Math 480, Spring 2013, Assignment 3

We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other. This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.

- Voltaire

Carefully define the following terms, then give one example and one non-example of each:

1. Ideal.
2. Ideal generated by the set \(S\).
3. Ideal of a variety.
4. Leading term (of a univariate polynomial).
5. Greatest common divisor of a set of polynomials.

Carefully state the following theorems (you need not prove them):

1. Theorem concerning the division algorithm (Proposition 1.5.2 in the text).
2. Classification of ideals in \(k[x]\) (Corollary 1.5.4 in the text).
3. Theorem on the GCD of many polynomials (Proposition 1.5.8 in the text).

Solve the following problems:

1. Consider the three-point set \(V=\{2,3,4\}\subset\mathbb{R}^1\). In the previous assignment you proved that \(V\) is an affine variety. Now find a generator for its ideal \(\mathbf{I}(V)\). (Hint: among all the non-zero polynomials vanishing at 2, 3, and 4, which ones have minimal degree?)
2. Use the algorithm described in class to decide whether the polynomial \(x^3+3x^2+4\) belongs to the ideal \(\left\langle x^2+2\right\rangle\).
3. Use the Euclidean algorithm to compute \(\mbox{GCD}\left(x^4-1, x^6-1\right)\).
4. Decide whether the polynomial \(x^2-4\) belongs to the ideal \(\left\langle x^3+x^2-4x-4, x^3-x^2-4x+4, x^3-2x^2-x+2\right\rangle\).

--------------------End of Assignment--------------------

Questions:

• I have a clarifying question to ask about question 4. If we could show that 1 over a polynomial (e.x \(\frac{1}{x^2-4}\)) multiplied by one of the polynomials in the ideal is equal to \(x^2-4\) is that sufficient to prove, by the property of ideals that says they must respect polynomial multiplication, that \(x^2-4\) is in the ideal? --Robert.Moray (talk) 21:01, 18 February 2013 (EST)

Sorry, I didn't see your question until now. The answer is no, it isn't. Ideals are required to absorb products by arbitrary polynomials, not by rational expressions. (If they did absorb products by rational expressions then every non-zero ideal would contain \(1\) and would hence equal the whole polynomial ring.) - Steven.Jackson (talk) 12:51, 22 February 2013 (EST)