AG on non-Noetherian rings I must apologize beforehand as this question is pretty basic, but I can't seem to find a satisfying answer in the introduction section of the book I'm currently reading (if there is a page on here, I could not find it).
As far as I know we can define $X=\text {Spec} R$ to be an affine scheme for any commutative ring $R$. If we choose $R=k[x_1,x_2,\ldots]$ do we still get a "reasonable" picture of the geometry of $X$, or is the non-noetherian case always somewhat more abstract than the "classical" picture of algebraic sets in finite affine space?
From what I was reading, any open cover of $X$ has a finite subcover ($\text {Spec} R$ is quasi-compact). Does this still hold if $R$ is not noetherian? (I thought the whole reason to consider noetherian rings in the first place was to deduce properties like this)
 A: Yeah every affine scheme $\text{Spec}\, R$ is quasicompact,whatever $R$ you choose. To show this, recall that a basis of open subsets in $\text{Spec}\,R$ is given by the  distinguished affine open subsets
$$ \text{Spec}\, R_f = \left\{ \mathfrak{p} \subseteq R \mid f\notin \mathfrak{p} \right\} \qquad \text{for} \,\, f\in R $$
So that it is enough to show that any open cover of $\text{Spec}\,R$ composed of distinguished affine subsets has a finite subcover: so let
$$ \text{Spec } R = \bigcup_{i\in I} \text{Spec } R_{f_i}  $$
Taking complements, this amounts to 
$$\emptyset = \bigcap_{i\in I} \mathbb{V}(f_i) = \mathbb{V}((f_i)_{i\in I})$$
and this is the same as saying that $(f_i)_{i\in I} = (1)$, i.e. we can write
$$ 1 = a_1f_{i_1} + \dots +a_n f_{i_n} $$
for certain $a_{j}\in R$ and certain indexes $i_j\in I$. But then, following the same reasoning of before, this implies
$$ \text{Spec } R = \bigcup_{j=1}^n \text{Spec } R_{f_j} $$ 

If instead $R$ is a Noetherian ring, you also have that $\text{Spec }R$ is a Noetherian topological space, meaning that every open subset is quasicompact. For example, this is not true for $X=\text{Spec } k[x_1,x_2,\dots]$: consider the ideal $\mathfrak{m}=(x_1,x_2,\dots)$ and the open subset $U=X\setminus \mathbb{V}(\mathfrak{m})$: as before, we see that $U=\bigcup_{n=1}^{+\infty}\text{Spec } k[x_1,x_2,\dots]_{x_i}$, however it is not true that $U=\bigcup_{i=1}^N \text{Spec } k[x_1,x_2,\dots]_{x_i}$ for any $N$, because otherwise $\mathfrak{m}=(x_1,\dots,x_N)$.

This is just from the side of the topological structure, but with $R$ Noetherian, you have also a lot of other properties on $\text{Spec } R$, basically because you can use more or less all the usual results of commutative algebra with finiteness hypotheses.
