Field Theory _ Dummit and Foote - Theorem 3

[Math Amateur]

I am reading Dummit and Foote, Chapter 13 - Field Theory.

I am currently studying Theorem 3 [pages 512 - 513]

I need some help with an aspect of the proof of Theorem 3 concerning congruence or residue classes of polynomials.

D&F, Chapter 13, Theorem 3 and its proof read as follows:https://www.physicsforums.com/attachments/2712
View attachment 2713In the above text D&F state the following:

" ... ... If $$ \overline{x} = \pi (x) $$ denotes the image of x in the quotient K, then

$$ p ( \overline{x} ) = \overline{p(x)} $$ ... ... ... (since $$ \pi $$ is a homomorphism) ... ... "

I do not completely understand this statement and further I am having some trouble interpreting the meaning of the notation ...

I hope someone can help .. ...

In order to ensure I understood congruence classes or residue classes in F[x] I went to Hungerford, Abstract Algebra: An Introduction, Chapter 5. In this chapter Hungerford gives the following definitions:View attachment 2714

View attachment 2715So, following Hungerford, if we want to find the residue class of $$ f(x) = x $$, we write:

$$ \overline{x} = [x] = \{ x + k(x)p(x) \ | \ k(x) \in F[x] \} $$

But given this ... how do we form $$ p( \overline{x})$$?

Do we substitute $$p(x)$$ for $$x$$, everywhere $$x$$ appears getting the following:

$$ p ( \overline{x} ) = p[x] = \{ p(x) + k(p(x))p(p(x)) \ | \ k(x) \in F[x] \} $$ ?

I know this seems clumsy ... but it seems to me to have some formal merit ... anyway, although I suspect it is not the way to go ... I am not sure why ...

So maybe the right interpretation is as follows:

$$ p ( \overline{x} ) = p[x] = \{ p(x) + k(x)p(x) \ | \ k(x) \in F[x] \} $$ ?

Can someone clarify this issue for me indicating not only which alternative is correct - but further why that alternative is correct and the other wrong.

I would appreciate some help in this matter.

Peter

 

[Deveno]

The notation is awkward, simply because "polynomial expresssions" are rather clumsy to write down in the first place.

Let's look at a simple example: let $F = \Bbb Q$, and let $p(x) = x^2 + 1$. Clearly, $x^2 + 1$ is irreducible over $\Bbb Q$, for if it were not, Gauss' lemma tells us it would be reducible over $\Bbb Z$, and thus would have an integer root. This integer root would be a factor of 1, hence could only be -1, or 1. Since neither one of these is a root, $x^2 + 1$ is irreducible over $\Bbb Q$.

So we can form the quotient ring: $\Bbb Q[x]/(x^2 + 1)$, which is a field, by dint of the fact that $x^2 + 1$ generates a maximal ideal (being irreducible).

Now elements of $I = (x^2 + 1)$ are rational polynomials of the form: $k(x)(x^2 + 1)$. For example, we have:

$x^3 + x, 2x^2 + 2, x^4 - 1 \in I$.

and a typical element $[f(x)] \in \Bbb Q[x]/I$ is comprised of elements of the form $f(x) + k(x)(x^2 + 1)$, for some $k$.

It behooves us to find a simpler form for $[f(x)]$, which will allow us to limit the complexity of computation in $\Bbb Q[x]/I$. Since $\Bbb Q[x]$ is a Euclidean domain, we may write:

$f(x) = q(x)(x^2 + 1) + r(x)$, where $r(x) \equiv 0$, or $0 \leq \text{deg}(r) < \text{deg}(x^2 + 1) = 2$

We then see that:

$[f(x)] = \{f(x) + k(x)(x^2 + 1)\} = \{q(x)(x^2 + 1) + r(x) + k(x)(x^2 + 1)\} = \{r(x) + (q(x) + k(x))(x^2 + 1)\}$;

that is: $[f(x)] = [r(x)]$.

It follows that we can thus write any $[f(x)]$ as: $[ax + b]$ for unique rational numbers $a,b$.

