$F$ be a field and $F(x)$ be the field of rational functions in $x$ over $F$. Then the element $x$ of $F(x)$ is transcendental over $F$ Let $F$ be a field and $F(x)$ be the field of rational functions in $x$ over $F$. Then the element $x$ of $F(x)$ is transcendental over $F$.
Proof: $F(x)$ is clearly a field extension of the field $F$. If $x$ is not transcendental over $F$, then there is a non-null polynomial $f (y) = a_0 + a_1y +···+ a_ny^n$ over $F$, of which $x$ is a root. This shows that $a_0 + a_1x +···+ a_nx^n = 0.$ Hence each $a_i = 0$.
implies that $f (y)$ is a null polynomial. This gives a contradiction.
I don't understand why $a_0 + a_1x +···+ a_nx^n = 0.$ $\implies$ $a_i = 0 \ \forall i$. If $x$ is a zero of this polynomial for some non zero coefficients then $x$ is algebraic, but it's consistent with the proof as we have assumed $x$ is not transcendental. Am I missing any point here?
 A: The confusion is really that polynomials are used in two differerent ways here. Given any ring $A$ and a polynomial $f(y)\in A[y]$, and any element $a\in A$, we can "evaluate" the polynomial $f$ at $a$ such that $f(a)\in A$. This is either due to the fact that polynomials are formed with addition and multiplications that can be performed in any ring, or the universal properties of polynomials.
Now, in this particular case, $A=F(x)$ and the coefficients are restricted to the subring $F$, hence $f(y)$ is chosen to be in $F[y]$ and $x$ is the formal variable from $F(x)$, we have the evaluation $f(x)$ is equal to itself as a polynomial. Now we have the identity $f(x)=0$ not in $F$ but in $F(x)$ (equivalently to in $F[x]$, as $F[x]$ is a domain), because $A=F(x)$, and it just happens that all coefficients are in $F$. And finally two polynomials are equal iff they have all the same coefficients.
In other words, the polynomial say $1+x+x^2\in\mathbb Q[x]$ can be regarded as an entity that is just an element of $\mathbb Q[x]$, but it can also be considered the result of $1$ plus $x$ plus $x$ times $x$ in the ring $\mathbb Q[x]$.
A: Writing “If $x$ is a zero of this polynomial” makes no sense here. Here, $x$ is the element $x$ of $F(x)$ and $F(x)$ is defined in such a way that$$a_0+a_1x+\cdots+a_nx^n=b_0+b_1x+\cdots+b_nx^n$$if and only if $a_i=b_i$ for each $i\in\{0,1,\ldots,n\}$. In particular, $a_0+a_1x+\cdots+a_nx^n=0$ if and only if each $a_i=0$.
