MHB Irreducible Polynomials and Quotient Rings - Rotman Proposition 3.116

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.8 Quotient Rings and Finite Fields ...

I need help with an aspect of the proof of Proposition 3.116

Proposition 3.116 and its proof reads as follows:View attachment 4696
View attachment 4697I need help with an aspect of the proof of (ii) above. In the proof of (ii) we read the following:

" ... ... $$k[x]/ (p(x))$$ is a field and, hence $$\text{ I am } \phi $$ is a field; that is, $$\text{ I am } \phi $$ is a subfield of $$K$$ containing $$k$$ and $$z$$, and so $$ k(z) \subseteq \text{ I am } \phi $$. ... ..."

Can someone please explain exactly why/how the above follows ... ...

... that is how, given that $$k[x]/ (p(x))$$ is a field, it then follows that $$\text{ I am } \phi $$ is a field; that is, $$\text{ I am } \phi $$ is a subfield of $$K$$ containing $$k$$ and $$z$$, and so $$k(z) \subseteq \text{ I am } \phi $$

Hope someone can help ... ...

PeterNOTE : The proof of (ii) above mentions Theorem 3.112, so in order for MHB readers to follow the above post, I am providing the statement of the theorem, as follows:View attachment 4698

 

[Deveno]

$\text{im }\phi$ is a field because it is isomorphic to $k[x]/(p(x))$ by the fundamental isomorphism theorem (that is: $\text{ker }\phi = (p(x))$).

Here is a "standard" example:

Let $p(x) = x^2 - 2$, with $k = \Bbb Q$ and $K = \Bbb R$.

Now $z = \sqrt{2} \in \Bbb R$, but $z \not\in \Bbb Q$, and $p(z) = 0$, that is, $z$ is a root of $p$. It is also clear that $p(x)$ is irreducible over $\Bbb Q$, for if not, then it would split into two linear factors in $\Bbb Q[x]$, one of which would have to be (an associate of) $x - z$, which contradicts the irrationality of $z = \sqrt{2}$.

Now the elements of $\Bbb Q[x]/(x^2 - 2)$ look like this:

$a + bx + (x^2 - 2)$, where $a,b \in \Bbb Q$.

Mapping the coset of $x$, which is $x + (x^2 - 2)$ to $z$, we get an isomorphism with $\Bbb Q(\sqrt{2})$, which has elements of the form:

$a + bz = a + b\sqrt{2}$ for $a,b \in \Bbb Q$, and said elements form a subfield of the real numbers.

 

[Euge]

In the proof of (ii) we read the following:

" ... ... $$k[x]/ (p(x))$$ is a field and, hence $$\text{ I am } \phi $$ is a field; that is, $$\text{ I am } \phi $$ is a subfield of $$K$$ containing $$k$$ and $$z$$, and so $$ k(z) \subseteq \text{ I am } \phi $$. ... ..."

Can someone please explain exactly why/how the above follows ... ...

... that is how, given that $$k[x]/ (p(x))$$ is a field, it then follows that $$\text{ I am } \phi $$ is a field; that is, $$\text{ I am } \phi $$ is a subfield of $$K$$ containing $$k$$ and $$z$$, and so $$k(z) \subseteq \text{ I am } \phi $$

The proof of part (ii) uses the map $\varphi$ without the mentioning how it's constructed, which is why, I think, you're having problems with (ii). Also, the condition $\varphi(a) = a$ for all $a\in k$ is technically false since elements of $k$ do not belong in $k[x]/(p(x))$. It should be $\varphi(a + (p(x))) = a$ for all $a\in k$.

