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Where $\mathbb{Z}[x]$ is the ring of polynomials in $x$ with integer coefficients. The book I am studying says the unity of this ring is $f(x) = 1$ so then if some $p \in \mathbb{Z}[x]$ is a unit, this then must mean that $p^{-1} \in \mathbb{Z}[x]$, where $p \cdot p^{-1} = 1$, correct?

Obviously then $f(x) = 1$ and $f(x) = -1$ are units, and I have seen elsewhere on the internet that people claim these are the only units of $\mathbb{Z}[x]$, but aren't simple one-term polynomials also units? That is, $\forall k \in \mathbb{Z}, f(x) = x^k$ is a unit because $(f(x) = x^k)^{-1} \Longleftrightarrow f(x) = x^{-k}$ and $x^k \cdot x^{-k} = 1$, no?

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    $\begingroup$ $x^{-k}$ is for positive $k$ is not an element of $\mathbb Z[x]$. $\endgroup$ – drhab Dec 28 '15 at 18:57
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$x^{-k}\notin \Bbb{Z}[x]$.

$x^{-k}$ is not a polynomial.

Definition- A polynomial in a single indeterminate over integers can be written in the form $$a_nx^n+\ldots+a_1x+a_0$$ where $a_i\in \Bbb{Z}$ and $n\in \Bbb{N} \cup 0$

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  • $\begingroup$ Ah ok, so then $f(x) = 1$ and $f(x) = -1$ must be the only units because $\forall a,b \in \Bbb{Z}$, the degree of $p \cdot q$ = sum of the degree of $p$ and the degree of $q$, but we need a polynomial of degree 0 to get $f(x) = 1$, correct? $\endgroup$ – rawa Dec 28 '15 at 19:04
  • $\begingroup$ Yes and also because only units of $\Bbb{Z}$ are $\pm1$ $\endgroup$ – Bhaskar Vashishth Dec 28 '15 at 19:09
  • $\begingroup$ If you put rational numbers instead of integers , then units are all non zero rationals $\endgroup$ – Bhaskar Vashishth Dec 28 '15 at 19:10
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The elements of $\mathbb{Z}[X]$ are of the form $\sum_{i=0}^n a_i X^i$ with $n \in \mathbb{N}$ and $a_0, \dotsc, a_n \in \mathbb{Z}$. So $X^{-k}$ is not an element of $\mathbb{Z}[X]$ for $k \geq 1$.

To understand the units in $\mathbb{Z}[X]$ notice that for all polynomials $p,q \in \mathbb{Z}[X]$ we have $\deg(p \cdot q) = \deg(p) + \deg(q)$ where $\deg$ denotes the usual degree, i.e. $\deg(\sum_{i=0}^n a_i X^i) = n$ if $a_n \neq 0$ and $\deg(0) = -\infty$.

Now suppose $p \in \mathbb{Z}[X]$ is a unit. Then there exists some $q \in \mathbb{Z}[X]$ with $pq = 1$. In particular $p, q \neq 0$, so $\deg(p), \deg(q) \geq 0$. Because $\deg(p) + \deg(q) = \deg(pq) = \deg(1) = 0$ it follows that $\deg(p) = \deg(q) = 0$. So $p, q \in \mathbb{Z}$ and $p$ is a unit in $\mathbb{Z}$. Because the only units in $\mathbb{Z}$ are $1$ and $-1$ it follows that $p = 1$ or $p = -1$.

(The same argument shows that for any (not necessarily commutative) domain $R$ the units in $R[X]$ coincide with the units in $R$.)

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More generally, the units of $D[x]$ are exactly the units of $D$, when $D$ is a domain.

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    $\begingroup$ A good exercise for the interested Reader is to exhibit additional units of $D[x]$ when there exist nilpotent elements of $D$. $\endgroup$ – hardmath Dec 28 '15 at 23:35

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