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I was a bit confused with what $F(a, b, y, \dotsc)$ means. The book assumes that $F$ is a subfield of $E$.

If $a, b, y, \dotsc$ are elements of $E$, then by $F(a, b, y, \dotsc)$ we shall mean the set of elements in $E$ which can be expressed as quotients of polynomials in $a, b, y, \dotsc$ with coefficients in $F$. It is clear that $F(a, b, y,\dotsc)$ is a field and is the smallest extension of $F$ which contains the elements $a, b, y, \dotsc$ . We shall call $F(a, b, y, \dotsc)$ the field obtained after the adjunction of the elements $a, b, y, \dotsc$ to $F$, or the field generated out of $F$ by the elements $a, b, y, \dotsc$ .

So are the elements $a, b, y, \dotsc$ polynomials? Also I'm not sure what "quotients of polynomials in $a, b, y, \dotsc$" means. What do they mean by "in"?

Thanks in advance.

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  • $\begingroup$ I have tried to improve the readability of your question by introducing Tex and a blockquote. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. $\endgroup$ – A.P. Mar 31 '13 at 11:27
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It means that $x \in F(a_1, a_2, \ldots)$ iff for some $n$ there are polynomials $P, Q \in F[x_1, x_2, \ldots, x_n]$ with coefficients in $F$, such that $x = \frac{P(a_1, a_2, \ldots, a_n)}{Q(a_1, a_2, \ldots, a_n)}$.

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    $\begingroup$ $P(a_1, \ldots, a_n)$ is a value of a polynomial $P \in F[x_1, \ldots, x_n]$ for argument $(a_1, \ldots, a_n)$. For instance, $\frac{1}{1+\sqrt{2}} \in \mathbb{Q}[\sqrt{2}]$, because for $P(x) = 1, R(x) = 1 + x$, we have $\frac{1}{1+\sqrt{2}} = \frac{P(\sqrt{2})}{R(\sqrt{2})}$ $\endgroup$ – xyzzyz Mar 31 '13 at 10:36
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    $\begingroup$ @Artus The book omits the essential condition that the denominator polynomial must evaluate nonzero (else, e.g. as in prior comment with $\,R(x) = x^2\!-\!2\,$ we deduce $\,a = 1/0\:\Rightarrow\:0 = 0\cdot a = 1\:$ in the field, contra field axioms). $\endgroup$ – Math Gems Mar 31 '13 at 17:40
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    $\begingroup$ @MathGems: this condition is not necessary. No element is of the form $\frac{P(a_1, \ldots, a_n)}{Q(a_1, \ldots, a_n)}$ where $Q(a_1, \ldots, a_n) = 0$, because this expression does not make sense. $\endgroup$ – xyzzyz Mar 31 '13 at 19:39
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    $\begingroup$ @xyzzyz Saying it "does not make sense" is simply an informal way of saying what I said. In any case, to be rigorous, any definition needs to imply this, one way or another. The point is that while $\rm\:F[a,b,c]\:$ is the image of the polynomial ring $\rm\:F[x,y,z]\:$ evaluated at $\rm\:(x,y,z)=(a,b,c)\:$ this does not generalize to rational "function" fields $\rm\,F(x,y,z)\,$ because the denominator may evaluate to $0.\ $ $\endgroup$ – Math Gems Mar 31 '13 at 19:59
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    $\begingroup$ @MathGems: sure, but the quoted definition just does not require to mention this. It all depends on wording. $\endgroup$ – xyzzyz Mar 31 '13 at 20:37

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