I'll do things differently, although the proof won't be terse. Consider the mapping $\varphi_z : k[x] \to K$ given by $\varphi_z(f(x)) = f(z)$. It is a ring homomorphism with kernel $I = \{f\in k[x]: f(z) = 0\}$. Since $I$ is an ideal of $k[x]$, $I = (m(x))$ for some monic polynomial $m$ of minimal degree. Since $p\in I$, $m(x) \, |\, p(x)$, but as $p(x)$ is irreducible, $p(x) = u\cdot m(x)$ for some unit $u$ of $k$. So $I = (p(x))$ and $\text{im}(\varphi_z)$ is isomorphic to $k[x]/I$, a field by Theorem 3.112. Thus $\text{im}(\varphi_z)$ is a subfield of $K$. Since every $a\in k$ satisfies $a = \varphi_z(a)$, $\text{im}(\varphi_z)$ contains $k$; as $z = \varphi_z(x)$, $\text{im}(\varphi_z)$ contains $z$. Therefore $\text{im}(\varphi_z)$ is a subfield of $K$ containing $k$ and $z$. By definition of $k(z)$, $k(z) \subseteq \text{im}(\varphi_z)$. On the other hand, elements of $\text{im}(\varphi_z)$ are finite sums of expressions of the form $a_iz^i$, where $a_i \in k$. So $\text{im}(\varphi_z) \subseteq k(z)$. Consequently, $\text{im}(\varphi_z) = k(z)$, and the mapping $\varphi_z$ induces an isomorphism $\varphi: k[x]/I \to k(z)$ such that $\varphi(f(x) + I) = \varphi_z(f(x))$, i.e., $\varphi(f(x) + I) = f(z)$. In particular, $\varphi(x + I) = z$ and $\varphi(a + I) = a$ for all $a\in k$.

 

[Deveno]

I'd like to point out a "bigger picture" sort of thing.

What we're after, is a study of fields-what are they like, how can we tell them apart, how can we characterize them (explicitly, or implicitly)?

The answer turns out to be intimately intertwined with the study of polynomials. Why should this be so?

Part of the reason has to do with the structure of $F[x]$ as a RING: namely, it is a quite "special" kind of ring-a Euclidean domain. Euclidean domains are more "special" than another, "looser" kinds of rings, including:

Integral domains (so, Euclidean domains have no zero divisors)
GCD domains (so every pair of polynomials has a greatest common divisor)
UFDs (so every polynomial can be factored uniquely into irreducible/prime polynomials)
PIDs (so every ideal of $F[x]$ is generated by a single polynomial. If we require this polynomial be monic, it is uniquely determined).

So with $F[x]$, because it IS Euclidean, we get a whole host of desirable properties.

$F[x]$ "inherits" a lot of its nice properties from $F$. The main thing it lacks is multiplicative inverses. We can try to "fix" this in two different ways:

1. Embed $F[x]$ in some larger ring that has inverses. The natural candidate here is $F(x)$, the field of fractions.

2. "Mod out" some prime element-this is the same approach we use to turn the integers into the field the integers modulo $p$.

The second way is more *interesting* because it leads to fields that are only "a little bit bigger" than $F$ (whereas $F(x)$ is rather large). In fact, we can quantify "how much bigger" in an at-first unobvious way: we can regard the larger field as a vector space over $F$, and record it's dimension.

 

[Math Amateur]

I'd like to point out a "bigger picture" sort of thing.

What we're after, is a study of fields-what are they like, how can we tell them apart, how can we characterize them (explicitly, or implicitly)?

The answer turns out to be intimately intertwined with the study of polynomials. Why should this be so?

Part of the reason has to do with the structure of $F[x]$ as a RING: namely, it is a quite "special" kind of ring-a Euclidean domain. Euclidean domains are more "special" than another, "looser" kinds of rings, including:

Integral domains (so, Euclidean domains have no zero divisors)
GCD domains (so every pair of polynomials has a greatest common divisor)
UFDs (so every polynomial can be factored uniquely into irreducible/prime polynomials)
PIDs (so every ideal of $F[x]$ is generated by a single polynomial. If we require this polynomial be monic, it is uniquely determined).

So with $F[x]$, because it IS Euclidean, we get a whole host of desirable properties.

