Factor Rings of Polynomials Over a Field - Nicholson, Lemma 1, Page 224

[Math Amateur]

I am reading W. Keith Nicholson's book: Introduction to Abstract Algebra (Third Edition) ...

I am focused on Section 4.3:Factor Rings of Polynomials over a Field.

I need some help with the proof of Part 1 of Lemma 1 on page 223-224.

The relevant text from Nicholson's book is as follows:
https://www.physicsforums.com/attachments/4631
View attachment 4632In the above text, we read the following:

"Proof. A typical element of R has the form \(\displaystyle \overline{f(x)}, f(x) \in R[x] \) ... ... "My question is as follows:

Does Nicholson actually mean \(\displaystyle f(x) \in R[x]\) in the above text ... ... or does he mean \(\displaystyle f(x) \in F[x]\) ... ?

If he does mean \(\displaystyle f(x) \in R[x]\) ... can someone please explain the nature of the ring \(\displaystyle R[x] = F[x]/ <h> [x]\) ... it seems a very awkward and bewildering ring ...Hope someone can help ...

Peter
***NOTE***

At the start of the above text, Nicholson refers to Example 1, so to help MHB members appreciate the context of the above extract from Nicholson, I am providing Example 1 as follows:
View attachment 4633

 

[Deveno]

It does appear to be a typo, the coset $\overline{f} = f+A$ should be for $f \in F[x]$. It also appears there is another typo when he writes:

$f(x) = q(x)h(x) + (a_0 + a_1x + \cdots + a_{m-1}x^{m-1}),\ a_i \in R$

which should read at the end: "$\dots a_i \in F$".

 

[Math Amateur]

It does appear to be a typo, the coset $\overline{f} = f+A$ should be for $f \in F[x]$.

Thanks, Deveno ... that gives me the confidence to go on ...

BUT ... just noticed in the next line we have \(\displaystyle a_i \in R\) ... presumably the author means \(\displaystyle a_i \in F\) ... or \(\displaystyle \overline{a_i} \in R\) ... is that right?Thanks again for your help ...

Peter

 

[Deveno]

Rings, just like groups, have a Fundamental Isomorphism Theorem:

If $f: R \to S$ is a ring-homomorphism, then $\text{ker }f = \{r \in R: f(r) = 0_S\}$ is an ideal of $R$ and:

$f(R) \cong R/\text{ker }f$.

Polynomial rings (for a commutative ring $R$) are "universal simple adjunction rings", if we have a ring $S$ with $R \subseteq S$, and we take any $a \in S\setminus R$, there is a *unique* homomorphism:

$\phi_a: R[x] \to S$ with $\phi_a(x) = a$, that is the identity on $R$.

Explicilty: $\phi_a(f(x)) = f(a)$.

The subring $\langle R \cup \{a\}\rangle$ of $S$ generated by $R$ and $a$, is denoted $R[a]$.

It should be clear that $\phi_a(R[x]) = R[a]$.

Clearly, $a \in \phi_a(R[x])$, since $x \in R[x]$. Also since we can regard constant polynomials in $R[x]$ as elements of $R$, we have $R \subseteq R[a]$, so:

$\langle R\cup\{a\}\rangle \subseteq \phi_a(R[x])$.

On the other hand, $f(a)$ (which is the image under $\phi_a$ of the polynomial $f(x)$) is a finite sum of terms $c_ia^i$, and such sums are in $R[a]$ if each term is, by the requirement of (additive) closure.

The terms, in turn, are products of $c_i$ (these constant polynomials are fixed by $\phi_a$) and powers of $a$, which lie in $R[a]$ by (multiplicative) closure.

Hence $\phi_a(R[x]) \subseteq R[a]$.

In short, our homomorphism $\phi_a$ is surjective, so $R[a] \cong R[x]/\text{ker }\phi_a$.

I urge you to convince yourself that $\text{ker }\phi_a = \{f(x) \in R[x]: f(a) = 0\}$

In the happy case that $R = F$ is a field, we know that $F[x]$ is a Euclidean domain, which is a *much stronger* condition than being a principal ideal domain. So we know that $\text{ker }\phi_a = \langle p(x)\rangle$, for some polynomial $p(x)$.

