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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 p(x) \in F[x] be an irreducible polynomial of degree n over a field F and let K be the field F[x]/(p(x)). Let \theta = x \ mod \ (p(x)) \in K. Then the elements
1, \theta, {\theta}^2, ... ... , {\theta}^{n-1}
are a basis for K as a vector space over F, so the degree of the extension is n i.e.
[K \ : \ F] = n. Hence
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 \}
consists of all polynomials of degree \lt n in \theta
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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 F = \mathbb{Q} and p(x) = x^3 - 2 which is irreducible by Eisenstein.
Denoting a root of p(x) by \theta we obtain the field
\mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}
with {\theta}^3 = 2 an extension of degree 3. ... ... etc
------------------------------------------------------------------------------------------------------------------------------
Now my problem is that in Theorem 4 we read 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 \} which becomes
K = \{a + b \theta + c {\theta}^2 in the situation of Example 4
But then in Example 4 we have
K = \mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}
?
It seems that in Theorem 4, we have \theta = x \ mod \ (p(x)) but in Example (4) we have \theta = \sqrt[3]{2} and we do not have equality but only an isomorphism, that is \mathbb{Q}[x]/(x^3 - 2) \cong \mathbb{Q}(\sqrt[3]{2}.
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?
PeterNotes:
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
I have a question regarding the nature of extension fields.
Theorem 4 (D&F Section 13.1, page 513) states the following (see attachment):
---------------------------------------------------------------------------------------------------------------
Theorem 4. Let p(x) \in F[x] be an irreducible polynomial of degree n over a field F and let K be the field F[x]/(p(x)). Let \theta = x \ mod \ (p(x)) \in K. Then the elements
1, \theta, {\theta}^2, ... ... , {\theta}^{n-1}
are a basis for K as a vector space over F, so the degree of the extension is n i.e.
[K \ : \ F] = n. Hence
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 \}
consists of all polynomials of degree \lt n in \theta
------------------------------------------------------------------------------------------------------------------------------
However, when we come to Example 4 on page 515 of D&F we read the following: (see attachment)
(4) Let F = \mathbb{Q} and p(x) = x^3 - 2 which is irreducible by Eisenstein.
Denoting a root of p(x) by \theta we obtain the field
\mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}
with {\theta}^3 = 2 an extension of degree 3. ... ... etc
------------------------------------------------------------------------------------------------------------------------------
Now my problem is that in Theorem 4 we read 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 \} which becomes
K = \{a + b \theta + c {\theta}^2 in the situation of Example 4
But then in Example 4 we have
K = \mathbb{Q}[x]/(x^3 - 2) \cong \{a + b \theta + c {\theta}^2 \ | \ a, b, c \in \mathbb{Q}
?
It seems that in Theorem 4, we have \theta = x \ mod \ (p(x)) but in Example (4) we have \theta = \sqrt[3]{2} and we do not have equality but only an isomorphism, that is \mathbb{Q}[x]/(x^3 - 2) \cong \mathbb{Q}(\sqrt[3]{2}.
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?
PeterNotes:
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