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I'm trying to understand this example:

Let $f(T_1,T_2)\subset k[T_1,T_2]$ be a non-constant irreducible polynomial. Let $X=Z(f)\subset \mathbb A^2$. We will see that $\dim(X)=1$. We have $k[X]=k[T_1,T_2]/(f)$ and

$$\dim(X)= \operatorname{tr.deg}_k k(X) \lneq \operatorname{tr.deg}_k k(T_1,T_2) = 2.$$

Since in $k(X)$ the generators $T_1, T_2$ follow to an algebraic relation $f$. On the other hand, $\dim(X)\ge 1$ since $X$ is not finite, thus $\dim (X)=1.$

I didn't understand why the $\lt$ part and why $\operatorname{tr.deg}_k(k(T_1,T_2))=2$. Could someone help me?

Thanks a lot.

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  • $\begingroup$ Why downvoted?? $\endgroup$
    – user75086
    Dec 21, 2013 at 19:49
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    $\begingroup$ You've been asking quite a few questions lately and in every instance it turns out you struggle with the concept of transcendence degree. This, I guess, caused the downvotes on this question, because you're essentially again asking "what is transcendence degree?" $\endgroup$
    – Magdiragdag
    Dec 22, 2013 at 3:44
  • $\begingroup$ @Magdiragdag sorry, but I'm not asking that. $\endgroup$
    – user75086
    Dec 22, 2013 at 18:39
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    $\begingroup$ @Magdiragdag I totally disagree with the downvotes, I know the rules of the site and I follow the meta site daily and I'm sure my questions meet the quality standards of the MSE. $\endgroup$
    – user75086
    Dec 22, 2013 at 18:45

1 Answer 1

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$T_1$ and $T_2$ satisfy an algebraic relation in $k[X] = k[T_1,T_2]/(f)$, but not in $k(T_1,T_2)$. The latter notation is specifically chosen to refer to the field in which $T_1$ and $T_2$ are formal variables with no relations, and the transcendence degree of such a field over $k$ equals $2$ by pretty much any definition of transcendence degree.

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    $\begingroup$ I guess I didn't really say why the transcendence degree of $k(X)$ is strictly less than $2$. For a fact like this, I recommend trying to write down the abstract, general statement that would imply the specific case. Here, we have a field, $k(X)$, generated by two elements, $T_1$ and $T_2$, with a relation between them: $f(T_1,T_2) = 0$. This alone should be enough to tell you that the transcendence degree is strictly less than 2—and it is! Otherwise, we can decompose $k(X) / k$ into two simple extensions, by $T_1$ and $T_2$, both with transcendence degree 1, implying no relations. $\endgroup$
    – Slade
    Dec 22, 2013 at 14:21

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