$F[x]$ "inherits" a lot of its nice properties from $F$. The main thing it lacks is multiplicative inverses. We can try to "fix" this in two different ways:

1. Embed $F[x]$ in some larger ring that has inverses. The natural candidate here is $F(x)$, the field of fractions.

2. "Mod out" some prime element-this is the same approach we use to turn the integers into the field the integers modulo $p$.

The second way is more *interesting* because it leads to fields that are only "a little bit bigger" than $F$ (whereas $F(x)$ is rather large). In fact, we can quantify "how much bigger" in an at-first unobvious way: we can regard the larger field as a vector space over $F$, and record it's dimension.

Deveno, Euge I cannot thank you enough for your help on this matter ... I was really struggling with it ... but am now seeing the light ... a bit anyway ...

I am now working through your posts ... line by line ...

Thanks again for all your help ... it is definitely appreciated ...

Peter

 

[Math Amateur]

$\text{im }\phi$ is a field because it is isomorphic to $k[x]/(p(x))$ by the fundamental isomorphism theorem (that is: $\text{ker }\phi = (p(x))$).

Here is a "standard" example:

Let $p(x) = x^2 - 2$, with $k = \Bbb Q$ and $K = \Bbb R$.

Now $z = \sqrt{2} \in \Bbb R$, but $z \not\in \Bbb Q$, and $p(z) = 0$, that is, $z$ is a root of $p$. It is also clear that $p(x)$ is irreducible over $\Bbb Q$, for if not, then it would split into two linear factors in $\Bbb Q[x]$, one of which would have to be (an associate of) $x - z$, which contradicts the irrationality of $z = \sqrt{2}$.

Now the elements of $\Bbb Q[x]/(x^2 - 2)$ look like this:

$a + bx + (x^2 - 2)$, where $a,b \in \Bbb Q$.

Mapping the coset of $x$, which is $x + (x^2 - 2)$ to $z$, we get an isomorphism with $\Bbb Q(\sqrt{2})$, which has elements of the form:

$a + bz = a + b\sqrt{2}$ for $a,b \in \Bbb Q$, and said elements form a subfield of the real numbers.

Hi Deveno,

Just having a bit of trouble with your first reply post ...

Can you help me to identify the ring homomorphism and the ring isomorphism of the First Isomorphism Theorem in your example ... these are the $$\phi$$ and $$\tilde{ \phi }$$ of Rotman's notation for Theorem 3.109 (First Isomorphism Theorem ... for rings) ... see below ...

Rotman's Theorem 3.109 (First Isomorphism Theorem ... for rings) reads as follows:View attachment 4699Can you please help me to identify the $$\phi$$ and $$\tilde{ \phi }$$ in your post ...

Peter

 

[Math Amateur]

The proof of part (ii) uses the map $\varphi$ without the mentioning how it's constructed, which is why, I think, you're having problems with (ii). Also, the condition $\varphi(a) = a$ for all $a\in k$ is technically false since elements of $k$ do not belong in $k[x]/(p(x))$. It should be $\varphi(a + (p(x))) = a$ for all $a\in k$.

I'll do things differently, although the proof won't be terse. Consider the mapping $\varphi_z : k[x] \to K$ given by $\varphi_z(f(x)) = f(z)$. It is a ring homomorphism with kernel $I = \{f\in k[x]: f(z) = 0\}$. Since $I$ is an ideal of $k[x]$, $I = (m(x))$ for some monic polynomial $m$ of minimal degree. Since $p\in I$, $m(x) \, |\, p(x)$, but as $p(x)$ is irreducible, $p(x) = m(x)$. So $I = (p(x))$ and $\text{im}(\varphi_z)$ is isomorphic to $k[x]/I$, a field by Theorem 3.112. Thus $\text{im}(\varphi_z)$ is a subfield of $K$. Since every $a\in k$ satisfies $a = \varphi_z(a)$, $\text{im}(\varphi_z)$ contains $k$; as $z = \varphi_z(x)$, $\text{im}(\varphi_z)$ contains $z$. Therefore $\text{im}(\varphi_z)$ is a subfield of $K$ containing $k$ and $z$. By definition of $k(z)$, $k(z) \subseteq \text{im}(\varphi_z)$. On the other hand, elements of $\text{im}(\varphi_z)$ are finite sums of expressions of the form $a_iz^i$, where $a_i \in k$. So $\text{im}(\varphi_z) \subseteq k(z)$. Consequently, $\text{im}(\varphi_z) = k(z)$, and the mapping $\varphi_z$ induces an isomorphism $\varphi: k[x]/I \to k(z)$ such that $\varphi(f(x) + I) = \varphi_z(f(x))$, i.e., $\varphi(f(x) + I) = f(z)$. In particular, $\varphi(x + I) = z$ and $\varphi(a + I) = a$ for all $a\in k$.

