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I'm looking for the dimension of $R=k[x,y,z]/(xy-z^2)$. In my correction it's written that $\{\bar x,\bar y\}$ is a transcendence basis of $R$, but I don't understand why (where $\bar x$ is the residue class of $x$). So my questions are the following :

1) Why $\{\bar x,\bar y\}$ are algebraically independent ?

2) Why $\{\bar x,\bar y,\bar z\}$ is not are algebraically independent ?

My attempts

1) I suppose by contradiction that there is a non-zero polynomial s.t. $f(\bar x,\bar y)=0$. But I can't get a contradiction.

2) No idea.

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    It is integral over $k[x,y]$, so the Krull dimension is the same as the one of $k[x,y]$, which is $2$. – 2017-01-27
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    Notice that asking "Why $\{\bar x,\bar y\}$ are algebraically independent" is very unclear ; you need to specify **over what base field** you are working, for instance algebraically independent over $k$. – 2017-01-27

1 Answers 1

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1)

Assume that $f(\bar x, \bar y) = 0_R = \bar 0$. Then $f(x,y) \in (xy-z^2)$, which means that $$f(x,y)=g(x,y,z) \cdot (xy-z^2)$$ for some $g \in k[x,y,z]$.

Consider this equality in $S[z]$, where $S = k[x,y]$. The degree (w.r.t. $z$) of $f(x,y) \in S[z]$ is $0$ (because $f \in S$), while the degree of $g(x,y,z) \cdot (xy-z^2)$ is $\mathrm{deg}(g) \cdot 2$. I let you think about why this fact implies that $f = 0 \in k[x,y,z]$.

2)

Hint: consider the non-zero polynomial $p(u,v,w) = w^2 - uv$. Does it give you an algebraic relation between $\overline x, \overline y$ and $\overline z$ ?

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    Thanks for your answer. In fact, I don't understand where is the polynomial $f$. Is it in $K[x,y]$ where $K=Frac(k[x,y,z]/(xy-z^2))$ ? – 2017-01-27
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    @user386627 : No, because $\overline x$ and $\overline y$ are algebraically dependent over $K$, just by taking $p(u,v)=\overline x v - \overline y u \in K[u,v]$. Here we consider algebraically independence over $k$, so $f \in k[x,y]$. However, $k$ embeds in $R$, which is a $k$-algebra (so that $f$ can be evaluated at elements of $R$). Similarly, $p \in k[u,v,w]$ has coefficients in $k$. – 2017-01-27
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    I missed something (sorry). We want to compute the transcendence basis of $Frac(k[x,y,z]/(xy-z^2))$ over $k$ right ? Then (in my definition), if $L/K$ is a field extension, and $\{x,y\}\subset L$ are algebraically independent over $K$, then there is no polynomial $f\in K[x,y]$ s.t. $f(x,y)=0$ right ? – 2017-01-27
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    @user386627 : here $L = \mathrm{Frac}(k[x,y,z]/(xy−z^2))$ and $K=k$. – 2017-01-27
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    Thanks for your (very) constructive answer. – 2017-01-27
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    Just a last thing, when you say $f(x,y)\in (xy-z^2)$, you wanted to say $f(\bar x,\bar y)\in (xy-z^2)$ no ? and same for $f(x,y)=g(x,y,z)(xy-z^2)$, is in fact $f(\bar x,\bar y)=g(\bar x,\bar y,\bar z)(\bar x\bar y-\bar z^2)$, no ? Otherwise, I don't really understand. – 2017-01-28
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    @user386627 : Say that $f(u,v) = u^2-3v \in k[u,v]$, and assume that $f(\bar x, \bar y) = 0$. It means $$ \bar x^2 - 3\bar y = 0_R = \bar 0 $$ which implies that $\overline{x^2-3y}=\bar 0$, i.e. $x^2-3y \in J$, where $R=k[x,y,z]/J ; J=(xy-z^2)$. Therefore $f(x,y) \in J$. – 2017-01-28