UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 6

Problem:

Solution:

Another long problem.

For part (a), let's review what is a vector space first.

A vector space is a set of vectors $ v $ such that we can add these vector together and do 'scalar' multiplication, where the scalar is an element of a field $ k $, and in this case we say it is a vector field over $ k $.

To see, how $ k[x] $ can be interpreted as a vector space, we already know the scalar are elements of $ k $, so the vectors must be power of $ x $, and therefore $ \langle I \rangle $ is a subspace with the 0th power missing.

To show that must have an infinite basis, we simply note that there is no way one can construct $ x^n $ from linear combination of $ x^k $ where $ k < n $, thus a basis must be infinite.

For part (b), it is trivial. So trivial that I am not sure if I will remember what I meant, so let's be specific. The form of the solution is:

$ c_1 x + c_2 y  = 0 $

So we pick the coefficients $ c_1 = y $ and $ c_2 = -x $, both of them are not zero and yet the result is 0.

Part (c) is equally trivial. We just simply do exactly the same as above:

$ \sum\limits_{i = 0}^{s} c_i f_i = 0 $

So we pick $ c_1 = f_2 $, $ c_2 = -f_1 $ and $ c_k = 0 $ for other $ k $.

For part (d), we need to be a little more creative, the two representations are

$ \begin{eqnarray*} x^2 + xy + y^2 &=& (x + y)(x) &+& (y)(y) \\ x^2 + xy + y^2 &=& (x)(x) &+& (x + y)(y) \end{eqnarray*} $

So there you go, two representations of the same polynomial.

For part (e), it suffice to show three facts:

$ \langle x + x^2, x^2 \rangle = \langle x \rangle $
$ \langle x + x^2 \rangle \ne \langle x \rangle $
$ \langle x^2 \rangle \ne \langle x \rangle $

All of them are pretty obvious.
$ x = (1)(x + x^2) + (-1)(x^2) \in \langle x + x^2, x^2 \rangle $, this implies $ \langle x \rangle \subset \langle x^2 + x, x^2 \rangle $.
$ x = (1)(x) \in \langle x \rangle $ and $ x + x^2 = (x + 1)(x) \in \langle x \rangle $, this implies $ \langle x + x^2, x \rangle \subset \langle x \rangle $.
Together we have shown $ \langle x + x^2, x^2 \rangle = \langle x \rangle $

Note that for the other two statements, any polynomial in the left hand side subsets necessarily have degree at least 2, so they cannot have the monomial $ x $, which is in $ \langle x \rangle $, so we establish the other two statements. These two statements together indicate it is a minimal basis.

In linear algebra, the minimal basis always have the same size, which is the dimension of the space. But apparently this does not hold with ideals.