Hi Euge,

Just working through your post above ... which, by the way is EXTREMELY helpful ...

In the above post you write:

" ... ... By definition of $k(z)$, $k(z) \subseteq \text{im}(\varphi_z)$. ... ... "Can you please explain exactly how the definition of $k(z)$, leads to the conclusion that $k(z) \subseteq \text{im}(\varphi_z)$?

Peter***EDIT***

I have been thinking about the above statement and my thoughts are as follows:

You have shown in your proof that:

" ... ... $\text{im}(\varphi_z)$ is a subfield of $K$ containing $k$ and $z$ ... ... "

So given that $\text{im}(\varphi_z)$ is a field, it is closed under addition and multiplication and hence contains elements like

$$a_0 + a_1z + a_2 z^2 + \ ... \ ... \ + a_n z^n$$ ...

... that is $\text{im}(\varphi_z)$ contains polynomials $$f(z), g(z)$$ with coefficients in $$k$$ ...

... so ... again, being a field, it contains rational functions of the form $$f(x)g(x)^{-1} = f(x)/g(x)$$ ...

hence it follows that $k(z) \subseteq \text{im}(\varphi_z)$ ...

Is that correct?*** Now if that is correct, then I think I understand your proof ... which incidentally is so much clearer than Rotman's proof ... but intuitively I am bothered by the following ... ...

The elements of elements of $\text{im}(\varphi_z)$, as you point out, are finite sums of expressions of the form $a_iz^i$, where $a_i \in k$ ... but then, how do members of $$k(x)$$ like $$f(z)/g(z)$$ belong ... they do not (intuitively at least) seem to fit the form $a_iz^i$ ... ...

Can you help?

Peter

Peter

 

[Euge]

Hi Peter,

Your analysis of $k(z) \subseteq \text{im}(\varphi_z)$ is correct, but what I meant is this. By definition of $k(z)$, $k(z)$ is the smallest subfield of $K$ containing $k$ and $z$. (Your book defines it this way, right?) So any subfield $F$ of $K$ containing $k$ and $z$ satisfies $k(z) \subseteq F$. In the proof, $F = \text{im}(\varphi_z)$.

To answer your last question, don't forget that denominators can equal $1$, i.e., you can have $g(z) = 1$. Since $f(z) = \frac{f(z)}{1}$ for all $f(x)\in k[x]$, we have more explicitly that $\text{im}(\varphi_z)$ consists of fractions $\frac{f(z)}{1}$ where $f(x) \in k[x]$. Each of those fractions lie in $k(z)$ (again, using the rational function representation you displayed), so $\text{im}(\varphi_z) \subseteq k(z)$.

If I'm not mistaken, your book defines $k(z)$ the same way I do, and later proves that $k(z)$ consists of rational functions. That being the case, you instinctively used the rational function representation every time you see $k(z)$. Is this correct?

 

[Math Amateur]

Hi Peter,

Your analysis of $k(z) \subseteq \text{im}(\varphi_z)$ is correct, but what I meant is this. By definition of $k(z)$, $k(z)$ is the smallest subfield of $K$ containing $k$ and $z$. (Your book defines it this way, right?) So any subfield $F$ of $K$ containing $k$ and $z$ satisfies $k(z) \subseteq F$. In the proof, $F = \text{im}(\varphi_z)$.

