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 $2 = 0 \in \mathbf{F}_2$.