Now consider the mapping $\Bbb Q \to \Bbb Q/I$ given by*:

$q \mapsto [0x + q] = q + I$. It is not hard to see this is a ring-homomorphism, and since $\Bbb Q$ is a FIELD, it's kernel must be an ideal of $\Bbb Q$. But $\Bbb Q$, as a field, only has TWO ideals: $(0)$ and $(1) = \Bbb Q$.

Since this map is evidently not the 0-map, the kernel must be trivial, so it is a monomorphism of rings, and thus an isomorphism of $\Bbb Q$ with its image. In THIS WAY, we can consider $\Bbb Q[x]/I$ an extension of $\Bbb Q$.

Now, clearly, taking $a = 1,b = 0$ yields a unique coset in $\Bbb Q[x]/I$, namely:

$[x] = x + I$.

So what might we mean by $f([x]) = f(x + I)$? If $\pi: \Bbb Q[x] \to \Bbb Q[x]/I$ is the canonical projection map, we just mean:

$f([x]) = f(\pi(x)) \equiv \pi(f(x))$, that is, if:

$f(x) = a_0 + a_1x + \cdots + a_nx^n$ , then:

$f(x + I) = (a_0 + I) + (a_1 + I)(x + I) + \cdots + (a_n + I)(x + I)^n$

$= (a_0 + I) + (a_1x + I) + \cdots + (a_nx^n + I)$

$ = (a_0 + a_1x + \cdots + a_nx^n) + I = f(x) + I$

(REMEMBER how we add and multiply cosets: $(r + I) + (s + I) = (r+s) + I$ and $(r + I)(s + I) = (rs) + I$).

Now what does this mean if $f(x) = p(x) = x^2 + 1$?

$p(x + I) = (x + I)^2 + (1 + I) = (x^2 + I) + (1 + I) = (x^2 + 1) + I = I = 0 + I$.

So in the ring $(\Bbb Q/I)[x+I] \cong \Bbb Q[x]/I$ (be sure you understand this isomorphism! $x + I$ is no longer an "indeterminate" on the left because $I$ annihilates multiples of $x^2 + 1$).

Let's look closer at how $\Bbb Q[x]/I$ behaves. As we saw above, we can write an element as:

$(ax + b) + I$.

Clearly, to add two elements, we get:

$(ax + b) + I + (cx + d) + I = ((a + c)x + (b + d)) + I$.

This suggests we have an abelian group homomorphism: $\Bbb Q \times \Bbb Q \to \Bbb Q[x]/I$ given by:

$(a,b) \mapsto (ax + b) + I$.

I recommend you PROVE this is, in fact, an isomorphism.

Now we can make $\Bbb Q \times \Bbb Q$ into a $\Bbb Q$-module in a natural way: we define, for $q \in \Bbb Q$:

$q\cdot (a,b) = (qa,qb)$.

Show that if we define a $\Bbb Q$-action on $\Bbb Q[x]/I$ by:

$q\cdot((ax + b) + I) = (q + I)((ax + b) + I) = ((qa)x + qb) + I$, this gives a $\Bbb Q$-module isomorphism.

This means that we can regard $\Bbb Q[x]/I$ as a 2-dimensional vector space over $\Bbb Q$, with basis $\{1+I,x + I\}$.

But since we know $\Bbb Q[x]/I$ is a field (so thus a ring), we have even MORE structure: a multiplication. What is this multiplication, in simplest terms? Let's find out:

$((ax + b) + I)((cx + d) + I) = (ax + b)(cx + d) + I = acx^2 + (ad + bc)x + bd + I$.

Now, note that this is of degree 2, so we ought to be able to "reduce" it (mod $I$). Let's do this.

$acx^2 + (ad + bc)x + bd + I = acx^2 + ac + (ad + bc)x + bd - ac + I = (ad + bc)x + bd - ac + ac(x^2 + I) + I$

Now, $ac(x^2 + I) \in I$ (since $I$ is an ideal), so we have:

$((ax + b) + I)((cx + d) + I) = ((ad + bc)x + (bd - ac)) + I$.

