Questions about the proof used in demonstrating the existence of a field extension

Question

In Chapter 27 of Pinter's "A Book of Abstract Algebra", a proof is provided for the existence of an extension field $E$ of $F$. I think the only necessary background is some framework that was omitted from the proof but provided on earlier pages:

1. $\sigma_c$ is the substitution function defined as $\sigma_c(a(x))=a(c)$ where $c \in E$, the extension field of $F$, which is a subfield of $E$. $\sigma_c$ has been proven to be a homomorphism.

2. $J_c$ (denoted as $J$ in the proof) is the kernel of $\sigma_c$ and has been demonstrated to be a principal ideal that is expressed as $J_c=\langle p(x) \rangle$, where $p(x)$ is the unique monic polynomial of lowest degree in $J_c$

3. $\operatorname{range} \sigma_c = \{a(c): a(x) \in F[x]\}$, and is denoted as $F(c)$

4. Using the fundamental homomorphism theorem, we conclude from 1,2 and 3 that $F(c) \cong F[x]/\langle p(x) \rangle$

With these established, here is the proof provided by Pinter:

I understand the majority of the proof; however, I have denoted in $\color{#c00}{red}$ two areas of the proof that have me a little confused.

Firstly, it is not immediately apparent to me that the range of $h$ is a proper subset of all cosets found in the quotient ring. Specifically, how do I know that the cosets that each element of $F$ map to do not match up $1$ to $1$ with all of the cosets found in $F[x]/\langle p(x) \rangle$?

It seems like there are obviously many different cosets that have constant polynomial representatives ...but I do not see how I know for certain that the "number" (probably not the best term) of cosets that elements from $F$ map into is "less than" (probably not the best term) the number of cosets that, in total, comprise $F[x]/\langle p(x)\rangle$. i.e. there are clearly cosets of $F[x]/\langle p(x)\rangle$ that do not have constant polynomial representatives but I do not see why.

Secondly, while I understand the construction that is taking place in the $\color{#c00}{bracketed}$ portion of the proof, I do not see the motivation to establish this. Is it just to confirm that the field $F[x]/\langle p(x) \rangle$, when interpreted from the perspective of a polynomial, is behaving like a polynomial? Also, this is the first time I have ever seen a variable itself (e.g. $J+x$) function as a root...as opposed to some constant...but I suppose this arises because $p(x)$ is known to map to $0$ when $x$ is substituted for $c$ via $\sigma_c$.

Any clarification is greatly appreciated!

Answer 1

Answering the first part of the OP's question...

The element (coset) $\langle p(x) \rangle + x \in F[x]/\langle p(x) \rangle$ can't be put represented by a constant polynomial.

The degree of $p(x)$ is greater than or equal to $2$ (we're looking at a non-trivial extension) and the product of $p(x)$ with any other polynomial has degree greater than or equal to $2$. Also, in general, if $g(x)$ and $h(x)$ are two polynomials

$\; \text{IF } \text{degree of } g(x) \ne \text{degree of } h(x) \text{ THEN } \text{degree of } \big[g(x) + h(x)\big] = \text{max}\big(deg(g),deg(h)\big)$

Note that the statement Basic Theorem of Field Extensions is vacuous if every polynomial in $F$ has a root - then every coset has a constant representative.

---

For the second part, I think we can agree that under the equivalence relation (abusing notation a bit - $J$ is not a polynomial)

$\quad (J+x)^{n} \big ( = \sum _{k=0}^{n}{n \choose k}J^{k}x^{n-k} \big ) \equiv J + x^n$

So under the equivalence relation (abusing notation a bit - $J$ is not a polynomial)

$\quad (J + a_n)(J+x)^{n} \equiv (J + a_n)(J+x^n) \equiv J +a_n x^n$

filling in some proof detail.

More to the point, if you like working with cosets, a representative for the coset $J + x$ is $0 + x = x$, and so raising the coset $J + x$ to the $n^{-th}$ power is equal to the coset $J + x^n$.

---

You can think of think of $J$ as one big zero. When you want to find a representative in the coset you can employ Euclidean division, dividing out your 'starter polynomial' by $p(x)$ (the big zero) to get a representation with a smaller degree than $p(x)$,

There is not much dividing you can do when the starting representative is a constant polynomial, or for that matter any polynomial with degree less than $p(x)$.

Answer 2

If $p(x)\in F[x]$ is irreducible then $K=F[y]/(p(y))$ is a field and $$p(x)=(x-y) g(x)\in K[x]$$ The root $y$ is not a variable, it is an element of $K$.

Then we take an irreducible factor $h(x)| g(x)\in K[x]$ and we repeat with $K[z]/(h(z))$, until $p(x)$ splits completely in the obtained splitting field.

