Question about algebraically equivalent divisors I'm trying to do exercise V.1.1.7 in Hartshorne, about algebraic equivalence of divisors, and the first problem is to show that the divisors algebraically equivalent to zero form a subgroup of $Div(X)$.
Here's a recap of Hartshorne's definition:

Let us define algebraic equivalence of divisors. We first do it for an effective divisor (i.e. a divisor that actually correspond to an ideal sheaf $\mathscr I$). Then it corresponds to some subscheme $C$ of $X$. A family of effective divisors is an effective divisor $D$ on $X \times T$, together with a flat map $\pi: X \times T \to T$, where $T$ is some smooth curve. If $0,1 \in T$ are two points, we say that $D_0$ (setting $t=0$) and $D_1$ are prealgebraically equivalent. Arbitrary divisors can be written as differences between effective divisors, and we say that these are prealgebraically equivalent if they can be written as differences between prealgebraically equivalent divisors. Two divisors $D, D'$ are algebraically equivalent if there is sequence $D=D_0D_1, \ldots, D_{n-1}D_n=D'$ with $D_i$ and $D_{i+1}$ prealgebraically equivalent.

So to do the exercise, I have to show that if $D \equiv 0$ and $E \equiv 0$, then $D+E \equiv 0$.
Since $D=D'-D'' \equiv 0$ (with $D'$ and $D''$ effective), we have a family $\pi: X \times T \to T$ as above, and similarly for $E$, possibly with a different $T$.
Here comes the problem. If I want to show that $D+E=(D'+E')-(D''+E'') \equiv 0$, then I have to find a family $X \times T^{''} \to T^{''}$ and an effective divisor $\mathscr D \subset X \times T^{''}$ with $\mathscr D_0=D'+E'$ and $\mathscr D_1=D''+E''$. The problem is I don't see how to do it. It would have been almost trivial if we could use the same curve $T$ in the all the families (then the answer would be just look at the ideal sheaves in the total space and tensor them).
Does anyone have a hint/explanation?
I sorta feel that I should be able to assume that $T= \mathbb A^1$, but I dont know how.
 A: It turns out that this is actually much easier than I made it out to be in the comments. The chief problem is actually understanding the definition:

*

*Two effective divisors $D_0$ and $D_1$ on $X$ are prealgebraically equivalent iff there's an effective divisor $D$ on $X\times T$ flat over $T$ and two points $0,1\in T$ such that $D\cap X\times\{0\}=D_0$ and $D\cap X\times\{1\} = D_1$, where $T$ is a smooth curve.

*Two general divisors $D_0$ and $D_1$ are prealgebraically equivalent iff there are effective divisors $E_0$ and $E_1$ so that $E_0$ and $E_1$ are prealgebraically equivalent and $E_0-E_1=D_0-D_1$.

*Two general divisors $D,D'$ are algebraically equivalent iff there's a finite sequence of divisors $D=D_0,\cdots,D_n=D'$ so that $D_i$ is prealgebraically equivalent to $D_{i+1}$.

The second bullet point implies that if $D_0$ and $D_1$ are prealgebraically equivalent, then $D_0+D$ and $D_1+D$ are prealgebraically equivalent for any divisor $D$: indeed, if $E_0$ and $E_1$ are the effective prealgebraically equivalent divisors so that $E_0-E_1=D_0-D_1$, then we also have $E_0-E_1=(D_0+D)-(D_1+D)$. This implies that if $D\sim_{alg} D'$ and $E\sim_{alg} E'$, $D+E\sim_{alg} D'+E'$, which proves the claim you're interested in, without getting bogged down in fiddling with $T$ at all.
