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The dimension of an affine algebraic variety $V\subset k^n$ is defined as the Krull dimension of its coordinate ring $k[V]=k[X_1,\cdots,X_n]/I(V)$ where $I(V)$ is the set of polynomials in $k[X_1,\cdots,X_n]$ vanishing on $V$.

In the case where $V$ is moreover a vector space over the field $k$, I don't succeed in showing that the dimension of $V$ as vector space is the same as the dimension of $V$ as affine algebraic variety. In particular, what can be said about the ideal $I(V)$ in this case ?

L. ZWALD
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1 Answers1

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It is true that the dimension of $V$ as a vector space is equal to it's dimension as a variety. This is because $I(V)$ can be generated by $n-\dim V$ linear forms, where we mean dimension as a linear space.

To prove this, select a basis $v_1,\cdots,v_m$ for $V$ and complete it to a basis of $k^n$ with the vectors $w_{m+1},\cdots,w_n$. Writing these vectors as the columns of a matrix $M$, we see that this matrix is invertible and takes the standard basis vectors $e_i\mapsto v_i$ if $i\leq m$ and $e_i\mapsto w_i$ if $i>m$. So $M$ and $M^{-1}$ define an isomorphism between $V$ and the subspace $V'$ consisting of vectors with final $n-m$ coordinates zero (this is an isomorphism as vector spaces and also as varieties). This means that as the coordinate functions $x_{m+1},\cdots,x_n$ generate the ideal of $V'$, their pullbacks under $M^{-1}$ generate the ideal of $V$, and we see the claim.

KReiser
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  • Thank you very much for your very clear answer. If I understand well, your change of variables induces a ring isomorphism between $k[V]$ and $k[V']$. Thus this two rings get the same Krull dimension and the Krull dimension of $k[V']$ is cearly dim V. Is it right ? – L. ZWALD Jan 27 '20 at 13:37
  • Yes, this is correct. – KReiser Jan 27 '20 at 19:48
  • What if $k$ is finite? Then $V=k^n$ has dimension $n$ over $k$ but Krull dimension $0$. – Jose Brox May 12 '25 at 16:31
  • @JoseBrox the vector space $k^n$ viewed as a variety has coordinate algebra $k[x_1,\cdots,x_n]$, which has Krull dimension $n$. You're missing that translation step. – KReiser May 12 '25 at 17:07
  • If I'm not mistaken, when $k=F_q$ the vanishing ideal of the variety $V=k^n$ is not $0$, but $I_q=\langle x_1^q-x_1,\ldots, x_n^q-x_n\rangle$, which gives a coordinate algebra of Krull dimension $0$. – Jose Brox May 12 '25 at 17:29
  • @JoseBrox Sure, but this question is written in a more classical setting, where $k$ is required to be algebraically closed. If one wanted to change to a more modern setting where $k$ is not algebraically closed, one should move from $k^n$ to $\operatorname{Spec} k[x_1,\cdots,x_n]$, where this whole thing becomes a bit tautological. – KReiser May 12 '25 at 20:16
  • I understand, but then which step of your proof needs that the field is algebraically closed? – Jose Brox May 12 '25 at 22:11
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    @JoseBrox The computation of the ideal of $V(x_{m+1},\cdots,x_n)$ depends on algebraic closedness of $k$. When $k$ is algebraically closed, we have in the classical setting $I(V(J))=\sqrt{J}$, and it is not so hard to see that $(x_{m+1},\cdots,x_n)$ is a radical ideal. – KReiser May 12 '25 at 22:20