I am reading a lemma on Dedekind domain but I have difficulties understanding it, I am bring it up here hoping for help. The original passage is in one big fat paragraph but I broke it down here for your easy reading. Thanks beforehand for all your time and help, let me know if I forget to include any underlying lemmas.
LEMMA: Dedekind domains have Krull dimension 0 or 1.
PROOF: (1) Let R be a Dedekind domain, let T be a nonzero prime ideal of R, and let U be a maximal ideal of R such that T $\subseteq$ U. We shall see that T = U.
(2) Since R is assumed to be a Dedekind domain, we obtain from T $\subseteq$ U an ideal V of R such that T = UV.
(3) Thus, as T is a prime ideal of R, U $\subseteq$ T or V $\subseteq$ T.
(4) Assume first that V $\subseteq$ T. Then, as T = UV, T $\subseteq$ UT.
[Here is my question from #(4): How do you go from T = UV to T $\subseteq$ UT ?]
(5) Thus, as T $\neq${0}, R $\subseteq$ U; see Lemma below.
[The lemma referred aboved is this: "Let R be a Dedekind domain, and let U and V be ideals of R. Assume that R possesses a nonzero ideal T with TU $\subseteq$ TV . Then U $\subseteq$ V." But even after arduous reading, I still don't know how R $\subseteq$ U is derived in #(5).]
(6) This contradicts the fact that U is a maximal ideal of R. Thus, U $\subseteq$ T.
(7) Thus, as T $\subseteq$ U, T = U.
[How does T = U relate to dimension of 0 or 1?]
As always, thank you so much for your time and help.
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EDIT: Here is how Dedekind domain defined in the text: An integral domain R is called Dedekind domain if, for any two ideals T and U of R with T $\subseteq$ U, there exists an ideal V of R such that T = UV.
(4) From $V\subseteq T$ we get $UV\subseteq UT$ and since $T=UV$ ...
(5) Set in your lemma $U=R$ and $V=U$. Note that $TR=T$.
(7) You have actually shown that every non-zero prime is maximal, so you can't have a chain of three primes in $R$.