If we arrange it so that the coefficient of the highest non-zero power of $p(x)$ is $1$, we can identify $p$ uniquely, and we call it the minimal polynomial of $a$ (over $F$).

There is an analogy here with *free groups*-given a set of generators for a group $G$:

$G = \langle g_1,g_2,\dots,g_k\rangle$, we can realize $G$ as a QUOTIENT of the free group on $k$ letters.

The generators of the kernel of the quotient homomorphism are called relators, for example, a cyclic subgroup of order $n$ is a quotient of the free group on one letter, say $X$ (which is isomorphic to the integers), with the single relator, $X^n$: sending the free group to the quotient means that we are identifying $X^n$ and $1$ (in the quotient).

So one often sees:

$C_n = \langle a: a^n = 1\rangle$.

Relators are more typically expressed as "relations": for example, the "abelian relation" is $ab = ba$, and the associated relator is $aba^{-1}b^{1}$, or the COMMUTATOR of $a$ and $b$.

So, how this relates to rings is:

The ring $R[a]$ can be seen as a QUOTIENT of $R[x]$, subject to certain "relations", namely:

$p(a) = 0$ (for whichever $p$ lie in the kernel).

For example, the ring $\Bbb R$ is just a quotient of $\Bbb R[x]$, with the relation: $i^2 = -1$ (the "relator" here is the polynomial $x^2 + 1$).

 

[Math Amateur]

Rings, just like groups, have a Fundamental Isomorphism Theorem:

If $f: R \to S$ is a ring-homomorphism, then $\text{ker }f = \{r \in R: f(r) = 0_S\}$ is an ideal of $R$ and:

$f(R) \cong R/\text{ker }f$.

Polynomial rings (for a commutative ring $R$) are "universal simple adjunction rings", if we have a ring $S$ with $R \subseteq S$, and we take any $a \in S\setminus R$, there is a *unique* homomorphism:

$\phi_a: R[x] \to S$ with $\phi_a(x) = a$, that is the identity on $R$.

Explicilty: $\phi_a(f(x)) = f(a)$.

The subring $\langle R \cup \{a\}\rangle$ of $S$ generated by $R$ and $a$, is denoted $R[a]$.

It should be clear that $\phi_a(R[x]) = R[a]$.

Clearly, $a \in \phi_a(R[x])$, since $x \in R[x]$. Also since we can regard constant polynomials in $R[x]$ as elements of $R$, we have $R \subseteq R[a]$, so:

$\langle R\cup\{a\}\rangle \subseteq \phi_a(R[x])$.

On the other hand, $f(a)$ (which is the image under $\phi_a$ of the polynomial $f(x)$) is a finite sum of terms $c_ia^i$, and such sums are in $R[a]$ if each term is, by the requirement of (additive) closure.

The terms, in turn, are products of $c_i$ (these constant polynomials are fixed by $\phi_a$) and powers of $a$, which lie in $R[a]$ by (multiplicative) closure.

Hence $\phi_a(R[x]) \subseteq R[a]$.

In short, our homomorphism $\phi_a$ is surjective, so $R[a] \cong R[x]/\text{ker }\phi_a$.

I urge you to convince yourself that $\text{ker }\phi_a = \{f(x) \in R[x]: f(a) = 0\}$

In the happy case that $R = F$ is a field, we know that $F[x]$ is a Euclidean domain, which is a *much stronger* condition than being a principal ideal domain. So we know that $\text{ker }\phi_a = \langle p(x)\rangle$, for some polynomial $p(x)$.

If we arrange it so that the coefficient of the highest non-zero power of $p(x)$ is $1$, we can identify $p$ uniquely, and we call it the minimal polynomial of $a$ (over $F$).

There is an analogy here with *free groups*-given a set of generators for a group $G$:

$G = \langle g_1,g_2,\dots,g_k\rangle$, we can realize $G$ as a QUOTIENT of the free group on $k$ letters.

