# Field Extensions - Lovett, Theorem 7.1.10 .... ....

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In summary, Lovett is proving that a polynomial of minimal degree is irreducible in a certain field, and that it has a linear factor if and only if the degree of the linear factor is the same as the degree of the polynomial. He shows that the polynomial is reducible by showing that there is a linear factor which has the same degree as the polynomial.
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I am reading Abstract Algebra: Structures and Applications" by Stephen Lovett ...

I am currently focused on Chapter 7: Field Extensions ... ...

I need help with an aspect of the proof of Theorem 7.1.10 ...Theorem 7.1.10, and the start of its proof, reads as follows:

In the above text from Lovett we read the following ...

" ... ... Let ##p(x)## be a polynomial of least degree such that ##p( \alpha ) = 0## ... ... "

Then Lovett goes on to prove that ##p(x)## is irreducible in ##F[x]## ... ...

... BUT ... I am confused by this since it is my understanding that if ##p( \alpha ) = 0## then ##p(x)## has a linear factor ##x - \alpha## in ##F[x]## and so is not irreducible ... ... ?Can someone please help clarify this issue ... ...

Peter

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You can use polynomial division here. One can always write ##p(x) = q(x) \cdot (x-\alpha) + s(x)## with ##\deg s(x) < \deg p(x)##.
Can you conclude the rest from here?

Math Amateur
fresh_42 said:
You can use polynomial division here. One can always write ##p(x) = q(x) \cdot (x-\alpha) + s(x)## with ##\deg s(x) < \deg p(x)##.
Can you conclude the rest from here?
If ##p(x) = q(x) \cdot (x-\alpha) + s(x)## and ##p( \alpha ) = 0## then we must have ##s( \alpha) = 0##

Given this ... I do not know how to square this with D&F Proposition 9, Section 9.4, page 307 which seems to imply that if ##p ( \alpha ) = 0## for ##\alpha \in F## then ##p(x)## has a linear factor ... and is therefore not irreducible ...

Peter

Math Amateur said:
If ##p(x) = q(x) \cdot (x-\alpha) + s(x)## and ##p( \alpha ) = 0## then we must have ##s( \alpha) = 0##

Given this ... I do not know how to square this with D&F Proposition 9, Section 9.4, page 307 which seems to imply that if ##p ( \alpha ) = 0## for ##\alpha \in F## then ##p(x)## has a linear factor ... and is therefore not irreducible ...

Peter
##p(x)## has been chosen to be a (non-zero) polynomial of minimal degree with ##p(\alpha)=0##. But now ##s(\alpha)=0## and ##\deg s(x) < \deg p(x)## which can only hold, if ##s(x) = 0##. Going back to the equation we have ##p(x) = q(x) \cdot (x-\alpha)## or ##(x-\alpha) \mid p(x)##.

Math Amateur
fresh_42 said:
##p(x)## has been chosen to be a (non-zero) polynomial of minimal degree with ##p(\alpha)=0##. But now ##s(\alpha)=0## and ##\deg s(x) < \deg p(x)## which can only hold, if ##s(x) = 0##. Going back to the equation we have ##p(x) = q(x) \cdot (x-\alpha)## or ##(x-\alpha) \mid p(x)##.
Well ... we seem to have shown that ##p(x)## is reducible as a consequence of ##p( \alpha ) = 0## ...

Exactly my problem with this ...

... hmmm ... how are we going to end up with ##p(x)## being irreducible when it has a linear factor ... ?

Peter

The equation ##p(x)= q(x)\cdot (x-\alpha) + s(x)## doesn't exist in ##F[x]##, because ##\alpha \notin F##. It only exists in ##F[\alpha][x]##.
Lovett shows that ##p(x)\in F[x]## is irreducible, because ##\alpha## isn't available in ##F##. However, it is reducible in ##F[\alpha][x]\;\;\;\;## so the argument with the linear factor (which we have shown is true) only holds over ##F[\alpha]##. Irreducibility depends on the ring under consideration, or likewise on the coefficients which are available in the corresponding domain of scalars.

##p(x) \in F[x]## irreducible (Lovett's proof)
##p(x) \in F[\alpha][x]## reducible, by your argument with the linear factor ##x-\alpha##

Sorry for the confusion, which I think I might have caused, by proving your assertion with the linear factor, instead of recognizing, that there are two different rings involved here: ##F[\alpha][x]## and ##F[x]##, which makes the difference.

Math Amateur
fresh_42 said:
The equation ##p(x)= q(x)\cdot (x-\alpha) + s(x)## doesn't exist in ##F[x]##, because ##\alpha \notin F##. It only exists in ##F[\alpha][x]##.
Lovett shows that ##p(x)\in F[x]## is irreducible, because ##\alpha## isn't available in ##F##. However, it is reducible in ##F[\alpha][x]\;\;\;\;## so the argument with the linear factor (which we have shown is true) only holds over ##F[\alpha]##. Irreducibility depends on the ring under consideration, or likewise on the coefficients which are available in the corresponding domain of scalars.

##p(x) \in F[x]## irreducible (Lovett's proof)
##p(x) \in F[\alpha][x]## reducible, by your argument with the linear factor ##x-\alpha##

Sorry for the confusion, which I think I might have caused, by proving your assertion with the linear factor, instead of recognizing, that there are two different rings involved here: ##F[\alpha][x]## and ##F[x]##, which makes the difference.
Thanks so much for the help in clarifying things ... really has been most helpful ...

Peter

## 1. What is a field extension in mathematics?

A field extension is a mathematical concept that involves extending a smaller field into a larger one by adding new elements. This is done to create a larger field that contains all of the elements of the smaller field, as well as additional elements that were not previously included.

## 2. How is a field extension related to Lovett's Theorem 7.1.10?

Lovett's Theorem 7.1.10 is a specific theorem that applies to finite field extensions, which are a type of field extension. This theorem states that the degree of a finite field extension is equal to the product of the degrees of its subextensions. In other words, it provides a way to calculate the degree of a finite field extension.

## 3. What is the significance of Lovett's Theorem 7.1.10?

Lovett's Theorem 7.1.10 is significant because it allows mathematicians to easily calculate the degree of a finite field extension. This can be useful in many mathematical contexts, such as in algebraic number theory and algebraic geometry.

## 4. Can Lovett's Theorem 7.1.10 be applied to infinite field extensions?

No, Lovett's Theorem 7.1.10 only applies to finite field extensions. This is because infinite field extensions have infinitely many subextensions, making it impossible to calculate their degree using this theorem.

## 5. What are some practical applications of field extensions in real-world problems?

Field extensions have many applications in various fields of mathematics, including cryptography, coding theory, and algebraic geometry. In the real world, they are used in applications such as error-correcting codes for data transmission, secure communication protocols, and in the study of elliptic curves and their applications in cryptography.

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