To answer your last question, don't forget that denominators can equal $1$, i.e., you can have $g(z) = 1$. Since $f(z) = \frac{f(z)}{1}$ for all $f(x)\in k[x]$, we have more explicitly that $\text{im}(\varphi_z)$ consists of fractions $\frac{f(z)}{1}$ where $f(x) \in k[x]$. Each of those fractions lie in $k(z)$ (again, using the rational function representation you displayed), so $\text{im}(\varphi_z) \subseteq k(z)$.

If I'm not mistaken, your book defines $k(z)$ the same way I do, and later proves that $k(z)$ consists of rational functions. That being the case, you instinctively used the rational function representation every time you see $k(z)$. Is this correct?

Hi Euge,

Thanks for the further clarification ...

Checked on k(x) and found that Rotman defines k(x) as the fraction field $$\text{Frac}(k[x]) $$ of $$k[x]$$ ... and calls it the function field over $$k$$ (page 241) as follows:https://www.physicsforums.com/attachments/4701

as you will note he 'shows' or notes that the elements of $$k(x)$$ are of the form $$f(x)/g(x)$$ ...Then ... on page 299 he defines $$k(x)$$ as the smallest subfield containing $$k$$ and $$z$$ as follows:https://www.physicsforums.com/attachments/4702I find it somewhat unnerving that an author defines an entity two ways and does not prove the two definitions to be defining the same entity ... but, if Rotman does this somewhere, I cannot find the proof ...

Do you have a reference that gives an clear proof that the smallest subfield containing k and z is actually $$\text{Frac}(k[x]) $$ of $$k[x]$$?

Peter

 

[Euge]

It's an abuse of notation. In the notation $k(x)$, $x$ is an indeterminate -- this notation refers to the function field over $k$. In the notation $k(z)$ (usually read "$k$ adjoined with $z$"), $z$ belongs to some extension field $K$ of $k$; it refers to the smallest subfield of $K$ containing $k$ and $z$. Let $k[z] := \varphi_z(k[x])$. Then $k(z) = \text{Frac}(k[z])$ (as opposed to $k(x) = \text{Frac}(k[x])$). More explicitly, $k(z) = \{f(z)/g(z) : f(x), g(x) \in k[x], g \neq 0\}$.

You've basically done half the argument for why $k(z) = \text{Frac}(k[z])$ -- just replace $\text{im}(\varphi_z)$ with an arbitrary subfield $F$ containing $k$ and $z$ in your argument to get $\text{Frac}(k[z]) \subseteq F$ for any such $F$. The other part is to simply note that the set $\text{Frac}(k[z])$ is a subfield of $K$; since we already know $\text{Frac}(k[z])$ contains $k$ and $z$, we deduce that its the smallest such field, and thus $k(z) = \text{Frac}(k[z])$.

The principle here is that, to prove a set $F$ equals $k(z)$, demonstrate that

1. Every subfield $S$ of $K$ containing $k$ and $z$ must also contain $F$, and

2. $F$ is itself a subfield of $K$ containing $k$ and $z$

 

[Deveno]

There is something else going on, here, too-something which often confuses students.

We would expect a root $z$ of a polynomial $p(x) \in k[x]$ to induce a homomorphism:

$k[x] \to k[z]$.

However, if $p$ is irreducible, we get an unexpected bonus: $k[z] = k(z)$.

Let's see how this plays out explicitly, for $k = \Bbb Q$, and $p(x) = x^2 - 2$, with $z = \sqrt{2}$

It is clear that the RING generated by $\Bbb Q$ and $z = \sqrt{2}$ consists of "polynomials in $\sqrt{2}$" (note this implicitly assumes we are working in some larger ring that already contains this root. For the rationals, this ring is usually taken to be a subring of the complex field $\Bbb C$, since that contains all roots of any real polynomial).

