# dimension of a variety with chains of irreducible varieties or Krull dimension

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 ?

• This correspondence holds if $k$ is supposed algebraically closed but, in my question, I am interested in the case of an arbitrary field $k$. Commented Jan 28, 2020 at 8:09
• 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. Commented Jan 28, 2020 at 8:25
• 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. Commented Jan 28, 2020 at 16:27

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 $$d=d_{1}$$.
• 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. Commented Jan 28, 2020 at 8:18