How are standard basis of polynomials linearly independent? Consider the set $\{1,z,z^2,...z^m\}$. As this is the standard basis for a vector space of polynomials, the list should span the space and also be linearly independent. 
One problem I'm having though is considering the equation 
$0 = a_{0}1 + a_{1}z + ... + a_{m}z^m$
If I let z = 0, then all the polynomials with the exception of the first one are equal to 0, should I should be able to pick scalars, not all 0 that lead to a representation of the 0 vector in this basis that does not correspond to the trivial solution. So this would contradict the linear independence of the set.
I'm not exactly sure where I'm getting confused. Is it because you are not supposed to let z be any particular value? I can see how the list would be linearly independent if the value of z was not fixed. 
 A: The statement
"$0 = a_01 + a_1z + \cdots + a_mz^m$ only holds when all $a_i = 0$"
is a statement about polynomials being equal.  You are talking about polynomials having a root.  Certainly $z - z^2$ has zero as a root, so the equation $z - z^2 = 0$ has a solution, but the polynomial $z - z^2$ is not the zero polynomial, just like it's not equal to the polynomial $z - z^3$ even though both have $0$ as a root.
A: If $a_0+a_1z+\cdots+a_nz^n$ is the constant zero function then $a_i=0$ for $0\le i\le n$. Remember that you are viewing the equation
$$P(z)=a_0+a_1z+\cdots+a_nz^n=0$$
as an equation of functions, meaning $P(z)=0$ for all $z$.
A: Two polynomials are equal if the coefficients of $x^i$ are equal for all $i$. For polynomials over $\mathbb{R}$ (but not necessarily over a finite field, for example), this is the same as taking the same values for all $x$. It is not the same thing as being equal for one particular value of $x$.
Also, can I point out that the basis for a polynomial ring is $\{x_1, x_2, ..., x_m, ...\}$, not $\{x_1, x_2, ..., x_m\}$.
A: There are different ways to interpret your question, depending on whether you are talking about polynomials or polynomial functions, and over which field.  
If you are talking about the vector space of polynomials over a field, then equality of polynomials means equality of coefficients, by definition, so being equal to the zero polynomial immediately implies all coefficients are $0$.  
If you are talking about polynomial functions, then you will want to use the fact that a polynomial of degree $n$ can have at most $n$ zeros in the field. Hence if you have a nonzero polynomial that is equal to the zero function on your field, then the degree of the polynomial is greater than or equal to the number of elements in your field.  If your field is infinite (like the field of real numbers), this implies that your polynomial must have all coefficients zero.

(In the question linked as a potential duplicate, it appears to be specifically a question about polynomial functions on $\mathbb R$, so these remarks seemed more appropriate here.)
