Do Irreducibles Induce Algebraic Extensions?

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SUMMARY

Given a field K and an irreducible monic polynomial f in K[x], the quotient K[x]/(f) is confirmed as an algebraic extension of K. This conclusion is supported by the properties of maximal ideals in commutative rings, where K[x] is a Euclidean ring and thus a principal ideal domain (p.i.d.). The degree of the extension corresponds to the degree of the polynomial f, establishing that every finite extension is algebraic. Therefore, K[x]/(f) is definitively a field extension of K.

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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]?
 
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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.
 
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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