If $(f, g)$ is a Galois connection between two bounded lattices, then if $T$ is an ideal we have $f^{-1}(T)$ is an ideal

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Let $\mathcal{I}(L)\:$ and $\mathcal{I}(N)\:$ be the ideal lattices of the bounded lattices $L$ and $N$ and let $(f, g)$ be a Galois connection between $L$ and $N$, then show that $\:\forall \:\: T \in \mathcal{I}(N)\,\,$ we have$\,\,\, f^{-1}(T) \in \mathcal{I}(L)$ (so that, it is well defined) and that $f^{-1}(T) = \downarrow g(T)$.

Showing that $f^{-1}(T)$ is a down set is quite easy: $\:\:\:u\in f^{-1}(T)\iff f(u)\in T\iff \exists v\in T\:$s.t.$\:\:f(u)\le v\iff\exists v\in T\:\:$s.t.$\:\:u\le g(v)\iff u\in\,\downarrow g(T)$.

Any idea about how to see that it is actually an ideal? I mean, that $\forall \: u, v \in f^{-1}(T)$ we have $u \vee v \in f^{-1}(T)$.