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Sunday, February 21, 2016

UTM Ideals Varieties and Algorithm - Chapter 2 Section 1 Exercise 4

Problem:


Solution:

For part (a), in order to show a set is an ideal, we check the two conditions for ideals.

Condition 1: If $ a \in I $, then $ h a \in I $ for all $ h \in R $.

This is true for I because if $ a \in I $, then $ a = x_{t_1}f_1 +  \cdots + x_{t_m}f_m $,
so $ ha = h(x_{t_1}f_1 +  \cdots + x_{t_m}f_m) = x_{t_1}(hf_1) +  \cdots + x_{t_m}(hf_m) \in I $.

Condition 2: If $ a \in I $ and $ b \in I $, then $ a + b \in I $.

This is also true because

$ a \in I $, then $ a = x_{t_1}f_1 +  \cdots + x_{t_m}f_m $,
$ b \in I $, then $ b = x_{u_1}g_1 +  \cdots + x_{u_m}g_m $,

The sum then must also be the same form and therefore $ a + b \in I $.

For part (b), if $ I $ has a finite generating set, then each element is a polynomial and therefore by definition above, a finite linear combination of monomials.

Therefore the number of variables involved in that finite generating set must be finite, but the ideal has infinite number of variables. So there cannot be a finite generating set.

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