$N^-D N^+$ is Zariski open dense in $\text{GL}(n, \mathbb R)$ Let $N^-$ (and resp. $N^+$) denote the upper (resp. lower) triangular unipotent subgroup of $\text{GL}(n,\mathbb R)$. Let $D$ denote the full diagonal subgroup of $\text{GL}(n,\mathbb R)$
I wonder how to prove that $N^-D N^+$ is Zariski open dense in $\text{GL}(n,\mathbb R)$. Since open sets in Zariski topology are always dense, it suffices to show that product is open, or its compliment is closed. But I have difficulty in expressing its complement in terms of polinomials.
An elementary proof would be great, but if you want to use some advanced tools like Bruhat decomposition, please clearly explain the details as well as the source of theorems you are using (The Bruhat decomposition itself doesn't give the Zariski openness).
 A: As a small caveat, Zariski open sets are not always dense inside varieties, you need the condition that the variety is irreducible. I will give two possible ways to see this, one is more elementary and the other uses the Bruhat decomposition.

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*The first way to see that a morphism $f:X\to Y$ between smooth(!) varieties is open, is by checking that it is injective and induces an isomorphism on tangent spaces, this is the inverse function theorem. Injectivity just follows from the fact that $D$, $N^+$ and $N^-$ are all pairwise disjoint, i.e. the map multiplying everything together
$$ f:N^+\times D\times N^-\to \operatorname{GL}_n(\mathbb{R})$$
is injective. Now choose $x=(1,1,1)$ which maps to $f(x)=1$. At the tangent space this is an isomorphism: The Lie algebras of $N^+$, $D$, $N^-$ are exactly the strictly upper triangular, diagonal and strictly lower triangular matrices inside $M_n(\mathbb{R})$. For any other point inside $N^+\times D\times N^-$, you can use the group action to shift it to $x$ (this is a bit subtle!) for which we just have verified this.
There is a more algberaic version of this approach, where you replace injective by monomorphism (in fact radicial would be sufficient) and isomorphism on tangent spaces by etale. But morally the proof is the same.

*More concisely (although you are implicitly using statements about Bruhat stratifications) you notice that $N^+ D N^-= BwBw$. Here $B$ is the Borel subgroup consisting of upper triangular matrices (which might have non-trivial diagonal entries) and $w$ is the so-called longest element of the Weyl group, in our case it is simply the anti-diagonal matrix with all its entries equal to $1$. Now you know that $BwB$ is the biggest Bruhat cell and it has to be open (in general $BwB$ is a locally closed subvariety isomorphic to $\mathbb{A}^{l(w)}$). Alright, but then $BwBw$ is just a shift of an open subvariety by $w$ and open itself.

I guess in the last case you can also give a more concrete description in terms of polynomials (see here): $BwB$ is the open subvariety such that all $k\times k$-minors are non-zero and thus its complement is the union of the closed subsets $V(f_k)$, where $f_k$ is the polynomial of the $k\times k$-minor. You should not forget to shift everything by multiplying by $w$ on the right then.