Of course, since $(\sqrt{2})^2 = 2 \in \Bbb Q$, any higher powers of $z\ (= \sqrt{2})$ in $p(z)$ can be "knocked down" to either an element of $\Bbb Q$ or a "linear polynomial expression in $z$", that is:

$a + bz$, with $a,b \in \Bbb Q$, which can write as:

$a + b\sqrt{2}$ (this actually covers BOTH cases, if we permit $b = 0$).

To show that $k[z] = k(z)$ IN THIS CASE, we need merely exhibit a multiplicative inverse for every non-zero element of $\Bbb Q[\sqrt{2}]$ that lies within $\Bbb Q[\sqrt{2}]$.

Such an inverse is:

$c + d\sqrt{2}$, where $c = \dfrac{a}{a^2 - 2b^2}$ and $d = \dfrac{-b}{a^2 - 2b^2}$.

There is one caveat: we must show $a^2 - 2b^2$ is non-zero, or else $c,d$ are ill-defined. But if:

$a^2 = 2b^2$ with $a,b$ rational, it follows that $2 = \left(\dfrac{a}{b}\right)^2$, contradicting the irrationality of $\sqrt{2}$, or else $a = b = 0$, which would mean $a+b\sqrt{2}$ is not non-zero.

The beauty of using the "general theorems" that:

$F[x]/I$ is a field, where $I$ is the ideal generated by a monic irreducible polynomial, is that we don't HAVE to compute inverses to know they exist-the drawback is that knowing they exist doesn't actually tell us how to COMPUTE them.

In short, if the element $z$ that we adjoin (in the RING sense) is *algebraic* over $F$ (it satisfies some polynomial in $F[x]$, which we can then factor into irreducible factors-and thus find the irreducible factor(s) it is a root of), then the "reciprocals" of "polynomials in $z$" are again: "polynomials in $z$", and so the ring-adjuction is the same as the field-adjunction.

In field theory, one sees the adjoined element written with both square and round brackets, as it makes no difference if we are talking about algebraic extensions.

Here is another commonly-cited example:

Consider the complex number:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)$

We have:

$\omega^2 = \left(\dfrac{1}{4} - \dfrac{3}{4}\right) + i\left(-\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}\right)= -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$

$= \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right)$.

Thus $\omega^2 + \omega = -1$, that is, $\omega$ is a root of $x^2 + x + 1$.

Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$ it follows that $\omega$ is a (complex) cube root of $1$ (if you know DeMoivre's Theorem, you can see that:

$\omega^3 = \cos(2\pi) + i\sin(2\pi) = 1 + i0 = 1$).

It can be shown that $x^2 + x + 1$ is irreducible over $\Bbb Q$, so we have:

$\Bbb Q(\omega) = \Bbb Q[\omega] \cong \Bbb Q[x]/(x^2 + x + 1)$.

Convince yourself that for $a,b \in \Bbb Q$, with $a^2 + b^2 \neq 0$, that:

$\dfrac{1}{a + b\omega} = \dfrac{1}{a^2 + b^2 - ab}((a-b) - b\omega)$

(again, the "tricky part" is showing that $a^2 + b^2 - ab \neq 0$. Consider the cases where $ab > 0$ and $ab < 0$ separately-why is $ab = 0$ trivial?).

 

[Math Amateur]

There is something else going on, here, too-something which often confuses students.

We would expect a root $z$ of a polynomial $p(x) \in k[x]$ to induce a homomorphism:

$k[x] \to k[z]$.

However, if $p$ is irreducible, we get an unexpected bonus: $k[z] = k(z)$.

Let's see how this plays out explicitly, for $k = \Bbb Q$, and $p(x) = x^2 - 2$, with $z = \sqrt{2}$

It is clear that the RING generated by $\Bbb Q$ and $z = \sqrt{2}$ consists of "polynomials in $\sqrt{2}$" (note this implicitly assumes we are working in some larger ring that already contains this root. For the rationals, this ring is usually taken to be a subring of the complex field $\Bbb C$, since that contains all roots of any real polynomial).

