# Is any subring of a Euclidean domain a unique factorization domain?

Let $R$ be a Euclidean domain. Then any subring $S$ of $R$ is a unique factorization domain. Is this statement true or not?

my idea: $R$ is a Euclidean domain implies that $R$ is a principal ideal domain. Hence $R$ is a unique factorization domain. Hence for any $r\in R$, there is a unique expression $r=u\cdot\Pi_{i=1}^n p_i$, where $u$ is a unit in $R$ and $p_i$ is irreducible in $R$. The expression is unique up to multiples of units in $R$ to each of the factor. This implies that for any $s\in S$, there is a unique expression $s=u\cdot\Pi_{i=1}^n p_i$, where $u$ is a unit in $R$ and $p_i$ irreducible in $S$. The expression is unique up to multiples of units in $R$ to each factor. But a unit in $R$ is not neccesarily a unit in $S$. What can I do?

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Any field is a Euclidean domain, any domain can be embedded in a field. – egreg Dec 27 '13 at 16:36
This question appears to be off-topic because the answer is clear – Phosphorus Dec 28 '13 at 7:51

The answer is no. Take $K[x]$ the ring of polynomials with coefficients in the field $K$. Let $R$ be the subring of $K[x]$ consisting of those polynomials with linear coefficient equal to $0$. Then consider the following two factorizations
$$x^6=x^3\cdot x^3 \textrm{ and } x^6=x^2\cdot x^2 \cdot x^2$$
You can prove that $x^3$ and $x^2$ are irreducible and distinct in $R$ so that you get two distinct factorizations of the same element of $R$, proving that $R$ is not UFD.
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