Let n be a positive integer and consider the ideals in Q[x]. Describe the elements of . What containment relations hold for these ideals? Explain why Q[x] is Noetherian. Explain why the ideals do not contradict the assertion that Q[x] is Noetherian.
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The ideal of course contains the ideal : is by definition all multiples of x by polynomials p(x) in Q[x], and is similarly all multiplies of x^2, and any such multiple of x^2 is q(x)x^2 = (q(x)x)x, so is also in . Similarly, if n < m, contains . However, does not contain x, and similarly does not contain for n < m. That is, we have the following sequence of strict containments:
(">" means "strictly contains" here)
> > > ... > > ...
Now, this is a chain of ideals. The most important condition for a ring to be Noetherian is that it satisfies the ascending chain condition, which says that any ascending chain must terminate after a finite number of proper containments. At first glance, since we have an infinite number of proper containments in the above chain, it might seem that Q[x] is not Noetherian. But the above chain is *descending* instead of ascending, so it's actually not a contradiction to Q[x] being Noetherian.
Also, Q[x] is in fact Noetherian. This follows in several ways. First, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is itself Noetherian; since Q is a field, it is obviously Noetherian, as it only has two ideals, {0} and Q. More basically, Q[x] is a Euclidean domain since the usual polynomial division algorithm applies. Any Euclidean domain is a principal ideal domain (which follows from applying the division algorithm to an element of minimum norm in the ideal; that element can be shown to generate the ideal). Any principal ideal domain is itself Noetherian, since for any ascending chain of ideals, you can just take the union of the ideals in the chain, which must be an ideal, and so must be principal; that union's generator is in some ideal, and so the chain stabilizes after that ideal, which occurs after only finitely many inclusions.
(">" means "strictly contains" here)
Now, this is a chain of ideals. The most important condition for a ring to be Noetherian is that it satisfies the ascending chain condition, which says that any ascending chain must terminate after a finite number of proper containments. At first glance, since we have an infinite number of proper containments in the above chain, it might seem that Q[x] is not Noetherian. But the above chain is *descending* instead of ascending, so it's actually not a contradiction to Q[x] being Noetherian.
Also, Q[x] is in fact Noetherian. This follows in several ways. First, Hilbert's basis theorem says that a polynomial ring over a Noetherian ring is itself Noetherian; since Q is a field, it is obviously Noetherian, as it only has two ideals, {0} and Q. More basically, Q[x] is a Euclidean domain since the usual polynomial division algorithm applies. Any Euclidean domain is a principal ideal domain (which follows from applying the division algorithm to an element of minimum norm in the ideal; that element can be shown to generate the ideal). Any principal ideal domain is itself Noetherian, since for any ascending chain of ideals, you can just take the union of the ideals in the chain, which must be an ideal, and so must be principal; that union's generator is in some ideal, and so the chain stabilizes after that ideal, which occurs after only finitely many inclusions.