Convince yourself that if $i = \sqrt{-1} \in \Bbb C$, that:

$\phi: \Bbb Q[x]/I \to \Bbb Q(i) \subset \Bbb C$, given by:

$\phi((ax + b) + I) = b + ai$ is an isomorphic of our field with the rational complex numbers (the field of quotients of the Gaussian integers).

Clearly: $\phi(x + I) = \phi((1x + 0) + I) = 0 + 1i = i$, and indeed, in $\Bbb C$, we have $i$ is a root of $x^2 + 1$, therefore (since we have an isomorphism), $(x + I)$ is a root of $(x + I)^2 + (1 + I)$ in $\Bbb Q[x]/I$.

This procedure is entirely general, and can be extended to:

Any polynomial in $F[x]$ has a root in an extension of $F$, and thus: for any field $F$, and any polynomial $f(x)$, there exists an extension field $E$ that contains EVERY root of $f$ (just repeat this procedure until all roots are accounted for).

* Note: this is rather an abuse of notation; $I$ is not an ideal of $\Bbb Q$, rather we mean the image of $\Bbb Q$ in $\Bbb Q[x]/I$, considering $\Bbb Q$ as a subring of $\Bbb Q[x]$.

 

[Math Amateur]

The notation is awkward, simply because "polynomial expresssions" are rather clumsy to write down in the first place.

Let's look at a simple example: let $F = \Bbb Q$, and let $p(x) = x^2 + 1$. Clearly, $x^2 + 1$ is irreducible over $\Bbb Q$, for if it were not, Gauss' lemma tells us it would be reducible over $\Bbb Z$, and thus would have an integer root. This integer root would be a factor of 1, hence could only be -1, or 1. Since neither one of these is a root, $x^2 + 1$ is irreducible over $\Bbb Q$.

So we can form the quotient ring: $\Bbb Q[x]/(x^2 + 1)$, which is a field, by dint of the fact that $x^2 + 1$ generates a maximal ideal (being irreducible).

Now elements of $I = (x^2 + 1)$ are rational polynomials of the form: $k(x)(x^2 + 1)$. For example, we have:

$x^3 + x, 2x^2 + 2, x^4 - 1 \in I$.

and a typical element $[f(x)] \in \Bbb Q[x]/I$ is comprised of elements of the form $f(x) + k(x)(x^2 + 1)$, for some $k$.

It behooves us to find a simpler form for $[f(x)]$, which will allow us to limit the complexity of computation in $\Bbb Q[x]/I$. Since $\Bbb Q[x]$ is a Euclidean domain, we may write:

$f(x) = q(x)(x^2 + 1) + r(x)$, where $r(x) \equiv 0$, or $0 \leq \text{deg}(r) < \text{deg}(x^2 + 1) = 2$

We then see that:

$[f(x)] = \{f(x) + k(x)(x^2 + 1)\} = \{q(x)(x^2 + 1) + r(x) + k(x)(x^2 + 1)\} = \{r(x) + (q(x) + k(x))(x^2 + 1)\}$;

that is: $[f(x)] = [r(x)]$.

It follows that we can thus write any $[f(x)]$ as: $[ax + b]$ for unique rational numbers $a,b$.

Now consider the mapping $\Bbb Q \to \Bbb Q/I$ given by*:

$q \mapsto [0x + q] = q + I$. It is not hard to see this is a ring-homomorphism, and since $\Bbb Q$ is a FIELD, it's kernel must be an ideal of $\Bbb Q$. But $\Bbb Q$, as a field, only has TWO ideals: $(0)$ and $(1) = \Bbb Q$.

Since this map is evidently not the 0-map, the kernel must be trivial, so it is a monomorphism of rings, and thus an isomorphism of $\Bbb Q$ with its image. In THIS WAY, we can consider $\Bbb Q[x]/I$ an extension of $\Bbb Q$.

Now, clearly, taking $a = 1,b = 0$ yields a unique coset in $\Bbb Q[x]/I$, namely:

$[x] = x + I$.

