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

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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 ... ... ?

Peter

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• Lovett - Theorem 7.1.10 and start of proof .....png
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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
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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?

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

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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
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##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

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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
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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