# Field Theory - Basic Theory - D&F Section 13.1

#### Peter

##### Well-known member
MHB Site Helper
I am reading Dummit and Foote (D&F) Section 13.1 Basic Theory of Field Extensions.

I have a question regarding the nature of extension fields.

Theorem 4 (D&F Section 13.1, page 513) states the following (see attachment):

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Theorem 4. Let [TEX] p(x) \in F[x] [/TEX] be an irreducible polynomial of degree n over a field F and let K be the field [TEX] F[x]/(p(x)) [/TEX]. Let [TEX] \theta = x \ mod \ (p(x)) \in K [/TEX]. Then the elements

[TEX] 1, \theta, {\theta}^2, ... ... , {\theta}^{n-1} [/TEX]

are a basis for K as a vector space over F, so the degree of the extension is n i.e.

[TEX] [K \ : \ F] = n [/TEX]. Hence

[TEX] K = \{ a_0 + a_1 \theta + a_2 {\theta}^2 + ... ... + a_{n-1} {\theta}^{n-1} \ | \ a_0, a_1, ... ... , a_{n-1} \in F \} [/TEX]

consists of all polynomials of degree [TEX] \lt n [/TEX] in [TEX] \theta [/TEX]

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However, when we come to Example 4 on page 515 of D&F we read the following: (see attachment)

(4) Let [TEX] F = \mathbb{Q} [/TEX] and [TEX] p(x) = x^3 - 2 [/TEX] which is irreducible by Eisenstein.

Denoting a root of p(x) by [TEX] \theta [/TEX] we obtain the field

[TEX] \mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q} [/TEX]

with [TEX] {\theta}^3 = 2 [/TEX] an extension of degree 3. ... ... etc

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Now my problem is that in Theorem 4 we read

[TEX] K = \{ a_0 + a_1 \theta + a_2 {\theta}^2 + ... ... + a_{n-1} {\theta}^{n-1} \ | \ a_0, a_1, ... ... , a_{n-1} \in F \} [/TEX] which becomes

[TEX] K = \{a + b \theta + c {\theta}^2 [/TEX] in the situation of Example 4

But then in Example 4 we have

[TEX] K = \mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q} [/TEX]

???

It seems that in Theorem 4, we have [TEX] \theta = x \ mod \ (p(x)) [/TEX] but in Example (4) we have [TEX] \theta = \sqrt[3]{2} [/TEX] and we do not have equality but only an isomorphism, that is [TEX] \mathbb{Q}[x]/(x^3 - 2) \cong \mathbb{Q}(\sqrt[3]{2} [/TEX].

In Field theory we seem to prove that an irreducible polynomial has a root in a field that is isomorphic to the actual field that contains the root.

Does what I am saying make sense? Can someone clarify this issue for me?

Peter

Notes:

1. I think Deveno was trying to clarify this for me in a previous post but since I was not quite sure of things I am trying to further clarify the issue

2. The above has also been posted on MHF

#### Deveno

##### Well-known member
MHB Math Scholar
Yes, $\Bbb Q[x]/(x^3 - 2)$ and $\Bbb Q(\sqrt[3]{2})$ are two distinct fields, one has elements composed of "cosets of polynomials", and the other has elements of $\Bbb Q$-linear combinations of $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, which are certain real numbers.

But ALGEBRAICALLY, this difference is purely "symbolic", the field isomorphism which sends:

$q \mapsto q + (x^3 - 2), q \in \Bbb Q$
$\sqrt[3]{2} \mapsto x + (x^3 - 2)$

shows these two fields "act" just the same, so the algebra is the same.

Technically speaking, the field $\Bbb Q[x]/(x^3 - 2)$ isn't an extension of $\Bbb Q$, but rather of the isomorphic field:

$\overline{\Bbb Q} = \{q + (x^3 + 2): q \in \Bbb Q\}$.

But keeping this notation becomes cumbersome, if all we are interested in is algebraic qualities.

As another example, the subset of "constant polynomial functions" of $\Bbb Q[x]$ isn't the same as the set of "rational numbers" ($\Bbb Q$) as one is composed of numbers, and the other is composed of FUNCTIONS. But the algebra is the same, and algebra is presumably what we are investigating.

