UTM Ideals Varieties and Algorithm - Chapter 1 Section 4 Exercise 13

Problem:

Solution:

Another interesting problem. Now we are in the arena of finite fields.

Part (a) is simple. First, we denote that $ f(x) = x^2 - x $, $ g(y) = y^2 - y $. We can simply check that $ f(0) = f(1) = 0 $, $ g(0) = g(1) = 0 $.

Any polynomial $ p \in \langle x^2 - x, y^2 - y \rangle $ has the form $ a(x, y)f(x) + b(x, y)g(y) $, so $ p(0, 0) = p(0, 1) = p(1, 0) = p(1, 1) = 0 $, so $ p \in I $.

Part (b) is the interesting part. Consider $ a $ as a polynomial in $ \mathbf{F}_2[x][y] $, then the division algorithm shows that $ a(x, y) = (y^2 - y)b(x, y) + c(x)y + d(x) $.

Next, we divide $ c(x) $ and $ d(x) $ by $ (x^2 - x) $ and get:

$ c(x) = e(x)(x^2 - x) + fx + g $, $ d(x) = h(x)(x^2 - x) + jx + k $.

Putting these all back together we get:

$ \begin{eqnarray*} a(x, y) &=& (y^2 - y)b(x, y) + c(x)y + d(x) \\ &=& (y^2 - y)b(x, y) + (e(x)(x^2 - x) + fx + g)y + (h(x)(x^2 - x) + jx + k) \\ &=& (y^2 - y)b(x, y) + e(x)(x^2 - x)y + fxy + gy + h(x)(x^2 - x) + jx + k \\ &=& (y^2 - y)b(x, y) + e(x)(x^2 - x)y + h(x)(x^2 - x) + fxy + gy + jx + k \\ &=& (y^2 - y)b(x, y) + (x^2 - x)(e(x)y + h(x)) + fxy + gy + jx + k \\ \end{eqnarray*} $

Now we show the required form.

For part (c), let $ f(x, y) = axy + bx + cy + d $, we have:

$ f(0, 0) = d= 0 $
$ f(1, 0) = b + d = b + 0 = 0 $
$ f(0, 1) = c + d = c + 0 = 0 $.
$ f(1, 1) = a + b + c + d = a + 0 + 0 + 0 = 0 $.

So we showed that $ a = b = c = d = 0 $.

For part (d), we know that any polynomial $ p \in I $ can be written as the form in part (b), and part (c) guarantee $ a = b = c = d = 0 $, so $ p \in \langle x^2 - x, y^2 - y \rangle $.

For part (e), we use the division algorithm we have above:

$ x^2y + y^2x = x(y^2 - y) + (x^2y + xy) = x(y^2 - y) + (x^2 - x)y + 2xy = x(y^2 - y) + (x^2 - x)y $.

The last term disappear because $ 2 = 0 \in \mathbf{F}_2 $.