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.