So what might we mean by $f([x]) = f(x + I)$? If $\pi: \Bbb Q[x] \to \Bbb Q[x]/I$ is the canonical projection map, we just mean:

$f([x]) = f(\pi(x)) \equiv \pi(f(x))$, that is, if:

$f(x) = a_0 + a_1x + \cdots + a_nx^n$ , then:

$f(x + I) = (a_0 + I) + (a_1 + I)(x + I) + \cdots + (a_n + I)(x + I)^n$

$= (a_0 + I) + (a_1x + I) + \cdots + (a_nx^n + I)$

$ = (a_0 + a_1x + \cdots + a_nx^n) + I = f(x) + I$

(REMEMBER how we add and multiply cosets: $(r + I) + (s + I) = (r+s) + I$ and $(r + I)(s + I) = (rs) + I$).

Now what does this mean if $f(x) = p(x) = x^2 + 1$?

$p(x + I) = (x + I)^2 + (1 + I) = (x^2 + I) + (1 + I) = (x^2 + 1) + I = I = 0 + I$.

So in the ring $(\Bbb Q/I)[x+I] \cong \Bbb Q[x]/I$ (be sure you understand this isomorphism! $x + I$ is no longer an "indeterminate" on the left because $I$ annihilates multiples of $x^2 + 1$).

Let's look closer at how $\Bbb Q[x]/I$ behaves. As we saw above, we can write an element as:

$(ax + b) + I$.

Clearly, to add two elements, we get:

$(ax + b) + I + (cx + d) + I = ((a + c)x + (b + d)) + I$.

This suggests we have an abelian group homomorphism: $\Bbb Q \times \Bbb Q \to \Bbb Q[x]/I$ given by:

$(a,b) \mapsto (ax + b) + I$.

I recommend you PROVE this is, in fact, an isomorphism.

Now we can make $\Bbb Q \times \Bbb Q$ into a $\Bbb Q$-module in a natural way: we define, for $q \in \Bbb Q$:

$q\cdot (a,b) = (qa,qb)$.

Show that if we define a $\Bbb Q$-action on $\Bbb Q[x]/I$ by:

$q\cdot((ax + b) + I) = (q + I)((ax + b) + I) = ((qa)x + qb) + I$, this gives a $\Bbb Q$-module isomorphism.

This means that we can regard $\Bbb Q[x]/I$ as a 2-dimensional vector space over $\Bbb Q$, with basis $\{1+I,x + I\}$.

But since we know $\Bbb Q[x]/I$ is a field (so thus a ring), we have even MORE structure: a multiplication. What is this multiplication, in simplest terms? Let's find out:

$((ax + b) + I)((cx + d) + I) = (ax + b)(cx + d) + I = acx^2 + (ad + bc)x + bd + I$.

Now, note that this is of degree 2, so we ought to be able to "reduce" it (mod $I$). Let's do this.

$acx^2 + (ad + bc)x + bd + I = acx^2 + ac + (ad + bc)x + bd - ac + I = (ad + bc)x + bd - ac + ac(x^2 + I) + I$

Now, $ac(x^2 + I) \in I$ (since $I$ is an ideal), so we have:

$((ax + b) + I)((cx + d) + I) = ((ad + bc)x + (bd - ac)) + I$.

Convince yourself that if $i = \sqrt{-1} \in \Bbb C$, that:

$\phi: \Bbb Q[x]/I \to \Bbb Q(i) \subset \Bbb C$, given by:

$\phi((ax + b) + I) = b + ai$ is an isomorphic of our field with the rational complex numbers (the field of quotients of the Gaussian integers).

Clearly: $\phi(x + I) = \phi((1x + 0) + I) = 0 + 1i = i$, and indeed, in $\Bbb C$, we have $i$ is a root of $x^2 + 1$, therefore (since we have an isomorphism), $(x + I)$ is a root of $(x + I)^2 + (1 + I)$ in $\Bbb Q[x]/I$.