Of course, since $(\sqrt{2})^2 = 2 \in \Bbb Q$, any higher powers of $z\ (= \sqrt{2})$ in $p(z)$ can be "knocked down" to either an element of $\Bbb Q$ or a "linear polynomial expression in $z$", that is:

$a + bz$, with $a,b \in \Bbb Q$, which can write as:

$a + b\sqrt{2}$ (this actually covers BOTH cases, if we permit $b = 0$).

To show that $k[z] = k(z)$ IN THIS CASE, we need merely exhibit a multiplicative inverse for every non-zero element of $\Bbb Q[\sqrt{2}]$ that lies within $\Bbb Q[\sqrt{2}]$.

Such an inverse is:

$c + d\sqrt{2}$, where $c = \dfrac{a}{a^2 - 2b^2}$ and $d = \dfrac{-b}{a^2 - 2b^2}$.

There is one caveat: we must show $a^2 - 2b^2$ is non-zero, or else $c,d$ are ill-defined. But if:

$a^2 = 2b^2$ with $a,b$ rational, it follows that $2 = \left(\dfrac{a}{b}\right)^2$, contradicting the irrationality of $\sqrt{2}$, or else $a = b = 0$, which would mean $a+b\sqrt{2}$ is not non-zero.

The beauty of using the "general theorems" that:

$F[x]/I$ is a field, where $I$ is the ideal generated by a monic irreducible polynomial, is that we don't HAVE to compute inverses to know they exist-the drawback is that knowing they exist doesn't actually tell us how to COMPUTE them.

In short, if the element $z$ that we adjoin (in the RING sense) is *algebraic* over $F$ (it satisfies some polynomial in $F[x]$, which we can then factor into irreducible factors-and thus find the irreducible factor(s) it is a root of), then the "reciprocals" of "polynomials in $z$" are again: "polynomials in $z$", and so the ring-adjuction is the same as the field-adjunction.

In field theory, one sees the adjoined element written with both square and round brackets, as it makes no difference if we are talking about algebraic extensions.

Here is another commonly-cited example:

Consider the complex number:

$\omega = -\dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} = \cos\left(\dfrac{2\pi}{3}\right) + i\sin\left(\dfrac{2\pi}{3}\right)$

We have:

$\omega^2 = \left(\dfrac{1}{4} - \dfrac{3}{4}\right) + i\left(-\dfrac{\sqrt{3}}{4} - \dfrac{\sqrt{3}}{4}\right)= -\dfrac{1}{2} - i\dfrac{\sqrt{3}}{2}$

$= \cos\left(\dfrac{4\pi}{3}\right) + i\sin\left(\dfrac{4\pi}{3}\right)$.

Thus $\omega^2 + \omega = -1$, that is, $\omega$ is a root of $x^2 + x + 1$.

Since $x^3 - 1 = (x - 1)(x^2 + x + 1)$ it follows that $\omega$ is a (complex) cube root of $1$ (if you know DeMoivre's Theorem, you can see that:

$\omega^3 = \cos(2\pi) + i\sin(2\pi) = 1 + i0 = 1$).

It can be shown that $x^2 + x + 1$ is irreducible over $\Bbb Q$, so we have:

$\Bbb Q(\omega) = \Bbb Q[\omega] \cong \Bbb Q[x]/(x^2 + x + 1)$.

Convince yourself that for $a,b \in \Bbb Q$, with $a^2 + b^2 \neq 0$, that:

$\dfrac{1}{a + b\omega} = \dfrac{1}{a^2 + b^2 - ab}((a-b) - b\omega)$

(again, the "tricky part" is showing that $a^2 + b^2 - ab \neq 0$. Consider the cases where $ab > 0$ and $ab < 0$ separately-why is $ab = 0$ trivial?).

Thanks for another gem of a post, Deveno ..

It is so helpful getting help with what is going on behind the scenes of a stream of definitions, theorem statements and proofs ...

Even more so ... it is helpful to have the explanation accompanied by well explained illustrative examples ...

Thanks so much for the help ...

Still reflecting on this post ...

Peter