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Do Irreducibles Induce Algebraic Extensions? 
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#1
Jan2911, 07:43 PM

P: 15

Given [tex]K[/tex] a field, and [tex] f\in K[x][/tex] an irreducible (monic) polynomial. Does it follow that the field [tex] K[x]/\left<f\right>[/tex] is an algebraic extension of [tex]K[/tex]?



#2
Jan2911, 08:20 PM

Sci Advisor
HW Helper
P: 9,470

i think so. in general, if M is a maximal ideal in a commutative ring R, then R/M is a field. If R contains a subfield k, then R/M is an extension of k. So in your case the ring K[X] contains the subfield K, and since K[X] is a Euclidean ring, hence also a p.i.d., an irreducible polynomial f generates a maximal ideal, so K[X]/(f) is a field extension of K.
Moreover the degree (vector dimension) of the extension equals the degree of f, hence is finite, and every finite extension is definitely algebraic. so YES! I had to think through all the details since I am old and losing my memory. hope this helps. 


#3
Jan2911, 08:23 PM

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Sci Advisor
PF Gold
P: 16,091

Isn't the answer obvious from the definition of "algebraic extension"? The interesting part is that it's a field and not just a ring.
(Hrm. I suppose there are equivalent definitions, and some would be less obvious than others. Which are you using?) 


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