# Show that (vector) subspaces of $\mathbb{A}^n$ are algebraic sets

i have just started to learn some algebraic geometry and there is a statement in the notes i am following that i do not understand: "Subvector spaces of $\mathbb{A}^n$ are algebraic sets. They are of the form $Z(f_1,\ldots,f_n)$ where $f_i$ are polynomials of degree 1." (from notes by Lothar Goettsche: http://users.ictp.it/~gottsche/)

My problem with this statement is the following: For every point $a =(a_1,\ldots, a_n)$ in $A^n$ i can give a set $S$ of polynomials such that their common zero set ist $\{a\}$: $S = \{x_1-a_1, \ldots, x_n - a_n\}$. But how can i find a set of polynomials that vanishes on a whole vector space ? Of course one can take the zero polynomial, but that would not be of degree one ?

Thank you for advice

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You have probably learned in your linear algebra course how to describe linear subspaces by equations... – Mariano Suárez-Alvarez Mar 9 '12 at 8:41
" But how can i find a set of polynomials that vanishes on a whole vector space ? " For example, in $\mathbb C^3$ with coordinates $x,y,z$ , all polynomials of the form $ay+bz$ vanish on the $x$-axis, which is a sub-vector space of $\mathbb A^3(\mathbb C)=\mathbb C^3$. – Georges Elencwajg Mar 9 '12 at 8:43
I thought affine space wasn't equipped with a vector space structure? The definition I'm used to is that $\mathbb A_k^n = k^n$ as sets, and that the structure we place on $\mathbb A_k^n$ is a topological one and, in particular, there is no 'distinguished point' i.e. no origin. – Matt Mar 9 '12 at 16:48