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Are rational functions between affine algebraic sets continuous with respect to the Zariski topology on their domain of definition?

Jiu
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1 Answers1

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A polynomial is continuous by definition of the Zariski topology. A rational function is a function that can be written locally (i.e. on an open neighborhood of each point) as a quotient $\frac{f}{g}$ where $f$ and $g$ are polynomial such that $g$ never vanishes on the open neighborhood. As a quotient of such (continuous) polynomials is continuous, you see that a rational function is locally continuous. As the notion of continuity is purely local on the source (which is a fancy way to say that "locally continuous" implies "continuous") it is also continuous.

Olórin
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  • Thanks for your answer. I need to think about it. I'm not familiar with the Zariski topology and I'm not sure if quotient of continuous functions is continuous or if a function with continuous components is continuous. – Jiu Mar 27 '18 at 10:29
  • The fact that a licit quotient of continuous functions is continuous and that, as you write, a fonction with continuous components is continuous are purely general properties of continuous functions and are are not proper to the Zariski topology. Now, what is this Zariski topology ? It is defined as the topology where the closed sets are zeros of families of polynomials. (According to your vocabulary I assumed you work on a polynomial ring, not on a general commutative ring with unit.) – Olórin Mar 27 '18 at 10:37
  • Now there is a useful (and completely general) characterization of continuity : a map is continuous if and only if the preimage of any closed set is a closed set. A polynomomial map is a map from the affine space to the base field. Try to apply this general characterization in this case and you will see. – Olórin Mar 27 '18 at 10:41
  • Well a function with continuous components is continuous for the product topology. But it's not obvious for me that for affine algebraic sets $X$, $Y$, a function $f: X\rightarrow Y$ with continuous components $f_i: X\rightarrow \mathbb{A}^1$ is continuous, because the target space is $Y$ with Zariski topology and not $\mathbb{A}^n$ with product topology. – Jiu Mar 27 '18 at 10:49
  • Exercise : the Zariski topology is finer that the product topology. – Olórin Mar 27 '18 at 10:54
  • and continuity of the quotient cannot be a general property because in a arbitrary topological space, division is not defined. Is it enough to show that division is continuous for the Zariski topology? – Jiu Mar 27 '18 at 10:54
  • No one said anything about dividing elements of an arbitrary topological space ... You divide values of functions (defined over an arbitrary topological space) that have value in "something" where division has a meaning and is continuous ... – Olórin Mar 27 '18 at 10:58
  • Yes Zariski topology is finer. Now $f$ seen as a map from $X$ to $A^n$ is continuous, but since the Zariski topology is finer, we need the pre image of more sets to be open. So being continuous with the target space equipped with the Zariski topology is a stronger condition. – Jiu Mar 27 '18 at 11:00
  • If $Z={h_1 = ... = h_k = 0}$ then $f^{-1}(Z) = { v \in V : f(v) \in Z }$ is the preimage of $Z$. The elements $h_i$ are the polynomials that define $Z$ as a closed subset of $W$. Since the composition of two polynomials is still a polynomial, each $h_i \circ f$ is a polynomial and that implies that $f^{-1}(Z)$ is a closed subset of $V$. – Olórin Mar 27 '18 at 11:05
  • There's something I don't understand here : you ask a question, one gives you an answer, and you don't seem at all to be willing to take yourself some part of the job consisting of showing yourself at least something. You want everything to be shown in the answer. – Olórin Mar 27 '18 at 11:07
  • your answer made me question things that I had taken for granted until just now. Of course I'll think about them myself. – Jiu Mar 27 '18 at 11:18