This procedure is entirely general, and can be extended to:

Any polynomial in $F[x]$ has a root in an extension of $F$, and thus: for any field $F$, and any polynomial $f(x)$, there exists an extension field $E$ that contains EVERY root of $f$ (just repeat this procedure until all roots are accounted for).

* Note: this is rather an abuse of notation; $I$ is not an ideal of $\Bbb Q$, rather we mean the image of $\Bbb Q$ in $\Bbb Q[x]/I$, considering $\Bbb Q$ as a subring of $\Bbb Q[x]$.

Thanks Deveno ... this post is significantly helpful ...

Working through what you have written carefully and reflecting on what you have said ...

Thanks again,

Peter

 

[Math Amateur]

The notation is awkward, simply because "polynomial expresssions" are rather clumsy to write down in the first place.

Let's look at a simple example: let $F = \Bbb Q$, and let $p(x) = x^2 + 1$. Clearly, $x^2 + 1$ is irreducible over $\Bbb Q$, for if it were not, Gauss' lemma tells us it would be reducible over $\Bbb Z$, and thus would have an integer root. This integer root would be a factor of 1, hence could only be -1, or 1. Since neither one of these is a root, $x^2 + 1$ is irreducible over $\Bbb Q$.

So we can form the quotient ring: $\Bbb Q[x]/(x^2 + 1)$, which is a field, by dint of the fact that $x^2 + 1$ generates a maximal ideal (being irreducible).

Now elements of $I = (x^2 + 1)$ are rational polynomials of the form: $k(x)(x^2 + 1)$. For example, we have:

$x^3 + x, 2x^2 + 2, x^4 - 1 \in I$.

and a typical element $[f(x)] \in \Bbb Q[x]/I$ is comprised of elements of the form $f(x) + k(x)(x^2 + 1)$, for some $k$.

It behooves us to find a simpler form for $[f(x)]$, which will allow us to limit the complexity of computation in $\Bbb Q[x]/I$. Since $\Bbb Q[x]$ is a Euclidean domain, we may write:

$f(x) = q(x)(x^2 + 1) + r(x)$, where $r(x) \equiv 0$, or $0 \leq \text{deg}(r) < \text{deg}(x^2 + 1) = 2$

We then see that:

$[f(x)] = \{f(x) + k(x)(x^2 + 1)\} = \{q(x)(x^2 + 1) + r(x) + k(x)(x^2 + 1)\} = \{r(x) + (q(x) + k(x))(x^2 + 1)\}$;

that is: $[f(x)] = [r(x)]$.

It follows that we can thus write any $[f(x)]$ as: $[ax + b]$ for unique rational numbers $a,b$.

Now consider the mapping $\Bbb Q \to \Bbb Q/I$ given by*:

$q \mapsto [0x + q] = q + I$. It is not hard to see this is a ring-homomorphism, and since $\Bbb Q$ is a FIELD, it's kernel must be an ideal of $\Bbb Q$. But $\Bbb Q$, as a field, only has TWO ideals: $(0)$ and $(1) = \Bbb Q$.

Since this map is evidently not the 0-map, the kernel must be trivial, so it is a monomorphism of rings, and thus an isomorphism of $\Bbb Q$ with its image. In THIS WAY, we can consider $\Bbb Q[x]/I$ an extension of $\Bbb Q$.

Now, clearly, taking $a = 1,b = 0$ yields a unique coset in $\Bbb Q[x]/I$, namely:

$[x] = x + I$.

So what might we mean by $f([x]) = f(x + I)$? If $\pi: \Bbb Q[x] \to \Bbb Q[x]/I$ is the canonical projection map, we just mean:

$f([x]) = f(\pi(x)) \equiv \pi(f(x))$, that is, if:

$f(x) = a_0 + a_1x + \cdots + a_nx^n$ , then:

$f(x + I) = (a_0 + I) + (a_1 + I)(x + I) + \cdots + (a_n + I)(x + I)^n$

$= (a_0 + I) + (a_1x + I) + \cdots + (a_nx^n + I)$

$ = (a_0 + a_1x + \cdots + a_nx^n) + I = f(x) + I$

(REMEMBER how we add and multiply cosets: $(r + I) + (s + I) = (r+s) + I$ and $(r + I)(s + I) = (rs) + I$).

