field $\Bbb E$ is an $\it{extension}$ $\it{field}$ of a field $\Bbb F$, i.e. $(\Bbb E/\Bbb F)$, when $\Bbb F$ is a $\Bbb E$ subring. Isomorphic fields are NOT necessarily equal. In the OP, $\Bbb F$ and $\Bbb H$ are not equal and should not be formally treated as such. It is also untrue that $\frac{\Bbb F[x]}{\langle p(x)\rangle}$ is an extension field of $\Bbb F$ or, equivalently, that $\Bbb F$ is a subfield of $\frac{\Bbb F[x]}{\langle p(x)\rangle}$ although $$\Bbb F\approx\widehat{\Bbb F}:=\{\alpha+\langle p(x)\rangle:\alpha\in\Bbb F\}$$ $\;\;\;\;\;$$\;\;\;\;\;$ As for the fundamental theorem from the OP, called Kronecker's theorem, given a $\Bbb F[x]$ irreducible $p(x)$ we can easily construct an extension field $\Bbb E$ of $\Bbb F$ in which $p(x)$ has a root by "identifying" $\Bbb F$ with $\widehat{\Bbb F}$. Let $$\Bbb E=(\frac{\Bbb F[x]}{\langle p(x)\rangle}\setminus \widehat{\Bbb F})\cup\Bbb F$$ Define a bijection $\psi:\Bbb E\to\frac{\Bbb F[x]}{\langle p(x\rangle)}$ by $\psi|_{\Bbb E\setminus\Bbb F}=\text{id}$ and $\psi|_{\Bbb F}:\Bbb F\rightarrow_{\small\approx}\widehat{\Bbb F}$. Finally, $$u\mathbf +v:=\psi^{-1}(\psi(u)+\psi(v))\;\;\;\;\;\;u\mathbf\cdot v:=\psi^{-1}(\psi(u)\cdot\psi(v))\;\;\;\;\;$$ defines on $\Bbb E$ a ring structure under which $\psi$ is a ring isomorphism and thus $\Bbb E/\Bbb F$.