What elements are in $\mathbb{Z}_p[x]$ that are not in $\mathbb{Z}_p(x)$ . I am having trouble understanding why is the former an integral domain while the latter is a field.
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It's the other way around: for any field $F$, $F(x)$ is "bigger" than $F[x]$. – Noah Schweber Dec 01 '21 at 17:25
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What does $\mathbb Z_p(x)$ represent? – ilaK Dec 01 '21 at 17:25
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There are no such elements. This does not contradict your second statement, the former is simply an integral domain contained in the latter (up to a canonical identification, depending on your definitions). – Douglas Molin Dec 01 '21 at 17:25
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2@ilaK The field of fractions of $\mathbb{Z}_p[x]$. The quotients of polynomials with coefficients in $\mathbb{Z}_p$. – plop Dec 01 '21 at 17:27
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TI99, what have you tried? Is $1/x$ a polynomial? – Dietrich Burde Dec 01 '21 at 17:29
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The relation $F[x]$ to $F(x)$ is the same as $\mathbb Z$ to $\mathbb Q.$ Every element of $F[x]$ is in $F(x),$ but some elements of $F(x)$ are not in $F[x].$ – Thomas Andrews Dec 01 '21 at 17:31
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Ok thank you everyone for the help & clarification! I had defined F[x] wrongly and had thought F(x) was contained in F[x]. I understand my mistake now! – TI99 Dec 01 '21 at 17:39
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$\mathbb Z_p[x]$ is the field of polynomials with coefficients in $\mathbb Z_p$. For example,
$$ 3x^7+2x+1\in \mathbb Z_4[x].$$
On the other hand, $\mathbb Z_p(x)$ is the field of fractions. You can think of it kind of like a fraction with numerator and denominator both in $\mathbb Z_p[x]$. So if $f(x), g(x) \in \mathbb Z_p[x]$, and $g(x)\neq 0$, then $\frac {f(x)}{g(x)}\in \mathbb Z_p(x)$. This is not exactly the case, but you can think of it like this for now.
As you may know, every field is an integral domain, and you can think of a field as an integral domain with inverses. So, the reason $\mathbb Z_p(x)$ is a field is because we now have inverses for every non-zero element. The inverse of any non-zero element $\frac {f(x)}{g(x)}\in \mathbb Z_p(x)$, i.e $f(x) \neq 0$, is simply $\frac {g(x)}{f(x)}$.
You can use the actual definition of $\mathbb Z_p(x)$ to make this argument more rigorous.
ilaK
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