Now what does this mean if $f(x) = p(x) = x^2 + 1$?

$p(x + I) = (x + I)^2 + (1 + I) = (x^2 + I) + (1 + I) = (x^2 + 1) + I = I = 0 + I$.

So in the ring $(\Bbb Q/I)[x+I] \cong \Bbb Q[x]/I$ (be sure you understand this isomorphism! $x + I$ is no longer an "indeterminate" on the left because $I$ annihilates multiples of $x^2 + 1$).

Let's look closer at how $\Bbb Q[x]/I$ behaves. As we saw above, we can write an element as:

$(ax + b) + I$.

Clearly, to add two elements, we get:

$(ax + b) + I + (cx + d) + I = ((a + c)x + (b + d)) + I$.

This suggests we have an abelian group homomorphism: $\Bbb Q \times \Bbb Q \to \Bbb Q[x]/I$ given by:

$(a,b) \mapsto (ax + b) + I$.

I recommend you PROVE this is, in fact, an isomorphism.

Now we can make $\Bbb Q \times \Bbb Q$ into a $\Bbb Q$-module in a natural way: we define, for $q \in \Bbb Q$:

$q\cdot (a,b) = (qa,qb)$.

Show that if we define a $\Bbb Q$-action on $\Bbb Q[x]/I$ by:

$q\cdot((ax + b) + I) = (q + I)((ax + b) + I) = ((qa)x + qb) + I$, this gives a $\Bbb Q$-module isomorphism.

This means that we can regard $\Bbb Q[x]/I$ as a 2-dimensional vector space over $\Bbb Q$, with basis $\{1+I,x + I\}$.

But since we know $\Bbb Q[x]/I$ is a field (so thus a ring), we have even MORE structure: a multiplication. What is this multiplication, in simplest terms? Let's find out:

$((ax + b) + I)((cx + d) + I) = (ax + b)(cx + d) + I = acx^2 + (ad + bc)x + bd + I$.

Now, note that this is of degree 2, so we ought to be able to "reduce" it (mod $I$). Let's do this.

$acx^2 + (ad + bc)x + bd + I = acx^2 + ac + (ad + bc)x + bd - ac + I = (ad + bc)x + bd - ac + ac(x^2 + I) + I$

Now, $ac(x^2 + I) \in I$ (since $I$ is an ideal), so we have:

$((ax + b) + I)((cx + d) + I) = ((ad + bc)x + (bd - ac)) + I$.

Convince yourself that if $i = \sqrt{-1} \in \Bbb C$, that:

$\phi: \Bbb Q[x]/I \to \Bbb Q(i) \subset \Bbb C$, given by:

$\phi((ax + b) + I) = b + ai$ is an isomorphic of our field with the rational complex numbers (the field of quotients of the Gaussian integers).

Clearly: $\phi(x + I) = \phi((1x + 0) + I) = 0 + 1i = i$, and indeed, in $\Bbb C$, we have $i$ is a root of $x^2 + 1$, therefore (since we have an isomorphism), $(x + I)$ is a root of $(x + I)^2 + (1 + I)$ in $\Bbb Q[x]/I$.

This procedure is entirely general, and can be extended to:

Any polynomial in $F[x]$ has a root in an extension of $F$, and thus: for any field $F$, and any polynomial $f(x)$, there exists an extension field $E$ that contains EVERY root of $f$ (just repeat this procedure until all roots are accounted for).

* Note: this is rather an abuse of notation; $I$ is not an ideal of $\Bbb Q$, rather we mean the image of $\Bbb Q$ in $\Bbb Q[x]/I$, considering $\Bbb Q$ as a subring of $\Bbb Q[x]$.

Hi Deveno ... Just a further note to say that I found this post EXTREMELY helpful ... particularly as it involved a concrete and instructive example ...

Thanks again,

Peter