The generators of the kernel of the quotient homomorphism are called relators, for example, a cyclic subgroup of order $n$ is a quotient of the free group on one letter, say $X$ (which is isomorphic to the integers), with the single relator, $X^n$: sending the free group to the quotient means that we are identifying $X^n$ and $1$ (in the quotient).

So one often sees:

$C_n = \langle a: a^n = 1\rangle$.

Relators are more typically expressed as "relations": for example, the "abelian relation" is $ab = ba$, and the associated relator is $aba^{-1}b^{1}$, or the COMMUTATOR of $a$ and $b$.

So, how this relates to rings is:

The ring $R[a]$ can be seen as a QUOTIENT of $R[x]$, subject to certain "relations", namely:

$p(a) = 0$ (for whichever $p$ lie in the kernel).

For example, the ring $\Bbb R$ is just a quotient of $\Bbb R[x]$, with the relation: $i^2 = -1$ (the "relator" here is the polynomial $x^2 + 1$).

Thanks for your generous help ... as you may have guessed I am having a bit of trouble with this section ... so your post will definitely help ...

Just working carefully through your post in detail now ...

Thanks again,

Peter

 

[Math Amateur]

Rings, just like groups, have a Fundamental Isomorphism Theorem:

If $f: R \to S$ is a ring-homomorphism, then $\text{ker }f = \{r \in R: f(r) = 0_S\}$ is an ideal of $R$ and:

$f(R) \cong R/\text{ker }f$.

Polynomial rings (for a commutative ring $R$) are "universal simple adjunction rings", if we have a ring $S$ with $R \subseteq S$, and we take any $a \in S\setminus R$, there is a *unique* homomorphism:

$\phi_a: R[x] \to S$ with $\phi_a(x) = a$, that is the identity on $R$.

Explicilty: $\phi_a(f(x)) = f(a)$.

The subring $\langle R \cup \{a\}\rangle$ of $S$ generated by $R$ and $a$, is denoted $R[a]$.

It should be clear that $\phi_a(R[x]) = R[a]$.

Clearly, $a \in \phi_a(R[x])$, since $x \in R[x]$. Also since we can regard constant polynomials in $R[x]$ as elements of $R$, we have $R \subseteq R[a]$, so:

$\langle R\cup\{a\}\rangle \subseteq \phi_a(R[x])$.

On the other hand, $f(a)$ (which is the image under $\phi_a$ of the polynomial $f(x)$) is a finite sum of terms $c_ia^i$, and such sums are in $R[a]$ if each term is, by the requirement of (additive) closure.

The terms, in turn, are products of $c_i$ (these constant polynomials are fixed by $\phi_a$) and powers of $a$, which lie in $R[a]$ by (multiplicative) closure.

Hence $\phi_a(R[x]) \subseteq R[a]$.

In short, our homomorphism $\phi_a$ is surjective, so $R[a] \cong R[x]/\text{ker }\phi_a$.

I urge you to convince yourself that $\text{ker }\phi_a = \{f(x) \in R[x]: f(a) = 0\}$

In the happy case that $R = F$ is a field, we know that $F[x]$ is a Euclidean domain, which is a *much stronger* condition than being a principal ideal domain. So we know that $\text{ker }\phi_a = \langle p(x)\rangle$, for some polynomial $p(x)$.

If we arrange it so that the coefficient of the highest non-zero power of $p(x)$ is $1$, we can identify $p$ uniquely, and we call it the minimal polynomial of $a$ (over $F$).

There is an analogy here with *free groups*-given a set of generators for a group $G$:

$G = \langle g_1,g_2,\dots,g_k\rangle$, we can realize $G$ as a QUOTIENT of the free group on $k$ letters.

The generators of the kernel of the quotient homomorphism are called relators, for example, a cyclic subgroup of order $n$ is a quotient of the free group on one letter, say $X$ (which is isomorphic to the integers), with the single relator, $X^n$: sending the free group to the quotient means that we are identifying $X^n$ and $1$ (in the quotient).

So one often sees:

$C_n = \langle a: a^n = 1\rangle$.

Relators are more typically expressed as "relations": for example, the "abelian relation" is $ab = ba$, and the associated relator is $aba^{-1}b^{1}$, or the COMMUTATOR of $a$ and $b$.

