Newton's Third Law and Net Force 
As shown in the diagram above, the object with thrust $T$ exerts a force downwards towards the Earth. The Earth also exerts a force with the same magnitude but in the opposite direction towards the object. Therefore we know that $T = F_R$. The lifting force of the object, $F_L$, depends on the reaction force from the Earth acting on the object. My question is, if we ignore frictional force, can we say that lifting force equals to the reaction force from the Earth acting on the object. But if that is true, the lifting force will be the same as the thrust. The net force will be equal if the lifting force is same as the reaction force and the object will stay in place instead of going upwards. How can I explain this?
 A: You can firstly think about this scenario without the earth at all.
Let's say you have a rocket in free space.
A chemical reaction inside the rocket can cause a force $\vec{T}$, which acts on the exhaust material and it will be directed downwards, away from the rocket as you showed. Newton's third law tells you then that there will be an equally large but opposite force on the rocket itself. The force T is not actually acting on the rocket but on the particles emitted by it since they are not physically connected to the rocket anymore.
The force $\vec{F_l} = - \vec{T}$ will then accelerate the rocket upwards.
Now on the earth, the rocket has to overcome the gravitational force pushing the rocket (as well as the exhaust material downwards). Therefore, if $\vec{F_l}$ is the total force on the rocket:  $\vec{F_l} = - \vec{T} + \vec{F_g})$.
If now $|T| > |F_g|$, then $\vec{F_l}$ will point upwards and accelerate the rocket away from earth.
A: The question is unclear. Perhaps you might want to include all the forces in including gravitational force and then reformulate the question.
Nevertheless, let me provide a complete picture of the earth and rocket system.
We have an earth that exerts a gravitational force on the rocket $F_{e-r}^g$ and by Newtons law the rocket exerts an exact equal but opposite in direction force on Earth, $F_{r-e}^g$ such that $F_{r-e}^g=-F_{e-r}^g$. Furthur we have the lifiting force of rocket exerted on earth $F_{r-e}^l$ and the reaction by earth on the rocket $F_{e-r}^l$, such that $F_{r-e}^l=-F_{e-r}^l$.
Now the net force on rocket is given by $$F_{rocket}^{net} = F_{e-r}^l-F_{e-r}^g$$ Here the upward direction is considered positive.
The rocket will have no net force if $F_{rocket}^{net}=0$ which happens when $F_{e-r}^l=F_{e-r}^g$. The rocket can accelerate/decelerate by varying $F_{e-r}^l$ which effects the $F_{rocket}^{net}$ by the relation shown above.
