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Let $k$ be an algebraically closed field, and let $$X = \operatorname{Spec} k[x,y,t]/(ty-x^2)$$ $$Y = \operatorname{Spec} k[t]$$

Hartshorne comments that both schemes $X$ and $Y$ are of finite type over $k$.

This means that the natural morphisms $X \to k$ and $Y \to k$ are of finite type.

I know that if $k$ is algebraically closed then the closed points of $A_k^1$ is in one-to-one correspondence with elements of $k$.

Is $k$ a scheme? How do I show that $X \to k$ a morphism of finite type?

user7090
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  • It is very common to write $k$ instead of $\operatorname{Spec} k$ at times. Surprisingly this doesn't cause a lot of confusion. – Hoot Jun 22 '16 at 20:07
  • It caused me some confusion! :p – user7090 Jun 22 '16 at 20:12
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    I should have used the word "eventually". Eventually it doesn't cause confusion. Unfortunately Hartshorne is filled with these moments. (Don't get me wrong, it's my favorite book in the world.) – Hoot Jun 22 '16 at 20:14
  • I was just kidding around, I completely understand what you meant. – user7090 Jun 22 '16 at 20:14

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When you write $k$ you should write $Spec \ k$. These schemes are of finite type over $Spec \ k$, because the $k$-algebras $k[x,y,t]/(ty-x^2)$ and $k[t]$ are obviously of finite type (that is, a quotient of a polynomial ring over $k$ in a finite number of variables).

syzygy
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  • I see! That was not clear to me at all from the description. Is it common convention to not write $\operatorname{Spec} k$ ? – user7090 Jun 22 '16 at 20:00
  • @asdvn: the notion of "finite type morphism" (and similar finiteness conditions) are derived from the corresponding ones for algebras! the basic thing to remember is: affine schemes over $Spec \ k$ = $k$-algebras! – syzygy Jun 22 '16 at 20:02
  • Ok sure. That definitely makes sense. Thank you! – user7090 Jun 22 '16 at 20:04