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The following is a text that I'm not quite understanding

"The set of rational expressions $K(t)$ is a transcendental extension of the subfield $K$ of $ℂ$.

PROOF Clearly $K(t)$ is a simple expression generated by $t$..."


How is it that $K(t)$ is a simple expression generated by $t$? Simple extensions are formed by adjoining an element to a certain field, $t$ is a variable, not an element, so how could $K(t)$ possibly be a simple field extension?


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


Does the latter sentence has anything to do to the fact that the only polynomial $p$ that will always equal $0$ regardless of the variable we plug in is $p=0$? Is the only sense I can get out of what the author is saying. Is is that we are looking each polynomial in $K(t)$ as an element? If this is the case, then still, how is it that $K(t)$ is formed from adjoining $t$ to $K$?


I would really appreciate any help/thoughts.

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  • $\begingroup$ Start from the polynomial ring $K[t]$ and construct the field $K(t)$ of rational expressions in the same way the rational numbers are constructed from the integers. The variable $t$ does not fulfill an equation $p(t)=0$ and so $p$ must be zero. $\endgroup$ – Wuestenfux Nov 3 '17 at 13:35
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    $\begingroup$ You say "$t$ is a variable, not an element", but these things are not mutually exclusive! A set can have variables as elements. In particular, $t$ is an element of the set $K(t)$. I'm not sure you understand the definition of $K(t)$...can you say what you think it means? $\endgroup$ – Eric Wofsey Nov 3 '17 at 20:02
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$K(t)$ is the set of rational function ${{p(t)}\over {q(t)}}$ where $p,q$ are polynomials. It is a trancendantal extension since $t$ is not algebraic over $K$, that is you cannot find a polynomial $p$ such that $p(t)=0$

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  • $\begingroup$ Though what is meant by a polynomial $p$ such that $p(t)=0$? Again, if $t$ was an actual number, say $e$, then a polynomial over $K$ such that $p(e)=0$ would make sense to me, yet I don't really grasp the meaning of $p(t)$ where $t$ is a variable. $\endgroup$ – Leo Nov 3 '17 at 13:38
  • $\begingroup$ An element $x$ of the extension $L$ of $K$ is algebraic if there exists a polynomial $a_0+a_1u+...+a_nu^n$ such that $a_0+a_1x+...+a_nx^n=0$, here $K(t)$ is a field, you cannot find a polynomial $p$ in $K[u]$ such that $p(t)=0$, i.e $a_0+a_1t+...+a_nt^n=0$. $\endgroup$ – Tsemo Aristide Nov 3 '17 at 13:49
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    $\begingroup$ @Leo: If $A$ is any $K$-algebra and $a\in A$, then you can evaluate a polynomial $p$ with coefficients in $K$ at $a$ to get an element $p(a)\in A$: you just plug in $a$ and evaluate using the operations of $A$. That's all that's going on here, with $A=K(t)$ and $a=t$. $\endgroup$ – Eric Wofsey Nov 3 '17 at 19:58

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