Before asking the question I should say that I don't know much about algebraic groups and I'm not sure if the question has the right level for MO. If not, please let me know and I will delete the question that emerged when I was trying to understand some theorems in a paper on the vanishing range of the cohomology of finite groups of Lie type.
Call an algebraic group (over an algebraically closed field) simple if it is simple as an abstract group and almost simple if it has finite center. Every almost simple algebraic group is of one of the following types:
$$A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4, G_2$$
Question: What are the simply connected simple algebraic groups (over an alg. closed field with no assumption on the characteristic) and of which type are they ?
Counter-examples: $SL_{n+1}$ is simply conntected almost simple of type $A_n$, but not simple while $PSL_{n+1}$ is simple but not simply connected of the same type.
Added: The mathematical problem is solved by Jay's answer. But I'm still confused about the terminology: In contrast to finite groups, where each simple group has a definite name, simple algebraic groups doesn't seem to have a name on their own (apart from the classical types). For example, by comments of Yves Cornulier and Jim Humphreys, for each type there is a unique simply connected simple alg. group. In my opinion it would be reasonable to give them a special name, for example $E_6^{sc}$. However the typical wording in the literature is like "let $G$ be a simply connected, connected complex algebraic group of type $E_6$".
Is there a particular reason for this convention ?
Best Answer
If $G$ is an adjoint algebraic group then it is always simple as an abstract group, (EDIT: This is because any proper normal subgroup of a simple algebraic group must be finite and lie in the centre). In general assume $G$ is connected reductive algebraic group and let $\pi : G \to G_{ad}$ be an adjoint quotient of $G$. Then $G$ is simple as an abstract group if the kernel of $\pi$ is trivial because $\pi$ is then an isomorphism of abstract groups. If $\pi$ does not have a trivial kernel then it is clear that $G$ is not simple as an abstract group.
For an example: in characteristic 2 the symplectic group $Sp_{2n}$ is simple as an abstract group as it is isomorphic (as an abstract group) to the corresponding adjoint group. However these two groups are not isomorphic as algebraic group (only isogenous).
EDIT: I suppose this didn't really explicitly answer your question. Explicitly we have a simple simply connected group is simple as an abstract group if and only if it is on the following list: