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Do Irreducibles Induce Algebraic Extensions?

  1. Jan 29, 2011 #1
    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. jcsd
  3. Jan 29, 2011 #2


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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.
  4. Jan 29, 2011 #3


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    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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