In general I'm having a hard time understanding what exactly $\mathbb{Q}(x)$ is.... i.e. () circular brackets.
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7$\mathbb Q[x]$ consists of polynomials with rational coefficients, whereas $\mathbb Q(x)$ consists of rational functions (quotients of two polynomials with rational coefficicents). – Nishant Jan 09 '15 at 18:54
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or in the formalism of algebraic geometry ${\bf{Q}}(x)$ is the function field of a straight line (defined by all rational functions $f(x)/g(x),g\not=0$ with coefficients from rationals), and $\bf{Q}[x]$ the coordinate ring. – 111 Jan 10 '15 at 02:15
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This notation is used when $Q \subseteq K$, where $K$ is some field and $x \in K$. Then $Q[x] = \{f(x) : f \text{ is a polynomial over } Q \}$, while $Q(x) = \{f(x) / g(x) : f,g \text{ are a polynomials over } Q \text{ and } g(x) \neq 0 \}$. If $x$ is algebraic over $Q$ then they are the same i.e. $Q[x] = Q(x)$. That's why you may be confused...
brick
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$\mathbf{Q}(x)$ is just the fraction field of the polynomial ring $\mathbf{Q}[x]$ over $\mathbf{Q}$.
Mister Benjamin Dover
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$\mathbb{Q}[x]$ denotes the intersection of all $\textbf {rings}$ containing $\mathbb{Q}$ and $x$, while $\mathbb{Q}(x)$ denotes the intersection of all $\textbf{fields}$ containing $\mathbb{Q}$ and $x$.
If $x$ is algebraic over $\mathbb{Q}$, you can prove that $\mathbb{Q}[x]=\mathbb{Q}(x)$.
Marco Flores
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4That doesn't properly explain the case of rings of polynomials and rational functions, i.e. when $x$ is an "indeterminate". – Bill Dubuque Jan 09 '15 at 20:00
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Well, but I think sometimes it is very convenient to think of the "indeterminate" $x$ as an element in some field extension. – Marco Flores Jan 09 '15 at 20:14
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1Right, but how do you obtain an appropriate such extension ring? You cannot define/construct a polynomial ring like that. – Bill Dubuque Jan 09 '15 at 20:53
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Do you mean I should explain the algebraic properties of $\mathbb{Q}[x]$ in addition to constructing it as a set? – Marco Flores Jan 09 '15 at 21:13
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How do you propose to define or construct the polynomial $,\Bbb Q[x],$ based on your answer? – Bill Dubuque Jan 09 '15 at 23:35