Exercise 4.4.1
Let be the field of rational numbers. Determine which of the following subsets of are ideals. When the set is an ideal, find its monic generator.
(a) all of even degree;
(b) all of degree ;
(c) all such that ;
(d) all such that ;
(e) all in the range of the linear operator defined by
Solution:
(a) This is not an ideal. Let be any polynomial of even degree. Let . Then which is odd and is therefore the set of polynomials of even degree does not satisfy the necessary property that is in the set whenever is in the set.
(b) This is not an ideal. Let and . Then and are in the set and therefore must be in the set if it is an ideal. But has degree equal to four and therefore is not in the set. In other words the set is not closed with respect to addition and therefore is not even a subspace.
(c) This is an ideal. Let . If and then thus is a subspace of . Now suppose and . Then Thus . Thus is an ideal. Let . Then . Thus if then necessarily . Let . Then and Thus generates .
(d) This is an ideal. Let . If and then and thus is a subspace of . Now suppose and . Then and Thus . Thus is an ideal. Let . We claim generates . Let . Then since , by Corollary 1 page 128 it follows that for some . Now since it follows that Thus there is a such that . Thus and therefore is in the ideal generated by .
(e) This is an ideal. In fact it is the same ideal as in part (c). Note that has no constant term, thus . Now let . Let Then . Thus is in the image of . Thus any polynomial with zero constant term is in the image of . Thus it is exactly the same ideal is in part (c).
Exercise 4.4.2
Find the g.c.d. of each of the following paris of polynomials
(a) , ;
(b) , ;
(c) , .
Solution:
This question on its face seems to depend on what the base field is. I’m going to assume it is .
(a) Let and . I have a feeling there’s a mistake and they really meant because then both polynomials are divisible by . But as it is, the g.c.d. of these two polynomials equals 1 – which does not seem that easy to prove using only what we know up to this point. If we knew about unique factorization in we could simply factor into linear factors and check none of the roots of are roots of . In fact the roots of are and .
But we can’t use that argument yet. Instead we are probably expected to argue using the comments on page 132 right after the proof of Theorem 7, commonly known as the “euclidean algorithm”.
Let be the ideal in generated by and . Then
Therefore .
Therefore .
Therefore .Therefore . And therefore . Therefore . Therefore the g.c.d. of and equals 1: gcd.
(b) Let and . Let . We have
Thus .
Thus .Now
Thus divides both and . In particular
and therefore it follows that(c) As in part (a) I have a feeling there’s a typo and they really meant because then both are divisible by . But as it is, is irreducible and the calculations are even worse than part (a). As I did in part (a), I’ll solve this in both the way they actually stated it and the was I think they intended it. First the way they stated it.
Let and and .
Thus .
Thus
And finally
Thus and it follows that and that gcd.
Exercise 4.4.3
Let be an matrix over a field . Show that the set of all polynomials in such that is an ideal.
Solution: Let be the set . Let and . Then Thus is a subspace of as a vector space over . Now suppose and . Then Thus has the required property to be an ideal.
Exercise 4.4.4
Let be a subfield of the complex numbers, and let
Find the monic generator of the ideal of all polynomials in such that .
Solution: Let . We know from the previous exercise that is an ideal. It’s clear that if . Thus if we find any such that then must be a generator. Let . Then
Exercise 4.4.5
Let be a field. Show that the intersection of any number of ideals in is an ideal.
Solution: Let be a set. Let be an ideal in for each . Let . By Theorem 2, page 36 (sec. 2.2), is a subspace of . We must just show satisfies the extra condition required to be an ideal. Specifically, let and . We must show . Since , it follows that . Since is an ideal , it follows that . Thus . Thus .
Exercise 4.4.6
Let be a field. Show that the ideal generated by a finite number of polynomials in is the intersection of all ideals containing .
Solution: Let be the ideal generated by . Let be any ideal containing . Let . Then for some in . Since is an ideal containing it must also contain . Thus . Thus . Now by the definition of intersection, is contained in every ideal which contains . Since is such an ideal, it follows that . Thus we’ve shown and . Thus .
Exercise 4.4.7
Let be a subfield of a field , and suppose , are polynomials in . Let be the ideal generated by and in and be the ideal they generate in . Show that and have the same monic generator.
Solution 1: We first note a general fact. Let be any field. Let such that gcd. Let be monic. We claim gcd in . Since gcd, the ideal generated by and is the same as the ideal generated by . In other words . Thus . Since is monic by assumption and is monic by definition of gcd, it follows that is monic – and therefore satisifes the definition of gcd in .
Now let in . So and for some . And such that and . Once you can write the three equations , and it follows that the ideal in generated by and is the same as the ideal generated by . That’s what it means for to be the g.c.d. of and . But these three equalities also hold in and consequently the ideal in generated by and is the same as the one generated by .
Solution 2: This might be easier if they had formalized what is discussed informally on page 132 after the proof of Theorem 7. What they are describing is the so-called “eucidean algorithm”. If you have two polynomials and and say . Then you divide into and take the remainder . That remainder is also in the ideal generated and . Then replace and with and and do the same thing, divide into and the remainder has degree less than . Eventually you must arrive at a remainder equal to zero or of degree equal to zero (i.e. a scalar). If the remainder is zero then the previous remainder generates the ideal in . If the remainder is a non-zero scalar then the ideal contains and therefore . The euclidean algorithm allows us to always find the gcd in a finite number of steps.
Once we have the euclidean algorithm, all we need to do to solve this problem is to note that all operations in the euclidean algorithm happen in and so if then in the same operations will lead to the same generator which will therefore still live in the smaller ring .
From http://greggrant.org