Let $I_1, I_2, I_3$ be three ideals in $\mathbb{C}[x_1, .., x_n]$. Let $V(J) = \{ \mathbf{x} \in \mathbb{C}^n : f(\mathbf{x}) = 0 \ for \ all \ f \in J \}$.
How can I prove that Codim $V(I_1 + I_2 + I_3) \leq$ Codim $V(I_1) + $ Codim $V(I_2)$ + Codim $V(I_3)$ ?
Here, Codim $V(I_j) = n - \dim V(I_j)$ and by dimension I mean as an algebraic set in $\mathbb{C}^n$. Any comments are appreciated. Thank you!
Let us fix once for all times in the sequel the polynomial ring $A=\mathbb C[X_1,X_2,\dots,X_n]$, which is noetherian, in it two ideals $$I,\ J$$ generated by $$f_1,\dots, f_i\in A\ ,$$ and respectively
$$g_1,\dots, g_j\in A\ .$$ We can and do assume that $i, j$ are minimal with these properties.
Let $K$ be the ideal generated by $I$ and $J$, $$ K = I+J\ ,$$ it has the generators $$f_1,\dots, f_i\ ;\ g_1,\dots, g_j\in A\ ,$$ but this list may not be minimal. (E.g. for $I=J$ non-trivial ideal.)
The question is now related and follows from "dimension theory". The dimension can be defined in various ways:Dimension_of_an_algebraic_variety
From the list, there are at least three "worlds", categories, mathematical domains, where the notion can be defined, and which are relevant for the question. These are:
For an ideal $I$ in a ring $R$ we can define its height. A notion of "dimension" is not used commonly for ideals, but it is used for commutative rings. In our case, this notion applies for $R$ and the quotient ring $R/I$, and it is called Krull dimension. Reference: Krull_dimension. This is a purely algebraic object.
In both cases, the definitions deal with the length of chains of prime ideals, ordered by strict inclusions. (The number of the strict inclusions in the chain is the length.)
(In order to relate $\dim A$ with $\dim (A/I)$ and the height of $I$, one has to find a natural way to "combine" and/or reduce chains. In the p.o.-set of ideals, one can consider the ideals that can be compared with $I$, and only chains going through this "strangulated" po-set. Then one needs a way to ensure same lenght of maximal chains - under favorable circumstances, we have such a situation.)
For an (affine) algebraic variety $X$ we have a dimension. This is defined in the algebraic geometry by passing to affine coverings, writing each affine piece as a Spec(trum) of a ring, then using the Krull dimension for rings.
For a differential manifold $M$ over $\mathbb C$ we also have a notion of dimension, this is defined by using charts (near regular points) to open domains in the one or other standard affine space $\mathbb C^k$ for some fixed $k$. (This is the dimension of $M$.) To an (affine) algebraic variety $X$ as in the posted question, $$ X = \text{Spec}(A/I)\ ,$$ we can associate the corresponding manifold $$M=X(\mathbb C)\ ,$$ which is as a set $V(I)$. Moreover, the projection $$ A\to A/I$$ induces the map (of ringed topological spaces) $$ \text{Spec} A \leftarrow \text{Spec} (A/I)\ ,$$ which is $$ \mathbb C^n\leftarrow V(I)\ .$$ The left arrows are inclusions. The codimension in the question is taken with respect to these canonical inclusions for $I,J;K$.
The inequality $\text{codim} V(I+J)\le \text{codim} V(I) +\text{codim} V(J)$ follows now as follows: \begin{align} \text{codim} V(I) &= \dim \mathbb C^n -\dim V(I) = \dim A -\dim(A/I) = \text{height}(I) \\ &= i\ , \\\\ \text{codim} V(J) &= \dim \mathbb C^n -\dim V(J) = \dim A -\dim(A/J) = \text{height}(J) \\ &= j\ , \\\\ \text{codim} V(I+J) &= \dim \mathbb C^n -\dim V(I+J) = \dim A -\dim(A/(I+J)) = \text{height}(I+J) \\ &\le i+j\ . \end{align}
Here, we have tacitly used the passage
which is Krull's principal ideal theorem as stated here: Consequences...
In most proofs, one needs Hilbert's Nullstellensatz.