Let $F$ be a field. Every (non-constant) polynomial $p \in F[X]$ is transcendental over $F$

Question

I will start with the Definition:

Let $L/K$ be a Field extension. The element $a \in L$ is called algebraic over $K$ if there exists a polynomial $p \in K[X]$ such that $p \neq 0$ and $p(a)=0$. If from $p(a)=0 \Rightarrow p=0$, then $a$ is called transcendental over K.

The following statement really confuses me:

Let $F$ be a field. Every (non-constant) polynomial $p \in F[X] \subset F(X)$ is transcendental over $F$

My Thoughts: For $a \in F[X]$ to be transcendental over $F$, considering the above definition, this means that we are looking at the field extension $F[X]/F$. The element $a \in F[X]$ is called transcendental if for $p \in F[X]$ if $p(a)=0 \Rightarrow p=0$

I think that $p(a)$ means a substitution with a polynomial. For example for $p=1+X^2$ and $a=X+1$, then $p(a)=1+(X+1)^2$. Question 1: Is my assumption correct?

Now I would like to show that the statement really is true: Let $a$ be an arbitrary element of $F[X]$. I will use the following theorem: Let $R'$ be a commutative extension ring of $R$ such that $1_{R'}=1_{R}$. Then for every $v \in R'$, the function $\phi: R[X] \rightarrow R'$, $P=\sum_{k \in \mathbb{N}_0} c_kX^k \mapsto P(v):= \sum_{k \in \mathbb{N}_0} c_k v^k$ is a ring homomorphism.

If I now choose $R=F[X]$, and $R'=F[X]$, then for an non constant $a \in F[X]$, $P(a)=\sum_{k \in \mathbb{N}_0} d_kX^k $. If now $P(a)=0_{polynomial}$ and since $a$ was arbitrary with degree $>0$ $\Rightarrow P=0_{polynomial}$

Question 2: Is my argumentation enough to show that the statement is true? (If not, what is it missing?)

Answer 1

I think your thinking is along the right lines.

To make our discussion more concrete, let's say that $F = \mathbb Q$. And let $L$ be the field $\mathbb Q(T)$, the field of rational functions with $\mathbb Q$ coefficients. Thus $L$ is an extension of $F$.

An element $a \in L$ is really a rational function with $\mathbb Q$ coefficients. For example, $a$ could be $(3T + 1)/(4T^2 + \tfrac 1 2)$, or $(5T^4 - 2)/(17T^9 + \tfrac 7 8 T + 1)$, or something like that.

An element $a \in L$ is algebraic over $F$ if there exists a polynomial $p(X) \in F[X]$ such that $p(a) = 0$.

Understandably, it can be difficult to wrap one's head around what $p(a)$ actually means. So let's work through an example.

Suppose that $a \in L$ is the element $$ a = \frac{3T + 1}{4T^2 + \tfrac 1 2} ,$$ and suppose $p(X) \in F[X]$ is the polynomial $$ p(X) = 7X^2 + 3X - \tfrac 2 3 .$$

Then $p(a)$ is another element in $L$; namely, it is the element $$ p(a) = 7\left(\frac{3T + 1}{4T^2 + \tfrac 1 2}\right)^2 + 3\left( \frac{3T + 1}{4T^2 + \tfrac 1 2} \right) - \tfrac 2 3.$$

As you can see, $p(a)$, viewed as an element in $L = \mathbb Q(T)$, is not the zero element.

Finally:

• $a \in L$ is algebraic over $F$ if there exists some non-zero $p(X) \in F[X]$ such that $p(a)$ is the zero element in $L$.
• $a \in L$ is transcendental over $F$ if for all non-zero $p(X) \in F[X]$, $p(a)$ is a non-zero element in $L$.

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Let's return to the statement you were reading.

The statement says that if $a \in L$ is (i) not a constant and (ii) is a polynomial, then $a$ is transcendental over $F$.

Let's be clear about which $a$'s this statement applies to.

• If $a = 42$, then this statement is not applicable, because $a$ is a constant.
• If $a = 9/(T + 4)$, then this statement is not applicable, because $a$ is not a polynomial. It's a rational function, but it is not a polynomial.
• If $a = T + 1$ or $a = T^2 + \tfrac 7 8 $, then this statement is applicable, since $a$ is a non-constant polynomial.

The statement says that if $a$ is an element like $T + 1$ or $T^2 + \tfrac 7 8 $, then there is no polynomial $p(X) \in F[X]$ such that $p(a) = 0$ (other than the zero polynomial).

Answer 2

The answer to your first question is yes.

I have to admit that I don't understand the second part. However, it is rather simple.

Suppose that $p(x) = p_n x^n + \dots + p_0 \in F[x]$ is a non-constant polynomial polynomial and that $0 \neq p_n \in F$. Showing that $p$ is transcendental over $F$, means that $p$ cannot be the root of a polynomial $F[y]$. By contradiction, if $p$ would be the root of $q(y) = q_m y^m + \dots +q_0$ with $q_m \neq 0$, we would have

$$q_m (p(x))^m + \dots + q_0 =0$$ hence considering the term of highest degree $q_m p_n^m = 0$. Which implies $q_m=0$, a contradiction.