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| Al2009... |
 Posted: Mon Nov 02, 2009 5:39 pm |
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Guest
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Hi,
I made several splitting fields x^q - x over Z_p ( where p is a prime number q=p^n) for the last couple of days.
When we factorise x^q - x over Z_p, the highest degree irreducible polynomial is always degree n.
For instance,
p=2, q=2^2=4, x^4-x = x * (x-1)(x^2 + 1) ---> The highest-degree irreducible polynomial is degree 2, which is x^2 + 1
p=3, q=3^2=9, x^9-x= x(x+1)(x-1)(x^2+1)(x^4 +1) ---> The highest-degree irreducible polynomial is degree 2, which is x^2 + 1, while (x^4+1) is reducible.
p=2, q=2^3 = 8 ,x^8 - x = x(x+1)(x^3+x^2+1)(x^3+x^2+1)---> The highest-degree irreducible polynomial is degree 3, which is x^3+x^2+1.
So all polynomial x^q - x over Z_p, the highest-degree irreducible polynomial is n, where q=p^n.
This does not look trivial to me. Why this pattern appears?
Thank you. |
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| Arturo Magidin... |
| Posted: Mon Nov 02, 2009 6:39 pm |
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Guest
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On Nov 2, 9:39 pm, Al2009 <algebra_whate... at (no spam) yahoo.ca> wrote:
[quote]Hi,
I made several splitting fields x^q - x over Z_p ( where p is a prime number q=p^n) for the last couple of days.
When we factorise x^q - x over Z_p, the highest degree irreducible polynomial is always degree n.
For instance,
p=2, q=2^2=4, x^4-x = x * (x-1)(x^2 + 1) ---> The highest-degree irreducible polynomial is degree 2, which is x^2 + 1
p=3, q=3^2=9, x^9-x= x(x+1)(x-1)(x^2+1)(x^4 +1) ---> The highest-degree irreducible polynomial is degree 2, which is x^2 + 1, while (x^4+1) is reducible.
p=2, q=2^3 = 8 ,x^8 - x = x(x+1)(x^3+x^2+1)(x^3+x^2+1)---> The highest-degree irreducible polynomial is degree 3, which is x^3+x^2+1.
So all polynomial x^q - x over Z_p, the highest-degree irreducible polynomial is n, where q=p^n.
This does not look trivial to me. Why this pattern appears?
[/quote]
Because the degree of the splitting field over Z_p is n; and the
extension is always simple: there is an element alpha such that Z_q Z_p(alpha). So the minimal polynomial of alpha must be a factor of x^q-
x over Z_p; this polynomial must be of degree n, since [Z_p
(alpha):Z_p] = degree(minimal polynomial of (alpha)).
--
Arturo Magidin
[quote]
Thank you.[/quote] |
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| Gerry Myerson... |
| Posted: Mon Nov 02, 2009 11:21 pm |
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Guest
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In article
<1703590632.852.1257219615788.JavaMail.root at (no spam) gallium.mathforum.org>,
Al2009 <algebra_whatever at (no spam) yahoo.ca> wrote:
[quote]Hi,
I made several splitting fields x^q - x over Z_p ( where p is a prime number
q=p^n) for the last couple of days.
[/quote]
That splitting field is the field of q elements.
[quote]When we factorise x^q - x over Z_p, the highest degree irreducible polynomial
is always degree n.
[/quote]
Because the field of q elements is of degree n over Z_p.
--
Gerry Myerson (gerry at (no spam) maths.mq.edi.ai) (i -> u for email) |
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| Al2009... |
| Posted: Tue Nov 03, 2009 11:08 am |
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Guest
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[quote]On Nov 2, 9:39Â pm, Al2009 <algebra_whate... at (no spam) yahoo.ca
wrote:
Hi,
I made several splitting fields x^q - x over Z_p (
where p is a prime number q=p^n) for the last couple
of days.
When we factorise x^q - x over Z_p, the highest
degree irreducible polynomial is always degree n.
For instance,
p=2, q=2^2=4, x^4-x = x * (x-1)(x^2 + 1) ---> The
highest-degree irreducible polynomial is degree 2,
which is x^2 + 1
p=3, q=3^2=9, x^9-x= x(x+1)(x-1)(x^2+1)(x^4 +1)
---> The highest-degree irreducible polynomial is
degree 2, which is x^2 + 1, while (x^4+1) is
reducible.
p=2, q=2^3 = 8 ,x^8 - x =
x(x+1)(x^3+x^2+1)(x^3+x^2+1)---> The highest-degree
irreducible polynomial is degree 3, which is
x^3+x^2+1.
So all polynomial x^q - x over Z_p, the
highest-degree irreducible polynomial is n, where
q=p^n.
This does not look trivial to me. Why this pattern
appears?
Because the degree of the splitting field over Z_p is
n; and the
extension is always simple: there is an element alpha
such that Z_q =
Z_p(alpha). So the minimal polynomial of alpha must
be a factor of x^q-
x over Z_p; this polynomial must be of degree n,
since [Z_p
(alpha):Z_p] = degree(minimal polynomial of (alpha)).
--
Arturo Magidin
Thank you.
[/quote]
That is a really helpful explanation for me to understand the basic ideas of finite fields.
Thank you so much. |
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