求同态φ:ℂ[x,y,z]→ℂ[t]的核ker(φ)生成元及表达式推导

Alright, let's break down how to express any element $k(x,y,z) \in \ker(\varphi)$ as a combination of x² - y and x³ - z. Remember, $\ker(\varphi)$ consists of all polynomials where substituting $x=t$, $y=t²$, $z=t³$ gives the zero polynomial in $\mathbb{C}[t]$.

Core Strategy: Iterative Reduction with Substitutions #

The key idea is to systematically eliminate high powers of $y$ and $z$ using the relations $y = x² - (x² - y)$ and $z = x³ - (x³ - z)$, then use the kernel condition to cancel out the remaining pure-$x$ polynomial terms.

Step 1: Reduce all high powers of $y$ #

Treat $k(x,y,z)$ as a polynomial in $y$:
$$k(x,y,z) = \sum_{n=0}^d k_n(x,z) y^n$$

For any $n \geq 2$, we can rewrite $y^n$ using recursion based on the relation $y = x² - (x² - y)$:
$$y^n = x² y^{n-1} - (x² - y) y^{n-1}$$
Repeating this until $y^n$ is expressed as $x^{2n}$ plus a multiple of x² - y:
$$y^n = x^{2n} + (x² - y) \cdot A_n(x,y,z)$$
where $A_n$ is some polynomial in $\mathbb{C}[x,y,z]$.

Substitute this back into $k$, and collect all multiples of x² - y:
$$k(x,y,z) = k_0(x,z) + k_1(x,z)y + (x² - y) \cdot F(x,y,z)$$
Here, $F$ is the sum of all the $A_n$ terms scaled by their coefficients $k_n(x,z)$.

Step 2: Reduce all high powers of $z$ #

Now look at $k_0(x,z)$ and $k_1(x,z)$ as polynomials in $z$. For any $m \geq 1$, rewrite $z^m$ using $z = x³ - (x³ - z)$:
$$z^m = x^{3m} + (x³ - z) \cdot B_m(x,z)$$
where $B_m$ is another polynomial in $\mathbb{C}[x,z]$.

Substitute these into $k_0$ and $k_1$:
$$k_0(x,z) = \sum a_m(x)x^{3m} + (x³ - z) \cdot G(x,z)$$
$$k_1(x,z) = \sum b_m(x)x^{3m} + (x³ - z) \cdot H(x,z)$$
where $a_m, b_m$ are coefficients from the original polynomials $k_0, k_1$.

Step 3: Use the kernel condition to cancel pure-$x$ terms #

Since $k \in \ker(\varphi)$, substituting $x=t$, $y=t²$, $z=t³$ gives:
$$\sum a_m(t)t^{3m} + t² \sum b_m(t)t^{3m} = 0$$
This is a polynomial in $\mathbb{C}[t]$ that vanishes for all $t \in \mathbb{C}$—which means it's the zero polynomial. Translating back to $x$, this gives:
$$\sum a_m(x)x^{3m} + x² \sum b_m(x)x^{3m} = 0$$

Now rewrite the $y \cdot \sum b_m(x)x^{3m}$ term using $y = x² - (x² - y)$:
$$y \sum b_m(x)x^{3m} = x² \sum b_m(x)x^{3m} - (x² - y) \sum b_m(x)x^{3m}$$
Substitute this into our reduced form of $k$, and use the zero polynomial condition above:
$$\sum a_m(x)x^{3m} + y \sum b_m(x)x^{3m} = - (x² - y) \sum b_m(x)x^{3m}$$

Step 4: Combine all terms to get the final decomposition #

Putting everything together, we substitute back into $k$:
$$
\begin{align*}
k(x,y,z) &= - (x² - y) \sum b_m(x)x^{3m} + (x³ - z)(G + Hy) + (x² - y)F \
&= (x² - y) \left(F - \sum b_m(x)x^{3m}\right) + (x³ - z)(G + Hy)
\end{align*}
$$

This gives us the explicit decomposition where:

• $f(x,y,z) = F(x,y,z) - \sum b_m(x)x^{3m}$
• $g(x,y,z) = G(x,z) + H(x,z)y$

Example: Verifying Redundancy of $y³ - z²$ #

Let's test this with the redundant element $k(x,y,z) = y³ - z²$:

1. Reduce $y³$: $y³ = x² y² - (x² - y)y²$, and $y² = x² y - (x² - y)y$. Substituting back gives $y³ = x⁴ y - (x² - y)(x² y + y²)$.
2. Reduce $z²$: $z² = x⁶ - (x³ - z)(x³ + z)$, and $x⁴ y = x⁴ \cdot x² = x⁶$.
3. Substitute into $k$: $y³ - z² = (x⁴ y - (x² - y)(x² y + y²)) - (x⁶ - (x³ - z)(x³ + z))$. Since $x⁴ y = x⁶$, this simplifies to $- (x² - y)(x² y + y²) + (x³ - z)(x³ + z)$.

Which confirms $y³ - z²$ is indeed redundant, as it's a combination of the two core generators.

内容的提问来源于stack exchange，提问作者Pascal's Wager