So, how this relates to rings is:

The ring $R[a]$ can be seen as a QUOTIENT of $R[x]$, subject to certain "relations", namely:

$p(a) = 0$ (for whichever $p$ lie in the kernel).

For example, the ring $\Bbb R$ is just a quotient of $\Bbb R[x]$, with the relation: $i^2 = -1$ (the "relator" here is the polynomial $x^2 + 1$).

Hi Deveno,

I am slightly perplexed (apologies... ) about the 'big picture' regarding the following text:"Polynomial rings (for a commutative ring $R$) are "universal simple adjunction rings", if we have a ring $S$ with $R \subseteq S$, and we take any $a \in S\setminus R$, there is a *unique* homomorphism:

$\phi_a: R[x] \to S$ with $\phi_a(x) = a$, that is the identity on $R$.

Explicilty: $\phi_a(f(x)) = f(a)$. "


So I am trying to figure out what is "going on" ... with apologies that I seem to be missing the point ...

It looks at first sight that what you are talking about is an evaluation map when you write:

$\phi_a(f(x)) = f(a)$

but what exactly is the role of \(\displaystyle S\) and why is \(\displaystyle a\) restricted from being an element of \(\displaystyle R\)? For example: Suppose \(\displaystyle R = \mathbb{Z}\) and \(\displaystyle S = \mathbb{R}\)

... then if we have \(\displaystyle f \in R[x] = \mathbb{Z} [x] \)

such that, say \(\displaystyle f(x) = 3x^2 + 2x + 1\)

then why can't we take \(\displaystyle a \in R = \mathbb{Z}\) where \(\displaystyle a = 3\), say

and get \(\displaystyle f(3) = 3.3^2 + 2.3 + 1 = 34\) ...Can you explain the role of S and why we a must belong to S\R ...

I feel I am not fully understanding the big picture here ...

Can you help?

Peter

 

[Deveno]

You *can* take an element in $R$, to "evaluate" a polynomial at an element of $R$.

For example, one can use $x^2 - 1 = (x + 1)(x - 1)$ to factor 8 as a product of 4 and 2. But what we are most often interested in, is leveraging our knowledge of $R$ into a "larger" ring structure, in other words in EXTENDING $R$.

So we look at "polynomials in $a$", where $a$ is some element NOT in $R$. The polynomials that annihilate $a$ (that is $f(a) = 0$) offer us some insight into how this larger ring behaves. In particular, they allow us to derive non-trivial algebraic relations that $a$ observes, often resulting in a much SMALLER ring than $S$ or $R[x]$.

For example, if $R = \Bbb Q$, we can look at the subring of $\Bbb R$ formed by "adjoining" the square root of $2$. If $a = \sqrt{2}$, the ring $\Bbb Q[\sqrt{2}]$ is real numbers of the form:

$\{q + r\sqrt{2}: q,r \in \Bbb Q\}$.

The fact that $a = \sqrt{2}$ is a root of $x^2 - 2$ essentially tells us we needn't look at polynomials of degree > 1, since we can use $a^2 = 2$ to "knock down their degree".

Using an element of $R$ doesn't tell us anything *new*-it's not that we can't do it, it's just not terribly enlightening. We could discover those things just by working within $R$.***********

Here is a simple exploration:

Suppose we wanted to know if we could make a field with $4$ elements. We will take it as given that it is known that $\Bbb Z_2$ is a field of two elements (this can be verified by the field axioms directly).

Let's call our $4$-element set: $F = \{0,1,a,b\}$ (we know we need the two identities).

We begin by asking: what is $1+1$? Since $1 \neq 0$, and $0$ is the additive identity, we know the additive order of $1$ is either $2$ or $4$. The first leads to a field of characteristic $2$, the second leads to the cyclic group $\Bbb Z_4$.

Suppose $1$ was of order $4$, so that it generates $F$. Then $1+1$ must be either $a$ or $b$. Since we have, at present, no way to distinguish these two elements, let us suppose it were $a$.

