Let \(R\) be the ring of all polynomials in \(\mathbb{Q}[x]\) having no \(x\)-term. Show that \(x^5\) and \(x^6\) have no GCD in \(R\).
If \(S\) is a Euclidean domain and \(T\) is a subring of \(S\), is it true that \(T\) is a Euclidean domain? Justify your answer.
Solution
Let \(R=\{f\in\QQ[x]: \text{the coefficient of } x \text{ in } f \text{ is } 0\}\). It is a subring of \(\QQ[x]\): for \(f=\sum f_ix^i\) and \(g=\sum g_ix^i\), the coefficient of \(x\) in \(fg\) is \(f_0g_1+f_1g_0=0\). Its units are the nonzero constants, since these are the units of \(\QQ[x]\) and they lie in \(R\).
1. For \(n\ge0\), an element \(d\in R\) divides \(x^n\) in \(R\) if and only if \(d=cx^k\) with \(c\in\QQ^\times\), \(0\le k\le n\), \(k\ne1\), and \(n-k\ne1\).
If \(de=x^n\) with \(d,e\in R\), unique factorization in \(\QQ[x]\) gives \(d=cx^k\) and \(e=c^{-1}x^{n-k}\) with \(c\in\QQ^\times\) and \(0\le k\le n\). A monomial \(x^j\) lies in \(R\) exactly when \(j\ne1\), so \(d,e\in R\) forces \(k\ne1\) and \(n-k\ne1\). Conversely, under these conditions \(e=c^{-1}x^{n-k}\in R\) and \(de=x^n\).
2. The common divisors of \(x^5\) and \(x^6\) in \(R\) are the elements \(cx^k\) with \(c\in\QQ^\times\) and \(k\in\{0,2,3\}\).
By step 1, the divisors of \(x^5\) are the \(cx^k\) with \(k\in\{0,2,3,5\}\), and the divisors of \(x^6\) are the \(cx^k\) with \(k\in\{0,2,3,4,6\}\). The common exponents are \(0,2,3\).
3. For part (a), \(x^5\) and \(x^6\) have no greatest common divisor in \(R\).
A greatest common divisor \(g\) would be a common divisor divisible in \(R\) by every common divisor, in particular by \(x^2\) and by \(x^3\). By step 2, \(g=cx^k\) with \(k\in\{0,2,3\}\), and \(x^3\mid g\) forces \(k=3\). Then \(x^2\mid cx^3\) in \(R\) would give \(cx^3=x^2e\) with \(e\in R\), so \(e=cx\notin R\). This contradiction shows that no such \(g\) exists.
4. For part (b), the answer is no: \(T=R\) is a subring of the Euclidean domain \(S=\QQ[x]\) that is not a Euclidean domain.
The ring \(\QQ[x]\) is Euclidean with the degree function. A Euclidean domain is a principal ideal domain [@DF04]. In a principal ideal domain, a generator \(d\) of the ideal \((a,b)\) is a greatest common divisor of \(a\) and \(b\): it divides \(a\) and \(b\), and writing \(d=ua+vb\) shows that every common divisor of \(a\) and \(b\) divides \(d\). By step 3, \(x^5\) and \(x^6\) have no greatest common divisor in \(R\), so \(R\) is not a principal ideal domain and hence not Euclidean.
5. Q.E.D.
Step 3 answers part (a) and step 4 answers part (b).