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have that $f=xy-z $ and $g=y-z$ and

$$\frac{k[x,y,z]}{xy-z,xz-z}$$

the s-polynomial of $f,g$ is $xz-z$

letting $i\geq 1 $ and $j \geq 1$ simplify $x^i z^j$

for $i

for $i > j$ (from Lecture)

$$ \begin{aligned} x^i z^j&= x^j z^j x^{i-j} \\ &= (xz)^jx^{i-j} \\ &=z^j x^{i-j} && \text{since } xz=z \\ &=z^j && \text{??} \end{aligned} $$

Not sure how $z^j$ was derived is $x^{i-j}=1$ ??

if it is unclear the question boils down to explain the $??$ steps

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Attempt 2] when reading $i$ and $j$ look the same to me sometimes switching to $i$ and $J$

for $i > j $ $$\begin{aligned} x^{i}z^{j}&= x^{i}z^{J} \\ &=x^{i}z^{J +i - i} \\ &=x^i z^{J-i} z^i \\ &= x^i z^i z^{J-i} \\ &= (xz)^i z^{J-i} && \text{ since } xz=z \\ &= (z)^i z^{J-i} \\ &= z^J \\ &= z^j \end{aligned} $$

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    What does $g$ have to do with anything? – 2017-02-20
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    it has evrything to do with it. $S(xy-z,g)=xz-z$. – 2017-02-20
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    That doesn't explain why $g$ has anything to do with the problem. It's true that $S(xy-z, x-1) = g$, but that doesn't mean $x-1$ is at all relevant! – 2017-02-20
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    I dont have time to argue. Got it . I get my $i,j $ mixed up. all is left is to type up case for $i – 2017-02-20

1 Answers 1

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The relations allow you to replace $xz$ by $z$, hence \begin{align*} \text{}\\[0pt] \text{For }&i < j,\\[6pt] x^i z^j &= (xz)^i z^{j-i} \\[6pt] &= z^i z^{j-i}\\[6pt] &= z^j\\[6pt] \text{}\\[0pt] \text{For }&i\ge j,\\[6pt] x^i z^j &= x^{i-j}(xz)^j\\[6pt] &= x^{i-j}z^j\\[6pt] &= z^j\qquad\text{[by the previous part]}\\[6pt] \end{align*}

Actually, my argument above is flawed.

I had the illusion that $i-j < j$.

To fix it, keep going ...

From $x^{i-j}$ to $x^{i-2j}$ to $x^{i-3j}$ ... until the exponent is less than $j$, at which point, you can then apply the previous part.

Alternatively, proceed by induction.