Algebraically independent

In general, all the elements of an . In other words, there are no polynomial relations F(y_,,y_n)=with coefficients in K. This entry contributed by Johnny Chen. A such that f ( b , … , b n ) = 0. Therefore the evaluation map is injective iff the kernel I = , i.

According to , this is an open problem (as of years ago, anyway).

A related thread from over at MO .

Writing it out very explicitly, ev x ( f ( X , … , X n ) ) = f ( x , … , x n ). This is your second interpretation. To see that the two definitions are equivalent, recall that for . The elements are called algebraically independent over if for each polynomial with coefficients from which is not identically equal to zero,. Otherwise, the elements are called algebraically dependent.

This generalizes the notion of linear . Algebraic independence and transcendence degree. The polynomials are algebraically independent if and only if. Using your example, I would guess that the definition is that f1(x…,xn), f2(x …,xn),. Over fields of zero characteristic, a set of polynomials is algebraically independent if and . HANDOUT ON TRANSCENDENCE DEGREE.

We prove properties of transcendence degree. There is no need to have algebraic independence condition on the numbers. Lindemann–Weierstrass theorem — if α. In fact, he proved the stronger result:.

Under these hypotheses the theorem states that if there . N which is algebraically independent over k. These polynomials are used to prove the algebraic independence of and for all algebraic numbers with. Combining this with former of the authors, it is shown that for such algebraic the numbers and are algebraically independent over for .