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 September 28th, 2012, 09:49 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 generating a maximal ideal, is this correct? Could someone verify If my way of thinking here is correct? let $\mathbb{Z}[x]$ and we know that maximal ideals of the ring has the form $m=(p,f(x))$ where p is prime and $f(x)\in\mathbb{Z}[x]$. 1) we choose an irreducible polynomial $f_{0}\in \mathbb{F}_{p}$ and then we lift $f_{0}$ any way we want to polynomial $f(x)\in\mathbb{Z}[x]$ say I choose $p=3$ and $x^3+2x+2\in\mathbb{Z}_{2}[x]$ then the pair $(3, 100x^3+3x^2+5x-1)$ is a maximal ideal of $\mathbb{Z}[x]$ and $\mathbb{Z}[x]/M=\mathbb{F}_{2}$ is a finite algebraic extension field. some other example $p=2$ and $x^2+x+1\in\mathbb{Z}_{2}[x]$ then I lift the polynomial mod 2 to $7x^2-x+3$ or $x^2+x+1$ to $x^2+3x-1$ so I can get 2 maximal ideals $m_{1}=(2,7x^2-x+3)$ $m_{2}=(2,x^2+3x-1)$ Is this right? Thanks
October 4th, 2012, 11:57 AM   #2
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Re: generating a maximal ideal, is this correct?

Unfortunately, it is not easy to see what exacly you are trying to show. Most of what you say is correct up to minor typos.

Quote:
 Originally Posted by rayman Could someone verify If my way of thinking here is correct? let $\mathbb{Z}[x]$ and we know that maximal ideals of the ring has the form $m=(p,f(x))$ where p is prime and $f(x)\in\mathbb{Z}[x]$.
It is correct that every maximal ideal of $\mathbb{Z}[x]$ is of that form, yes.
Quote:
 1) we choose an irreducible polynomial $f_{0}\in \mathbb{F}_{p}$ and then we lift $f_{0}$ any way we want to polynomial $f(x)\in\mathbb{Z}[x]$
yes, this construction will give exacly the maximal ideals.
Quote:
 say I choose $p=3$ and $x^3+2x+2\in\mathbb{Z}_{2}[x]$ then the pair $(3, 100x^3+3x^2+5x-1)$ is a maximal ideal of $\mathbb{Z}[x]$ and $\mathbb{Z}[x]/M=\mathbb{F}_{2}$ is a finite algebraic extension field.
I suppose all the 2's in the index should be 3's, then it is true, that you will get a finite extension of $\mathbb{F}_{3}$, but it will be a proper extension, i.e. the "="-sign in $\mathbb{Z}[x]/M=\mathbb{F}_{3}$ is wrong.
Quote:
 some other example $p=2$ and $x^2+x+1\in\mathbb{Z}_{2}[x]$ then I lift the polynomial mod 2 to $7x^2-x+3$ or $x^2+x+1$ to $x^2+3x-1$ so I can get 2 maximal ideals $m_{1}=(2,7x^2-x+3)$ $m_{2}=(2,x^2+3x-1)$ Is this right? Thanks
Well it is true, however it is worth noting, that the ideals you get by chosing diffrent lifts of the polynomial will allways be the same. In particular m_1=m_2. To get different ideals you can take either different p's or take different irreducible polynomials in $\mathbb{F}_{p}[x]$.

 October 4th, 2012, 11:06 PM #3 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 Re: generating a maximal ideal, is this correct? thank you for your reply I just wanted to find out some systematical way of producing these ideals. So these polynomials can be of any degree as long as they are irreducible right?
 October 6th, 2012, 09:17 AM #4 Senior Member   Joined: Sep 2008 Posts: 150 Thanks: 5 Re: generating a maximal ideal, is this correct? correct. As long, as by irriduceable you imply ireducible as a polynomial over $\mathbb{F}_p$. The degree of the polynomial (mod p) will then be the degree of the (finite algebraic) field extension $(\mathbb{Z}[X]/M)|\mathbb{F}_p$.

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