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Find all units of $S$, where $S$ is the set of polynomials in $\mathbb{Q}[x]$ whose coefficient of $x$ is $0$. I think the units are $\mathbb{Q} \setminus \{0\}$. Is that correct?

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What is $Q$? Do you mean the rational numbers, $\mathbb{Q}$? –  Brandon Carter Apr 4 '12 at 17:16
    
Q means the set of rational numbers –  max Apr 4 '12 at 17:27
    
@TheChaz: That was an edit made after Brandon's comment. –  joriki Apr 4 '12 at 17:29
    
"units of $S$, where $S$ is the set of ..." is a somewhat unusual use of the terminology -- usually in "units of $X$", $X$ is a ring; otherwise one would say something like "the units of $\mathbb Q[x]$ that lie in $S$". But it so happens that $S$ is indeed a ring, so you could say "Find all units of $S$, where $S$ is the ring of ...". –  joriki Apr 4 '12 at 17:32
    
so, what are the units then? –  max Apr 4 '12 at 17:38
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2 Answers 2

We have $R=\mathbb{Q}[x]$ and $S=\mathbb{Q}[x^2,x^3]$. Since $U(R)=\mathbb{Q}\setminus\{0\}\subseteq S$, it follows that $U(S)=\mathbb{Q}\setminus\{0\}$.

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Hint $\ $ Let $\rm\:T = \mathbb Q[x].\:$ By $\rm\:U(T) = U(\mathbb Q)\:$ and unit inheritance $\rm\:R\subset R'\:\Rightarrow\:U(R)\subset U(R')\:$ follows

$$\rm\: \mathbb Q\subset S\subset T\ \ \Rightarrow\ \ U(\mathbb Q) \subset U(S) \subset U(T)\subset U(\mathbb Q)\:\ \Rightarrow\ \ U(\mathbb Q) = U(S) = U(T) $$

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