$ S:=\{(x,a_3,a_2,a_1,a_0) \in \Bbb R^5 : x^4+a_3x^3+a_2x^2+a_1x+a_0=0\}$ is a connected manifold Consider the following subset of $\Bbb R^5$:
$$ S:=\{(x,a_3,a_2,a_1,a_0) \in \Bbb R^5 : x^4+a_3x^3+a_2x^2+a_1x+a_0=0\}$$
I want to show that $S$ is a connected (topological) manifold and compute its dimension.  I think the dimension will be automatically obtained during the procedure showing that $S$ is a manifold. So I think the first question is significant. I know that $S$ is second-countable and Hausdorff, being a subspace of $\Bbb R^n$, so we are left to show that it is locally-Euclidean, but how do I have to start? 
Thanks in advance
 A: In order to show that it is a manifold, you could use the submersion theorem applied to function $f: \mathbb{R}^5 \to \mathbb{R}$, such that  $ f(x,a_3,a_2,a_1,a_0) =x^4 + a_3 x^3 +a_2x^2 +a_1 x + a_0$
Since the Jacobian matrix is $(4x^3+ 3a_3 x^2+ 2a_2 x+ a_1,x^3,x^2,x, 1)$ it is of maximal rank, hence $f$ is a submersion and $S=f^{-1} (\{0\})$ is a smooth manifold of dimension 4.
To show that it is connected, maybe it's better to see if it is path connected, I guess it is. 
In an non rigorous way, if you want to find a path between $(x, a_3, a_2, a_1, a_0)$ and $(y,b_3,b_2,b_1,b_0)$, first see if 'the polynomial in x' and in 'y' are split, if not smoothly move $a_0$ or $b_0$ negatively (and $x$ and $y$ accordingly) until they are.
When you have only split polynomials, since the coefficients of the polynomial are symmetric polynomials in the roots of it, then you just have to smoothly move each roots of the polynomial in 'x' to the ones of the polynomial in 'y'. 
This is just a rough idea, perhaps it can be done more rigorously.
