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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 ?

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  • $\begingroup$ What example do you have in mind? $\endgroup$
    – Youngsu
    Jan 24, 2020 at 18:21
  • $\begingroup$ I have no example in mind. It is just to understand the definition of the dimension of the variety $V$ by the Krull dimension of $k[V]$. $\endgroup$
    – L. ZWALD
    Jan 27, 2020 at 13:39

1 Answer 1

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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.

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  • $\begingroup$ 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 ? $\endgroup$
    – L. ZWALD
    Jan 27, 2020 at 13:37
  • $\begingroup$ Yes, this is correct. $\endgroup$
    – KReiser
    Jan 27, 2020 at 19:48

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