Stewart's Galois Theory doesn't make sense

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The discussion centers on Lemma 3.15 from a book on Galois Theory, which states that if an irreducible polynomial f over a subfield K divides the product of two polynomials g and h, then f must divide at least one of them. The user initially misinterprets the lemma, believing it implies exclusivity between the divisions. However, the correct interpretation clarifies that "either A or B" includes the possibility of both A and B being true. The lemma effectively asserts that irreducibility and primality are equivalent in the polynomial ring K[x].

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I am going through this book, and on page 38, there is

LEMMA 3.15
Let K be a subfield of C, f an irreducible polynomial over K, and g, h polynomials over K. If g divides gh, then either f divides h or f divides h.

OK, so I have proven that f must divide over g or h - i.e., if f doesn't divide g, it must divide h - but it seems that f could still divide both, which is not what the text says.

f = ( x - 1 )

g = ( x - 1 )2 ( x - 2 )

h = ( x - 1 )3 ( x - 3 )

g h = ( x - 1 )5 ( x - 2 ) ( x - 3 )

Clearly, f divides ( g h ), g & h, so the LEMMA is wrong.

What am I missing here?
 
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I assume it is f divides gh. But the either is not necessarily exclusive. But, yes, it could be made more clear, I agree.
 
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swampwiz said:
I am going through this book, and on page 38, there is

LEMMA 3.15
Let K be a subfield of C, f an irreducible polynomial over K, and g, h polynomials over K.
If g
f
divides gh, then either f divides h or f divides h.
g

This is simply the fact that irreducibility and primality are the same thing in ##K[x]##. Lemma 3.15 if written correctly says, that any irreducible polynomial is prime.
 
in mathematics the phrase "either A or B" always means "either A or B or both".
 

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