MHB Is that correct so far ... ?Yes, that is correct. Good job!

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need help to get started on Exercise 26 of Section 15.1 ...Exercise 26 of Section 15.1 reads as follows:
https://www.physicsforums.com/attachments/4764Help with this exercise will be much appreciated ...

Peter
***EDIT***

Thoughts so far are as follows:

... ...

I am assuming that when Dummit and Foote ask us to show that $$\phi$$ is a morphism that, given the context, they are referring to what they have defined as a morphism or polynomial map between two algebraic sets ... in which case we would have to be sure we were actually dealing with a map between algebraic sets ... now $$V$$ is obviously by its definition, an algebraic set ... but ... how do we show that $$\mathbb{A}^1$$ is an algebraic set ... ?If, however we assume that both $$V$$ and $$\mathbb{A}^1$$ are indeed algebraic sets then we have to show that:

$$\phi \ : \ \mathbb{A}^1 \rightarrow V$$

is a morphism or polynomial map of algebraic sets ... that is we have to show that there exists polynomials $$\phi_1, \phi_2, \phi_3$$ such that:

$$\phi(a) = ( \phi_1 (a) , \phi_2 (a) , \phi_3 (a) )$$

But, this is obviously true with

$$\phi_1 (a) = a^3 \ , \ \phi_2 (a) = a^4 \ , \ \phi_3 (a) = a^5$$ ...

Is that correct so far ... ?

But... not sure of how to show the surjectivity ... even with the hint ...

Further ... I need help to make a significant start on parts (b) and (c) ...Hope someone can help ... particularly with parts (b) and (c) ...

Peter

 

[Deveno]

Let's go one step at a time.

Suppose $(0,y,z)$ is in the zero set.

Then $xz - y^2 = 0 \implies y^2 = 0 \implies y = 0$.

Similarly, $z^2 - x^2y = 0 \implies z^2 = 0 \implies z = 0$.

Thus either ALL of $x,y,z$ are $0$, or $x \neq 0$.

Now it is clear that $(x,y,z) = (t^3,t^4,t^5)$ is in the zero set, for any $t \in k$:

$xz - y^2 = t^3t^5 - (t^4)^2 = t^8 - t^8 = 0$
$yz - x^3 = t^4t^5 - (t^3)^3 = t^9 - t^9 = 0$
$z^2 - x^3y = (t^5)^2 - (t^3)^2t^4 = t^{10} - t^6t^4 = t^{10} - t^{10} = 0$.

So we see that $\{(t^3,t^4,t^5): t\in k\} \subseteq \mathcal{Z}$. The question becomes: is EVERY element of the zero set of that form?

Now if $(x,y,z) \neq 0$, we see that $x \neq 0$. Thus $x$ is a unit of $k$. Hence $y = ax$ for some $a \in k$ (namely $a = \dfrac{y}{x}$).

Then:

$xz - y^2 = 0 \iff xz - a^2x^2 = 0 \iff x(z - a^2x) = 0 \implies z - a^2x = 0$ (since $x \neq 0$).

So $(x,y,z) = (x,ax,a^2x)$.

Consequently, re-writing $yz - x^3 = 0$:

$ax(a^2x) - x^3 = 0 \implies a^3x^2 - x^3 = 0 \implies x^2(a^3 - x) = 0 \implies a^3 = x$

so $(x,y,z) = (a^3,a^4,a^5)$, or if you prefer: $(x,y,z)\neq 0 \in \mathcal{Z} \implies (x,y,z) = \left(\dfrac{x^3}{y^3},\dfrac{x^4}{y^4},\dfrac{x^5}{y^5}\right)$

Now the zero set point $(0,0,0)$ is clearly of the form: $(0^3,0^4,0^5)$, so this shows the zero set is contained in the image of $\phi$, so $\phi$ is surjective.