Now $a^2 = (1 + 1)(1 + 1) = (1)(1 + 1) + (1)(1 + 1)$ by the distributive law (which we must have, if we are to have a field). Continuing:

$a^2 = (1)(1 + 1) + (1)(1 + 1) = (1)(1) + (1)(1) + (1)(1) + (1)(1) = 1 + 1 + 1 + 1 = 0$.

In other words, if $F$ is a $4$-element field, and $a = 1+1$, then $a$ is a zero-divisor, which is a contradiction (fields don't have zero-divisors). This means that $1$ does not generate $(F,+)$, so we must have:

$1 + 1 = 0$.

We can then see, that if we are to have a field:

$a + a = (1)(a) + (1)(a) = (1 + 1)(a) = 0a = 0$, and similarly for $b$.

Thus the additive group is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$ (it is not cyclic, and there are only two "types" of groups of order $4$).

At this point, we can see that we can replace $b$ by $a+1$, since:

$a + 1 = a \implies 1 = 0$ (very bad)
$a + 1 + 1 \implies a = 0$ (just as bad)
$a + 1 = 0 \implies a + 1 = 1 + 1 \implies a = 1$ (also not good).

If we wanted to, we could make the assignment:

$(0,0) \leftrightarrow 0$
$(1,0) \leftrightarrow 1$
$(0,1) \leftrightarrow a$
$(1,1) \leftrightarrow a+1$, but we will not.

So now we have a suitable addition (at least it *appears* so), and we turn to multiplication.

Now $F^{\ast} = \{1,a,a+1\}$ is a group (hopefully!) of order $3$, with $a$ a generator (since $1$ only generates $1$ under multiplication), and as such we have:

$a^2 = a+1$.

This pretty much determines the entire multiplication table:

$a(a+1) = a^2 + a = (a+1) + a = a + (1 + a) = a + (a + 1) = (a + a) + 1 = 0 + 1 = 1$
$(a+1)(a+1) = a^2 + a + a + 1 = a^2 + 0 + 1 = a + 1 + 1 = a$

Now, we could check that:

$x(y + z) = xy + xz$ for all $64$ possible choices of $x,y,z$, but maybe there's a better way...

Note that $(a + 1) + (a + 1) = (a + a) + (1 + 1) = 0 + 0 = 0$.

This means we can write:

$a + 1 = -a - 1$, and so:

$a^2 = a + 1 = - a - 1$, and thus $a^2 + a + 1 = 0$.

So we see that $a$ is a ROOT of the polynomial $x^2 + x + 1 \in \Bbb Z_2[x]$.

Note that neither element $0,1$ of $\Bbb Z_2$ (which is a SUBSET of $F$) is such a root.

Now show that $\Bbb Z_2[x]/(x^2 + x + 1)$ yields a four element set with the same addition and multiplication tables (which coset are we mapping to $a$?).

Conclude that $F \cong \phi_a(\Bbb Z_2[x])$ and is therefore a ring, and since $U(F) = F^{\ast}$ (why?), we have a field.

 

[Math Amateur]

You *can* take an element in $R$, to "evaluate" a polynomial at an element of $R$.

For example, one can use $x^2 - 1 = (x + 1)(x - 1)$ to factor 8 as a product of 4 and 2. But what we are most often interested in, is leveraging our knowledge of $R$ into a "larger" ring structure, in other words in EXTENDING $R$.

So we look at "polynomials in $a$", where $a$ is some element NOT in $R$. The polynomials that annihilate $a$ (that is $f(a) = 0$) offer us some insight into how this larger ring behaves. In particular, they allow us to derive non-trivial algebraic relations that $a$ observes, often resulting in a much SMALLER ring than $S$ or $R[x]$.

For example, if $R = \Bbb Q$, we can look at the subring of $\Bbb R$ formed by "adjoining" the square root of $2$. If $a = \sqrt{2}$, the ring $\Bbb Q[\sqrt{2}]$ is real numbers of the form:

$\{q + r\sqrt{2}: q,r \in \Bbb Q\}$.

The fact that $a = \sqrt{2}$ is a root of $x^2 - 2$ essentially tells us we needn't look at polynomials of degree > 1, since we can use $a^2 = 2$ to "knock down their degree".

