Producing a field with 343 elements

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Producing a field with $7^3=343$ elements.



Okay, so if I can find an irreducible polynomial over $mathbb Z_7$ of degree 3 then I'll have done it.



Now, since it's of degree three all I have to do is check for linear factors by finding a degree 3 polynomial with no roots. I could just guess and check, but I was wondering if there was a more methodical way to do this, perhaps there is an insight that i'm missing. Thanks!






abstract-algebra field-theory irreducible-polynomials







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  • Kummer extension?
    – Lord Shark the Unknown
    19 mins ago










  • Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
    – Michael Vaughan
    15 mins ago










  • Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
    – Jyrki Lahtonen
    1 min ago














up vote
4
down vote

favorite












Producing a field with $7^3=343$ elements.



Okay, so if I can find an irreducible polynomial over $mathbb Z_7$ of degree 3 then I'll have done it.



Now, since it's of degree three all I have to do is check for linear factors by finding a degree 3 polynomial with no roots. I could just guess and check, but I was wondering if there was a more methodical way to do this, perhaps there is an insight that i'm missing. Thanks!






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question























  • Kummer extension?
    – Lord Shark the Unknown
    19 mins ago










  • Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
    – Michael Vaughan
    15 mins ago










  • Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
    – Jyrki Lahtonen
    1 min ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











Producing a field with $7^3=343$ elements.



Okay, so if I can find an irreducible polynomial over $mathbb Z_7$ of degree 3 then I'll have done it.



Now, since it's of degree three all I have to do is check for linear factors by finding a degree 3 polynomial with no roots. I could just guess and check, but I was wondering if there was a more methodical way to do this, perhaps there is an insight that i'm missing. Thanks!






abstract-algebra field-theory irreducible-polynomials







share|cite|improve this question















Producing a field with $7^3=343$ elements.



Okay, so if I can find an irreducible polynomial over $mathbb Z_7$ of degree 3 then I'll have done it.



Now, since it's of degree three all I have to do is check for linear factors by finding a degree 3 polynomial with no roots. I could just guess and check, but I was wondering if there was a more methodical way to do this, perhaps there is an insight that i'm missing. Thanks!






abstract-algebra field-theory irreducible-polynomials




abstract-algebra field-theory irreducible-polynomials






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edited 7 mins ago









Parcly Taxel

34k136789




34k136789










asked 21 mins ago









Michael Vaughan

54611




54611











  • Kummer extension?
    – Lord Shark the Unknown
    19 mins ago










  • Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
    – Michael Vaughan
    15 mins ago










  • Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
    – Jyrki Lahtonen
    1 min ago
















  • Kummer extension?
    – Lord Shark the Unknown
    19 mins ago










  • Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
    – Michael Vaughan
    15 mins ago










  • Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
    – Jyrki Lahtonen
    1 min ago















Kummer extension?
– Lord Shark the Unknown
19 mins ago




Kummer extension?
– Lord Shark the Unknown
19 mins ago












Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
– Michael Vaughan
15 mins ago




Nah I'm supposed to do it this way by finding an irreducible polynomial >.<
– Michael Vaughan
15 mins ago












Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
– Jyrki Lahtonen
1 min ago




Related 1, 2. Not gonna use my dupehammer. Only gonna make the calculus rep farmers look better.
– Jyrki Lahtonen
1 min ago










1 Answer
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$x^3equivpm1bmod7$ for all $xin1,dots,6$. We thus choose $x^3+2$ as the irreducible polynomial.






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  • Probably simplest.
    – Jyrki Lahtonen
    8 mins ago










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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
6
down vote













$x^3equivpm1bmod7$ for all $xin1,dots,6$. We thus choose $x^3+2$ as the irreducible polynomial.






share|cite|improve this answer




















  • Probably simplest.
    – Jyrki Lahtonen
    8 mins ago














up vote
6
down vote













$x^3equivpm1bmod7$ for all $xin1,dots,6$. We thus choose $x^3+2$ as the irreducible polynomial.






share|cite|improve this answer




















  • Probably simplest.
    – Jyrki Lahtonen
    8 mins ago












up vote
6
down vote










up vote
6
down vote









$x^3equivpm1bmod7$ for all $xin1,dots,6$. We thus choose $x^3+2$ as the irreducible polynomial.






share|cite|improve this answer












$x^3equivpm1bmod7$ for all $xin1,dots,6$. We thus choose $x^3+2$ as the irreducible polynomial.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 10 mins ago









Parcly Taxel

34k136789




34k136789











  • Probably simplest.
    – Jyrki Lahtonen
    8 mins ago
















  • Probably simplest.
    – Jyrki Lahtonen
    8 mins ago















Probably simplest.
– Jyrki Lahtonen
8 mins ago




Probably simplest.
– Jyrki Lahtonen
8 mins ago

















 

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