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.
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2Euclid's theorem on infinitely many primes generalizes as follows: $\tag{}$ Theorem* $\ $ An infinite ring $\rm R$ has infinitely many prime ideals if it has fewer units than it has elements. $\tag*{}$ – Bill Dubuque May 14 '15 at 23:40
4 Answers
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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 $k>j$.
ajotatxe
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Works the same for $ R[X] $ for any (unital) ring $ R $. – Berrick Caleb Fillmore May 14 '15 at 18:57
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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 $n\in\mathbb N$.
Gregory Grant
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It even has an infinite number of prime ideals (for each degree $n$ there is an irreducible polynomial of degree $n$).
Bernard
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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.
David Holden
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