infinitely many ideals

Question

does the ring

$\Bbb Z_2[x]$ have infinitely many ideals like $\Bbb Z[x]$?

How do you know if a ring has a finite number of ideal. particularly asking about seemingly large rings.

Answer 1

This case is easy:

$$\{(x),(x^2),\ldots,(x^n),\ldots\}$$

The ideals of this set are all different because $x^j\notin (x^k)$ when $k>j$.

Answer 2

Yes, it does have infinitely many, basically if $f$ and $g$ have different degrees then $(f)$ is different from $(g)$. So at least one for every $n\in\mathbb N$.

Answer 3

It even has an infinite number of prime ideals (for each degree $n$ there is an irreducible polynomial of degree $n$).

Answer 4

the homomorphism $\phi:\Bbb Z \to \Bbb Z_2$ extends to a homomorphism $\phi':\Bbb Z[x] \to \Bbb Z_2[x]$ by setting $\phi'(x)=x$. the kernel of this map is the ideal $2\Bbb Z[x]$ which is prime, so $\Bbb Z_2[x]$ is an integral domain. moreover since $\phi$ is surjective it maps ideals in the domain to ideals in its image so any ideal of $\Bbb Z[x]$ which is not contained in $2\Bbb Z[x]$ will map to a non-zero ideal in $\Bbb Z_2[x]$. in informal language, although $\phi$ certainly reduces the diversity of ideals, there are still plenty left.