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Suppose we have a variety, $Z$, in $\mathbb{A}^3$ that is given by the equations $st=v^2, s^3=vt, t^2=s^2v$. I want to show that this is the same as the image of a map from $\mathbb{A}^1$ to $\mathbb{A}^3$ given by $(x^3,x^4,x^5)$.

I know if we check that when we let $s=x^3$, $v=x^4$, and $t=x^5$ all three of our equations are satisfied, but I'm not sure what this means. That cannot prove anything because I could just have taken an equation away and we still satisfy the other two equations. I am not sure how to do this problem. I plotted $st=v^2, s^3=vt, t^2=s^2v$ but that gave zero-intuition (It was very hard to see what the intersection looked like.)

From here though, we would have that $Z$ is irreducible since it is the image of an irreducible variety, namely all of $\mathbb{A}^1$.

I am not sure whether we need the field to be algebraically closed, but if it does make the problem easier,let's just assume it is.

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If you let $s = x^3$, $y = \cdots$, etc and check that these satisfy the equations of your variety then what you are showing is that the image of your map is a subset of the given variety. What you need now is to show that every element of the variety is in the image of the map. So assume $(s, v, t)$ satisfies the equations. How could we pick an $x$ so that $(s, v, t) = (x^3, x^4, x^5)$? Start by arguing that if $s = 0$ then also $v = t = 0$. So if $s = 0$ then $(s, v, t) = (0, 0, 0)$ and we pick $x = 0$.

Now assume $s \neq 0$. If $(s, v, t) = (x^3, x^4, x^5)$ for some $x$ then because $\frac{x^4}{x^3} = x$ we know exactly what our choice of $x$ should be in terms of $s$ and $v$. What is it? Assuming we choose that for $x$ can you show that $s = x^3$, $v = x^4$, and $t = x^5$?

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Sorry, but I may be being silly here, but is the choice for $x$ just $v/s$ since we know $s=x^3$ and $v=x^4$. So know we choose our $x$ as $s/v$. I am we have $s=v^3/s^3$ $v=v^4/s^4$ and $t=v^5/s^5$. It just seems like that doesn't do anything different. Or am I supposed to start with the equations $st=v^2, s^3=vt, t^2=s^2v$ and deduce those? I am not sure why this is what we want? Sorry! –  Steven-Owen Feb 10 '13 at 19:53
    
You are correct that you should choose $x = \frac{v}{s}$, but not becuase we know $s = x^3$ and $v = x^4$. We don't know that. We only know that if such an $x$ exists then it satisfies $x = \frac{v}{s}$. So we try and show that $\frac{v}{s}$ is the $x$ that we want. This means that what you have to do is use the equations $st = v^2$, $s^3 = vt$, and $t^2 = s^2v$ to prove that $s = \left(\frac{v}{s}\right)^3$, $v = \left(\frac{v}{s}\right)^4$, and $t = \left(\frac{v}{s}\right)^5$. Then you will know that $x = \frac{v}{s}$ is the correct choice. –  Jim Feb 10 '13 at 20:04
    
Is the first equation $st=v^2$ redundant? Starting with $t^2=s^2v$ we get $v=t^2/s^2$ which we can with $v$ and it becomes $v^2=ttv/s^2$ which by us of the second equation becomes $v^2=s^3t/s^2=st$ which is the first equation. Are these the kind of manipulations to be doing? –  Steven-Owen Feb 10 '13 at 20:24
    
I'm not sure what you're doing there, but it doesn't matter if the equations of the variety are redundant. You get to assume those hold. It's the equations $s = \left(\frac{v}{s}\right)$ and so on that you need to derive. –  Jim Feb 10 '13 at 20:56
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