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

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:

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.

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:

• Please link to the previous threads so we don't waste time repeating answers that didn't help you. – Bill Dubuque Feb 11 at 17:10
• Thank you, I have linked the previous threads in the post :) – Malcolm Feb 11 at 17:17
• It might help to review the motivation and construction of polynomial rings, e.g. see here. – Bill Dubuque Feb 11 at 17:17
• @BillDubuque, I have read your response and from what I understood of it, I could surmise that rather than $K[t]$ being notation for a polynomial over $K$ with indeterminate $t$, it rather is the set of polynomials obtained from carrying out the operations within the ring with an element $t$. So $\mathbb{Q}(\sqrt{2})=\{p+q\sqrt{2}+r(\sqrt{2})^2:p,q,r \in \mathbb{Q}\}$ rather than any polynomial. However I do not see how that aids with my difficulties above, maybe clarifies some things (if I understood them correctly) – Malcolm Feb 11 at 17:50
• The proof applies only to the fraction field of a polynomial ring $K[t]\,$ because it uses the fact that $t$ is transcendental over $K$. So it doesn't apply to $\,\Bbb Q[\sqrt2]\,$ since $\sqrt 2\,$ is algebraic over $\Bbb Q.\,$ You need to determine from the context if the adjoined element is algebraic or transcendental because the notation itself doesn't specify that (though there may be ambient conventions that do, e.g. certain symbols reserved for indeterminates = transcendentals, e.g. $\,x,y,z).\$ See also here on ring adjunctions. – Bill Dubuque Feb 11 at 23:54

## 1 Answer

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 $$\Bbb Z$$.

• What's wrong with t being algebraic over K then? Why does t have to be necessarily trascendental over K? Can't t be algebraic and it would forfeit the result? – Malcolm Feb 11 at 23:33