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I am looking for irreducible polynomials in $\mathbb{F}_{11}$ with the the form h(y) = $y^7+$. My considerations are: the group order is 10 and is relatively prime to the degree 7 of the polynomial. Therefore no irreducible polynomials exist of this form. Is this correct ?

MathPowerUser
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  • this is wrong. In fact there exists irreducible polynomials of every degree over every finite field. – oxeimon Feb 26 '15 at 20:46
  • The exercise says to list all irreducible polynomials. It's a shame, I thought this would be the solution. Is there another trick behind it ? – MathPowerUser Feb 26 '15 at 21:10
  • Can you clarify what are the precise restrictions on your polynomial. Surely you are not asked to list all monic irreducible degree seven polynomials over $F_{11}$ as there are more than a million of them. – quid Feb 26 '15 at 21:13
  • Unfortunately only this information is given. Maybe a typo. – MathPowerUser Feb 26 '15 at 21:16
  • Maybe. Or your initial interpretation is in fact correct. – quid Feb 26 '15 at 22:23

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Your argument is correct only for showing that there is no irreducible polynomial of the form $X^7 + a$ with $a$ in the field. If this is the form you are looking for it is correct. However, not at all each degree seven polynomial is of that form, so that you can not conclude that there are no irreducible polynomials of degree seven.

quid
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