The key is this: (field) isomorphism is an equivalence relationship amongst fields, and it is common practice to regard equivalence CLASSES of fields, rather than the actual fields themselves (it makes things easier to talk about). We do this all the time with other things, and rarely notice it:

The "fraction" $\frac{1}{2}$ is actually an equivalence class of LOTS of fractions:

$\frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \cdots$

But rather than think of $\frac{1}{2}$ as "an infinite amount of equivalent ratios", we think of it as some definite value: half of one. We often parenthetically reference this by referring to a fraction $\frac{p}{q}$ in "lowest terms" (that is choosing a particular well-defined representative of the equivalence class).

A careful reading of most algebra texts on field extensions will probably indicate somewhere that when a symbolic letter like $F(a)$ is being used, one of two situations is the case:

A) $a$ lies in some explicitly defined extension $E$ of $F$ (like the real number $\sqrt{2}$).

B) $a$ is a symbolic adjunction of a formal root of some polynomial $p(x)$. The adjunction polynomial doesn't actually exists in $F[x]$, really, it lies in the isomorphic ring: $\overline{F}[x]$ which has coefficients of the form $c_j + (p(x))$, which "acts" just like $F[x]$, since in a quotient ring, we have (for any ideal $I$):

$(a + I) + (b + I) = (a+b) + I$
$(a + I)(b + I) = ab + I$

and it is needless clutter to keep carrying around the "$+I$".

This process of regarding two isomorphic fields as "the same" is called IDENTIFICATION, and is meant to keep brevity and clarity in the exposition.

#### Peter

##### Well-known member
MHB Site Helper
Yes, $\Bbb Q[x]/(x^3 - 2)$ and $\Bbb Q(\sqrt[3]{2})$ are two distinct fields, one has elements composed of "cosets of polynomials", and the other has elements of $\Bbb Q$-linear combinations of $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$, which are certain real numbers.

But ALGEBRAICALLY, this difference is purely "symbolic", the field isomorphism which sends:

$q \mapsto q + (x^3 - 2), q \in \Bbb Q$
$\sqrt[3]{2} \mapsto x + (x^3 - 2)$

shows these two fields "act" just the same, so the algebra is the same.

Technically speaking, the field $\Bbb Q[x]/(x^3 - 2)$ isn't an extension of $\Bbb Q$, but rather of the isomorphic field:

$\overline{\Bbb Q} = \{q + (x^3 + 2): q \in \Bbb Q\}$.

But keeping this notation becomes cumbersome, if all we are interested in is algebraic qualities.

As another example, the subset of "constant polynomial functions" of $\Bbb Q[x]$ isn't the same as the set of "rational numbers" ($\Bbb Q$) as one is composed of numbers, and the other is composed of FUNCTIONS. But the algebra is the same, and algebra is presumably what we are investigating.

The key is this: (field) isomorphism is an equivalence relationship amongst fields, and it is common practice to regard equivalence CLASSES of fields, rather than the actual fields themselves (it makes things easier to talk about). We do this all the time with other things, and rarely notice it:

The "fraction" $\frac{1}{2}$ is actually an equivalence class of LOTS of fractions:

$\frac{1}{2} = \frac{2}{4} = \frac{3}{6} = \cdots$

But rather than think of $\frac{1}{2}$ as "an infinite amount of equivalent ratios", we think of it as some definite value: half of one. We often parenthetically reference this by referring to a fraction $\frac{p}{q}$ in "lowest terms" (that is choosing a particular well-defined representative of the equivalence class).

A careful reading of most algebra texts on field extensions will probably indicate somewhere that when a symbolic letter like $F(a)$ is being used, one of two situations is the case:

A) $a$ lies in some explicitly defined extension $E$ of $F$ (like the real number $\sqrt{2}$).

B) $a$ is a symbolic adjunction of a formal root of some polynomial $p(x)$. The adjunction polynomial doesn't actually exists in $F[x]$, really, it lies in the isomorphic ring: $\overline{F}[x]$ which has coefficients of the form $c_j + (p(x))$, which "acts" just like $F[x]$, since in a quotient ring, we have (for any ideal $I$):

$(a + I) + (b + I) = (a+b) + I$
$(a + I)(b + I) = ab + I$

and it is needless clutter to keep carrying around the "$+I$".

This process of regarding two isomorphic fields as "the same" is called IDENTIFICATION, and is meant to keep brevity and clarity in the exposition.
Thanks Deveno, Just working through this now

Peter

Yes, definitely helped a lot Deveno ... think I am following ... going on to splitting fields and the other topics in D&F Chapter 13 and then on to Galois Theory ... now have more confidence to go forward

Peter

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