Using an element of $R$ doesn't tell us anything *new*-it's not that we can't do it, it's just not terribly enlightening. We could discover those things just by working within $R$.***********

Here is a simple exploration:

Suppose we wanted to know if we could make a field with $4$ elements. We will take it as given that it is known that $\Bbb Z_2$ is a field of two elements (this can be verified by the field axioms directly).

Let's call our $4$-element set: $F = \{0,1,a,b\}$ (we know we need the two identities).

We begin by asking: what is $1+1$? Since $1 \neq 0$, and $0$ is the additive identity, we know the additive order of $1$ is either $2$ or $4$. The first leads to a field of characteristic $2$, the second leads to the cyclic group $\Bbb Z_4$.

Suppose $1$ was of order $4$, so that it generates $F$. Then $1+1$ must be either $a$ or $b$. Since we have, at present, no way to distinguish these two elements, let us suppose it were $a$.

Now $a^2 = (1 + 1)(1 + 1) = (1)(1 + 1) + (1)(1 + 1)$ by the distributive law (which we must have, if we are to have a field). Continuing:

$a^2 = (1)(1 + 1) + (1)(1 + 1) = (1)(1) + (1)(1) + (1)(1) + (1)(1) = 1 + 1 + 1 + 1 = 0$.

In other words, if $F$ is a $4$-element field, and $a = 1+1$, then $a$ is a zero-divisor, which is a contradiction (fields don't have zero-divisors). This means that $1$ does not generate $(F,+)$, so we must have:

$1 + 1 = 0$.

We can then see, that if we are to have a field:

$a + a = (1)(a) + (1)(a) = (1 + 1)(a) = 0a = 0$, and similarly for $b$.

Thus the additive group is isomorphic to $\Bbb Z_2 \times \Bbb Z_2$ (it is not cyclic, and there are only two "types" of groups of order $4$).

At this point, we can see that we can replace $b$ by $a+1$, since:

$a + 1 = a \implies 1 = 0$ (very bad)
$a + 1 + 1 \implies a = 0$ (just as bad)
$a + 1 = 0 \implies a + 1 = 1 + 1 \implies a = 1$ (also not good).

If we wanted to, we could make the assignment:

$(0,0) \leftrightarrow 0$
$(1,0) \leftrightarrow 1$
$(0,1) \leftrightarrow a$
$(1,1) \leftrightarrow a+1$, but we will not.

So now we have a suitable addition (at least it *appears* so), and we turn to multiplication.

Now $F^{\ast} = \{1,a,a+1\}$ is a group (hopefully!) of order $3$, with $a$ a generator (since $1$ only generates $1$ under multiplication), and as such we have:

$a^2 = a+1$.

This pretty much determines the entire multiplication table:

$a(a+1) = a^2 + a = (a+1) + a = a + (1 + a) = a + (a + 1) = (a + a) + 1 = 0 + 1 = 1$
$(a+1)(a+1) = a^2 + a + a + 1 = a^2 + 0 + 1 = a + 1 + 1 = a$

Now, we could check that:

$x(y + z) = xy + xz$ for all $64$ possible choices of $x,y,z$, but maybe there's a better way...

Note that $(a + 1) + (a + 1) = (a + a) + (1 + 1) = 0 + 0 = 0$.

This means we can write:

$a + 1 = -a - 1$, and so:

$a^2 = a + 1 = - a - 1$, and thus $a^2 + a + 1 = 0$.

So we see that $a$ is a ROOT of the polynomial $x^2 + x + 1 \in \Bbb Z_2[x]$.

Note that neither element $0,1$ of $\Bbb Z_2$ (which is a SUBSET of $F$) is such a root.

Now show that $\Bbb Z_2[x]/(x^2 + x + 1)$ yields a four element set with the same addition and multiplication tables (which coset are we mapping to $a$?).

Conclude that $F \cong \phi_a(\Bbb Z_2[x])$ and is therefore a ring, and since $U(F) = F^{\ast}$ (why?), we have a field.

Thanks for your help and support, Deveno ...

Just working through your post in detail and reflecting on what you have said ...

Peter