Concretely with $p(x)=x^2+1\in \Bbb{R}[x]$ we get $$K=\Bbb{R}[y]/(y^2+1),\qquad p(x)=(x+y)(x-y)\in K[x]$$

Answer 3

That notation obscures the structure preserving action of the natural map into the quotient. Write this ring hom as $\,h: F[x]\to F[x]/(p(x)),\,$ where $\,f\mapsto h(f) = \bar f = f + J,\,$ with $\,J = p(x)F[x].\,$ Then

$$\begin{align} \bar 0\, &\,= h(p(x))\ \ \ {\rm by}\ \ p(x)\in J\\[.2em] &=\, h(a_0 \ \ \ + \ \ a_1\ x\ \ +\ \ \cdots\ \ +\ \ a_n\ x^n)\\[.2em] &=\, h(a_0)\!+\! h(a_1)h(x) + \cdots + h(a_n) h(x)^n\ \ \ {\rm by}\ h\ {\rm a\ ring\ hom}\\[.2em] & =\, \ \ \bar a_0\ \ \ +\ \ \ \bar a_1\ \overline{x}\ \ +\ \ \cdots\ \ +\ \ \bar a_n \ \overline{x}^n\\[.2em] & =\ \bar p(\bar x) \end{align}\qquad$$

Therefore we conclude that $\,\bar x\,$ is root of $\,\bar p\,$ in the quotient ring.

Conceptually it is essential to ignore the internal structure (representation) of the elements of the quotient (here as cosets). Rather, we should focus on the essential properties of this generic root adjunction construction. Namely, for a commutative ring $R$ with $1,\,$ and a polynomial $\,p(x)\in R[x]\,$ we wish to adjoin to $R$ a "generic" root $\,\omega\,$ of $\,p(x),\,$ i.e. we wish this adjunction to be as general as possible, so that adjoining any specific root $\,r\,$ will be a special case, obtainable by specializing the generic root to the specific root (yielding a ring image of the generic ring adjunction).

For example, in $\,R[x]/(x^n-2) = R[\omega]\,$ we have adjoined $\,\omega\,$ being a generic $n$'th root of $\,2.\,$ Any ring calculations in this generic ring (unexceptionally) persist to hold true if we specialize to specific roots, i.e. they have natural images in any ring containing both an $n$'th root of $\,2\,$ and an image $R$ (or $R/I),\,$ i.e. the generic root adjunction $\,R[\omega]\,$ has a natural image in any specific root adjunction.

Analogous is the fact that arithmetic $\!\bmod 1001 = 7(11)13\,$ persists to hold true $\!\bmod 7,\,11\,$ & $\,13,\,$ because $\,\Bbb Z/1001\,$ is the most general ring containing an image of $\,\Bbb Z\,$ where $1001 = 0\,$ (see the method of simpler multiples for analogous common elementary instances).

Informally, by working in the most general way possible - assuming only that the adjunction satisfies ring laws and that $\omega\,$ is a root of $\,p(x)\,$ we obtain the maximally general ("universal") root adjunction. This universality will be made more precise if you study universal algebra - where this can be viewed as a special case of the universal properties of polynomial and quotient rings.

Answer 4

Indeed it is super-super confusing. So we both still don't get this, and I hope by knowing there is someone else out there doesn't get it too, would make both of us feeling better. Hence this answer.

The confusion starts when it says:

We will prove that $F[x]/J$ is an extension of $F$ by identifying each element $a$ in $F$ with its coset $J + a$.

It doesn't mention at all the motivation behind this! Why by "identifying" will prove that it is an extension? While $F[x]/J$ has been showed before it is already a field and by definition its isomorphism $F(c)$ contains $c$.

By reading the answers and comments above, it is suggested that it wants to show that $F \to F[x] \to F[x]/J$ is injective, but still why we need the "identifying" and "showing it is injective"? Both looks the same explanation and still cannot find out the motivation.

The explanation after that is pretty much straightforward, but then it comes to the conclusion:

This subfield is therefore an isomorphic copy of $F$, which may be identified with $F$, so $F[x]/J$ is an extension of $F$.

Which is pretty much just a repeat of what has been said before, so again, nothing new could be found out here.

Then comes the most amusing part of all like trying to infuse a confusion of the maximum order. It introduces the new variable $x$ out of thin air:

Finally, if $p(x) = a_0 + a_1 x + ⋯ + a_n x^n$, let us show that the coset $J + x$ is a root of $p(x)$ in $F[x]/J$.

What we ("I" if you are not part my group) have understood so far, $p(x)$ is a polynomial that if we replace $x$ with $c$, not just any random values, then it would spit out the result $J = \{0\}$, hence it should use $c$ instead of $x$, be it $p(c)$, not $p(x)$? I mean it should be written $p(c)$ which is clearly an element of the kernel $J$ of a homomorphism $\sigma_c : F[x] \to F(c)$.

Thus as $F(c) \cong F[x]/J$, it does make sense that if we put $J + c$ or $\bar{c}$ into polynomial $p(x)$ we will get $J$ as the result.

In which also brings out my last confusion, why do we still need to prove (AGAIN) that:

$\bar{a_0} + \bar{a_1} \bar{c} + ⋯ + \bar{a_n} \bar{c}^n = J$

(or in the text written as $\bar{a_0} + \bar{a_1} \bar{x} + ⋯ + \bar{a_n} \bar{x}^n = J$)

??