Let N = {f in Z[x] | f has constant term 0}, clearly N is an ideal of Z[x], show that N is a maximal ideal of Z[x].

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- Apr 19th 2009, 08:59 AMCoda202Maximal Ideal
Let N = {f in Z[x] | f has constant term 0}, clearly N is an ideal of Z[x], show that N is a maximal ideal of Z[x].

- Apr 19th 2009, 09:20 AMTheEmptySet
I don't beleive it is a maximal Ideal... Here is why

Define a Homomorphism from $\displaystyle \phi:\mathbb{Z}[x] \to \mathbb{Z}$ by $\displaystyle \phi(f) \to f(0)$

Note that the kernel of $\displaystyle \phi$ is N

$\displaystyle \phi $ is onto so we can use the first isomorphism theorem to get

$\displaystyle \mathbb{Z}/N \cong \mathbb{Z}$

Since $\displaystyle \mathbb{Z}$ is an Integral Domain this implies that The Ideal is Prime and not maximal.

This ideal in $\displaystyle \mathbb{Z}[x]$ is contained in the Ideal gentered by (2, and the same set f) and this ideal is maximal