$\dim V=\dim F[t_1,\dots,t_k]/\mathcal{I}(V)$ I was trying to prove that for an irreducible algebraic set $V\subseteq F^k$ we have the equality $\dim V=\dim F[t_1,\dots,t_k]/\mathcal{I}(V)$, and this is what I came up with. It looks right to me but I am not entirely sure about a few things and I was hoping someone could shine some light.
For ease of notation we let $R:=F[t_1,\dots,t_k]/\mathcal{I}(V)$. We firstly show that $\dim V\leq \dim R$.
Let $\emptyset \neq V_n\subset V_{n-1}\subset\dots\subset V_0=V $ be a maximal chain of irreducible algebraic sets. Then by applying  $\ \mathcal{I}$ we have the chain $\mathcal{I}(V)=\mathcal{I}(V_0)\subset\mathcal{I}(V_1)\subset\dots\subset\mathcal{I}(V_n)\ \ $ of prime ideals of $F[t_1,\dots,t_k]$. Hence by the Correspondance Theorem we have the following chain of prime ideals of $R, \ \ $ $0=\mathcal{I}(V_0)/\mathcal{I}(V_0)\subset\mathcal{I}(V_1)/\mathcal{I}(V_0)\subset\dots\subset\mathcal{I}(V_n)/\mathcal{I}(V_0)$ where we note that the zero ideal in this case is prime since $V$ being irreducible implies that $\mathcal{I}(V)=\mathcal{I}(V_0)$ is prime and hence $R$ is a domain. So in particular we have dim $R\geq$ dim $V \\$. We will now show that dim $V\geq$ dim $R$. Let $P_0\subset\dots\subset P_r$ be a maximal chain of prime ideals of $R$. Again by the Correspondance Theoreom, for each $i$ we have, $P_i=Q_i/\mathcal{I}(V)$ for some prime ideal $Q_i$ of $F[t_1,\dots,t_k]$ such that $\mathcal{I}(V)\subseteq Q_i$ and we have the chain $\mathcal{V}(Q_r)\subset\dots\subset\mathcal{V}(Q_0)\subseteq\mathcal{V}(\mathcal{I}(V))=V$ where the last equality follows from $V$ being algebraic.
Here is where I face some issues. Ideally I would like to ensure that $\mathcal{V}(Q_r)\neq\emptyset$ but that requires $F$ to be algebraicaly closed which is something I haven't assumed so far, so is there a way to get around this or is it a necessary asumtption?
And most importantly I would like to show that if $Q\subset F[t_1,\dots,t_k]$ is a prime ideal then $\mathcal{V}(Q)$ is irreducible, which I think would finish the proof but I am having trouble proving this and I haven't been able to find a proof of this fact anywhere. I would greatly appreciate any insights.
 A: If $V(Q)$ is reducible, then $V(Q)=V(I)\cup V(J)$ for some ideals $I$ and $J$ in $F[t_1,\dots, t_k]$.
Now $V(I)\cup V(J)=V(IJ).$ Thus in particular, $Q\supset IJ$. You can check that $Q\supset I$ or $Q\supset J$.
I wrote this in the context of the prime spectrum of a ring, which you can ignore.
In terms of the field valued points (which is the set up you are considering):
Let $V(Q)=V(I_1)\cup V(I_2)$ a union on non-empty, distinct proper closed subsets. Assume that $V(I_1)\nsubseteq V(I_2)$ so there is $a=(a_1,\dots, a_k)\in V(I_1)$ but not in $V(I_2)$. Thus there is $g\in I_2$ such that $g(a)\neq 0$.  Now $$Q=I(V(Q))=I(V(I_1))\cap I(V(I_2))=rad(I_1)\cap rad(I_2)$$ by the Nullstellenzatz over algebraically closed field.
Now $rad(I_1)\cap rad(I_2)=rad(I_1I_2)$. For every $f\in I_1, fg\in I_1I_2$. So $(fg)^m\in Q.$ But $a\in V(Q)$, so $(fg)^m(a)=0$. Since $g(a)\neq 0$, $f(a)=0$. So $f\in Q$. So $I_1\subset Q$.
Hence $V(Q)\subset V(I_1)$ and $V(I_2)=\emptyset$ a contradiction.
Hence $V(Q)$ is irreducible.
Hopefully this answers your last question.
