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Let $k$ be an arbitrary field and $V\subset k^n$ an irreducible variety. I have seen the two following definitions of the dimension of $V$. At first, the dimension of $V$ is the Krull dimension $d$ of the coordinate ring $k[V]=k[X_1,\cdots,X_n]/I(V)$ where $I(V)$ is the set of polynomials vanishing on $V$. Next, the dimension of $V$ is $d_1$ the maximal length of a chain $V_1\subsetneq V_2\subsetneq\cdots\subsetneq V_k\subsetneq V$ of irreducible varieties included in $V$.

I have shown that $d_1\leq d$. Is it possible to prove that $d\leq d_1$ without using scheme ?

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  • $\begingroup$ This correspondence holds if $k$ is supposed algebraically closed but, in my question, I am interested in the case of an arbitrary field $k$. $\endgroup$
    – L. ZWALD
    Commented Jan 28, 2020 at 8:09
  • $\begingroup$ Ah, I misread your definition of a variety. In that case, this is blatantly false: consider $k=\Bbb R$ and $V$ to be the vanishing set of $x^2+y^2$ inside $\Bbb R^2$. Then the coordinate algebra $\Bbb R[x,y]/(x^2+y^2)$ is dimension 1, while the $V$ is just the origin and has dimension $0$. One very big part of moving to more extensible definitions is to make correspondences like this still hold - once we move to a scheme-based viewpoint, we get access to the "hidden" complex points of this variety and everything works. $\endgroup$
    – KReiser
    Commented Jan 28, 2020 at 8:25
  • $\begingroup$ I don't agree with your counter-example. Indeed, in my question, the coordinate ring is defined by the quotient by $I(V)$. With your variety, $I(V)$ is the ideal generated by X and Y. Thus the coordinate ring is the field $\mathbb{R}$ which is of Krull dimension $0$, as the dimension of your variety. Thus, this is not a counter-example. $\endgroup$
    – L. ZWALD
    Commented Jan 28, 2020 at 16:27

1 Answer 1

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We know that an algebraic set $X \in \mathbb{A}^{n}$ is irreducible iff. the corresponding ideal $I(X) \in k[x_{1},\cdots,x_{n}]$ is a prime ideal. I'll prove it one way and leave the other way as an exercise. If $p$ is a prime ideal, consider the algebraic set $V(p)$ and assume $V(p)=X_{1} \cup X_{2}$ Then using the Nullstellensatz,$rad(p)=I(V(p))=I(X_{1} \cup X_{2})=I(X_{1}) \cap I(X_{2})$. But $rad(p)=p$. If $p \neq I(X_{1})$ and $p \neq I(X_{2})$, then there exists some $x_{1} \in I(X_{1})$ and $x_{2} \in I(X_{2})$ such that $x_{1},x_{2} \notin p$ but $x_{1}x_{2} \in I(X_{1}) \cap I(X_{2})=p$, a contradiction. So, either $p=I(X_{1})$ or $p=I(X_{2})$.So, $V(p)=X_{i}$ for some $i$, hence it is irreducible. I leave the other direction.

Now, consider the affine co-ordinate ring of a variety $X$, $A(X)=k[x_{1},\cdots,x_{n}]/I(X)$. A variety can be uniquely written as $X=X_{1} \cup X_{2} \cdots \cup X_{n}$ where $X_{i}$ is irreducible and no $X_{i}$ contains another $X_{j}$. So, the closed irreducible subsets of an algebraic set $X$ correspond exactly to prime ideals of $k[x_{1},\cdots,x_{n}]$ that contain $I(X)$, in other words, the prime ideals of $A(X)$. This immediately gives $d=d_{1}$.

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  • $\begingroup$ Thanks for your attempt. However, in your proposal, you use the Nullstellensatz which holds if $k$ is supposed algebraically closed. In my question, I am interested in the case where $k$ in an arbitrary field. $\endgroup$
    – L. ZWALD
    Commented Jan 28, 2020 at 8:18

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