Rational expression $K(t)$ is a transcendental extension of $K$?

Question

I know this question was asked before, but none of the previous threads end up answering the question satisfactorily enough for me. So let me try to summarize my problems succinctly:

1. The notation $\mathbb{Q}(\sqrt{2})$ is commonly used to denote the smallest sub-field generated by $\mathbb{Q} \cup \{\sqrt{2}\}$. However, for a general sub-field $K$, $K[t]$ is defined to be the ring of polynomials over $K$ which then follows $K(t)$ which is the field of polynomials over $K$ (or rational expressions). However, is $K(t)$ simply a way of notation or does it follow the convention of $\mathbb{Q}(\sqrt{2})$? If the latter, how does this relate with rational functions over $K$?
2. I am not understanding the proof for how $K(t)$ is a transcendental extension of $K$ which goes as follows:

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If $p$ is a polynomial over $K$ s.t. $p(t)=0$ then $p=0$ by definition of $K(t)$, so the extension is transcendental.

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I understand that to show transcendence over $K$ we shall assume some element $t =\frac{r(s)}{q(s)}\in K(s)$ satisfies $p(t) =0, p\in K[t]$ and show that p must be identically $0$ as a result. However, where are we using the definition of $K(t)$ to show this is so?

The other threads are linked here:

• What is $K(t)$ and why is a transcendental extension of $K$?

• Show a rational function is transcendental over a field.

Answer 1

Well, if $t$ is algebraic over $K$, then $K[t]=K(t)$, i.e., ring extension equals field extension.

If $t$ is transcendental over $K$, then $K[t]$ is a polynomial ring and $K(t)$ the field of rational functions of $t$ over $K$. The construction of $K(t)$ from $K[t]$ is the same as the construction of $\Bbb Q$ from $\Bbb Z$.