UTM Ideals Varieties and Algorithm - Chapter 2 Section 2 Exercise 7

Problem:

Solution:

It took me a good while to solve this problem, but looking in retrospect, this problem is easy. It is just that the notation tripped me up.

(a)

First, note that $ \alpha \in \mathbf{Z}_{\ge 0}^{n} $, so $ \alpha = 0 $ means it is an n-tuple of all 0, which means all the powers are 0, which means the monomial is actually 1, not 0.

With that in mind, the solution for this is not hard, we do infinity descent here as follow. Suppose the contrary, we have :

$ 0 > \beta $ for some $ \beta $
$ \beta = 0 + \beta  > \beta + \beta =2\beta $
$ 2\beta = 0 + 2\beta > \beta + 2\beta = 3\beta $

and so on.

Now we have a sequence of $ 0, \beta, 2\beta, 3\beta, \cdots $ and this is an infinitely decreasing sequence with no minimum, contradicting the well ordering property required for monomial order.

(b)

Now we know 1 as a term is the minimum monomial for all order, in particular, we know

$ 1 \le x^{\beta - \alpha} $

Multiplying $ x^{\alpha} $ on both side we get the inequality we wanted

$ x^{\alpha} \le x^{\beta} $.

The converse is not true, in particular, we know in lex order

$ x < y $.

But obviously neither divide each other.

(c)

We do a similar infinite descent for this one, support the contrary, we have

$ \alpha + \beta < \alpha $

We reach the same conclusion as above, to be more concrete, we can write

$ \alpha + \beta +k\beta < \alpha + k\beta $.

So we have the sequence $ \alpha, \alpha + \beta, \alpha + 2\beta, \cdots $. This is an infinitely decreasing sequence with no minimum, so again it contradicts the well ordering property.