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For a serendipity element, the shape function of a middle pt on a edge can be formed using the tensor product in each direction, e.g. point 5 on Quad8

$$ \phi_5(\xi,\eta)=\frac{1}{2}(1-\xi^2)(1-\eta) $$

but the shape function on corner node (point 1) is $$ \phi_1(\xi,\eta)=\frac{1}{4}(1-\xi)(1-\eta)-\frac{1}{4}(1-\xi^2)(1-\eta)-\frac{1}{4}(1-\eta^2)(1-\xi) $$

Why the shape function of the corner point from a serendipity element (Q8) can not be built by the tensor products like this

$$ \phi_1(\xi,\eta)=\frac{1}{4}\xi\eta(1-\xi)(1-\eta) $$

Does this mean I just pick up the wrong basis?

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Your proposed corner shape function is the same as for the Q9 element. So your question is why isn't it allowed to simply "resuse" these shape functions for the Q8?

One of the several criteria for shape functions is that at any point $\xi$ and $\eta$ in the element, the sum of all shape functions evaluated at that point must equal one.

Consider the center point, $\xi=0, \eta=0$. The Q9 corner point shape functions are all equal zero there. The Q8 edge shape functions (e.g. your first equation) all equal $1/2$ so their sum is two, violating this criterion.

The shape functions for the Q8 element can be found in many FE texts, of course. I find this set of notes by Felippa (Shape Function Magic) that describes a systematic process for deriving shape functions in general and those for the Q8, in particular, to be very readable.

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