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| + | There are 6 different polynomials that need to be checked since we are in Z_3 = {0,1,2}. I did what Josh did and used the Reducibility Test for Degrees 3 and 3 to determine which ones are the irreducible ones. The only ones that give roots [f(x)=0] are the answers shown above. The only way to get the roots though is to be in Z_3. So for example f(x) = x^2 + 2 when I put in 1 I get 3 but in Z_3 equals 0 and thus why it is irreducible. | ||
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| + | --[[User:Robertsr|Robertsr]] 09:57, 13 November 2008 (UTC) | ||
Latest revision as of 04:57, 13 November 2008
- Find all monic irreducible polynomials of degree 2 over Z3.
x^2 + 1, x^2 + x + 2,x^2 + 2x + 2, this is the answer from the book. But I have no idea what it actually means. Can someone explain?
-Wooi-Chen Ng
Monic means the leading coefficient is 1. Degree two means the highest power is 2. And irreducible means it doesn't factor interestingly. So each polynomial has x^2 as the leading term. Then we just add all the possible combinations of linear polynomials in Z3 and see which do not factor.
-Allen
Since the Reducibility Test for Degrees 2 and 3 (Theorem 17.1) is an if-and-only-if statement, it can also be used instead of trying to factor to determine which ones are the irreducible ones.
-Josh
There are 6 different polynomials that need to be checked since we are in Z_3 = {0,1,2}. I did what Josh did and used the Reducibility Test for Degrees 3 and 3 to determine which ones are the irreducible ones. The only ones that give roots [f(x)=0] are the answers shown above. The only way to get the roots though is to be in Z_3. So for example f(x) = x^2 + 2 when I put in 1 I get 3 but in Z_3 equals 0 and thus why it is irreducible.
--Robertsr 09:57, 13 November 2008 (UTC)
