Consider a projective variety $X$, and let $Y$ be a closed subvariety. Consider the blow-up of $X$ along Y: we obtain a new variety $\tilde{X}\subset X\times \mathbb{P}^{\dim Y}$, together with a birational map $$b:\tilde{X}\to X$$ which is an isomorphism outside the exceptional locus $\tilde{Y}= b^{-1}(Y) \simeq \mathbb{P}(\mathcal{N}_{Y\mid X})$.
Question: Is it true that $\dim\tilde{X}=\dim X$?
My idea: The blow-up is a birational map, hence an isomorphism on an open (dense) subset; since the dimension of a variety is defined as the trascendence degree of the function field, which is the same on open set, then we can conclude.
Is my idea correct? I'm asking this becuase, while I'm convinced of this for the case of blowing-up a point, for the general case of a subvariety I'm not sure, becuase intuitively it looks like to me I'm adding quite a big space (I know, it's not rigorous, but I've just started studying this topic).
There are a couple of things to keep in mind. One is that I think you want the blowup construction to live in $X \times \mathbb P^{\operatorname{codim} Y}$; at the very least this works when $Y$ is a complete intersection.
In general though, as you said, the exceptional locus arises as the projectivization of the normal bundle $\mathcal N_{Y|X}$. Set $\dim X = n$ and $\dim Y = m$. The rank of the normal bundle is always the codimension of $Y\subset X$, hence $\mathbb P(\mathcal N_{Y|X})$ has relative dimension $n - m - 1$ (meaning that this is the dimension of the fibers of $ \mathbb P(\mathcal N_{Y|X}) \to Y$, the structure map of the projective bundle). Now adding to this the dimension of $Y$, we get the dimension of the projective bundle as a variety, namely $(n-m-1) + m = n-1$, showing that it has codimension $1$ in $X$. It is for this reason that the exceptional locus of a blowup is typically referred to as the exceptional divisor.