Answer:
The equations for
the formation of the two oxides are:
4/3AI(s) + O2(g) 
2/3AI2O3(s)
2Mg(s) + O2(g) 2MgO(s)
If we look at the
plots for the formation of the two oxides on the Ellingham diagram, we find
that the 
value
of Al2O3, at temperatures below 1623 K, is less negative
than that of MgO.
Thus, 
of the reaction
2Mg + 2/3Al2O3
2MgO + 4/3 Al
is negative.
Therefore, below 1623 K, Mg can reduce Al2O3 to Al.
However above 1623 K, the value
for Al2O3 is more negative than that of MgO. Thus, of the reaction;
4/3 Al + 2 MgO
2/3Al2O3 + 2Mg
is negative.
Therefore above 1623 K, Al can reduce MgO to Mg.
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