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Pick out the integral domains from the following list of rings

a. $\{a+b\sqrt{5} : a,b\in\mathbb{Q}\}$

b. The rings of Continuous functions from $[0,1]$ into $\mathbb{R}$

c. The ring of complex analytic functions on the disc $\{z\in\mathbb{C} : |z|<1\}$

d. The polynomial ring $\mathbb{Z}[x]$

My attempt: options a,d is correct but not sure about options c,b

Halima.Khatun
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2 Answers2

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Hints:

a. This ring is $\mathbb{Q}(\sqrt{5})$. If you want an elementary approach, assume that $(a+b\sqrt{5})(c+d\sqrt{5})=(ac+5bd)+(ad+bc)\sqrt{5}=0$. Then it must be that $ad+bc=0$. Consider a few cases:

  • $a=0$, then $b=0$ or $c=0$. If $b=0$, then $a+b\sqrt{5}=0$. If $c=0$, then $5bd=0$ so one of $b$ or $d$ is zero and so one of the factors is zero. A similar situation happens if $c=0$.

  • $a\not=0$ and $c\not=0$, then $d=-\frac{bc}{a}$. Then, $c+d\sqrt{5}=c-\frac{bc}{a}\sqrt{5}=\frac{c}{a}(a-b\sqrt{5})$. Then, you can write down the product and see that it is never $0$.

b. If you use the pointwise multiplication operation, then think about a continuous function $f$ which is $0$ on $[0,\frac{2}{3}]$ and nonzero on $(\frac{2}{3},1]$ and a continuous function $g$ which is $0$ on $[\frac{1}{3},1]$ and is nonzero on $[0,\frac{1}{3})$.

c. How many zeros can a nonzero complex analytic function have on a disk? The unit disk is compact. What happens if the zeros have an accumulation point?

d. Observe that for the product of polynomials, degrees add. Therefore, the only way for this not to be a domain is if a product of two integers is $0$, but $\mathbb{Z}$ is a domain.

Michael Burr
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  • I understood the explanations in (a),(b) and (d). (c) I know the fact that zeros of a complex having accumulation point $\implies$ $f$ is identically zero.. How do I use the fact here to prove that statement (c) is true? Please explain sir. –  Oct 18 '17 at 04:22
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    If $f$ and $g$ are nonzero, then they have finitely many zeros in the disk, otherwise there would be an accumulation point of zeros by compactness. The product of $f$ and $g$ also has finitely many zeros (the union of the zero sets). – Michael Burr Oct 18 '17 at 09:34
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Regarding c, suppose $f(z)g(z)=0$ for analytic functions $f$ and $g$. Suppose $f(z)\not\equiv 0$. The set of zeros of $f(z)$ is a closed set. So, the complement of the set $S$ of zeros of $f(z)$ is a non-empty open set and so contains a closed disk and $g(z)$ has to be zero on this closed disk. So, $g(z)$ has to be zero by principle of analytic continuation which says that if two analytic functions defined in a region agree on a set with a limit point, the functions are equal. Note that, the connectedness of $\vert z\vert <1$ plays an important role. It allows us to use principle of analytical continuation. If we take the union of two disjoint disks, the ring of analytic functions will not be an integral domain.

S. Venkataraman
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  • I didn't understand the logic at the end part. If the complement of $S$ (the zero set of $f$) is empty, then $f$ is identically zero, which is not by your assumption. How the "connectedness" is coming into picture here? – Krish Sep 01 '17 at 15:44
  • You are right. Connectedness comes into the picture in the application of principle of analytic continuation. I have changed the answer accordingly. – S. Venkataraman Sep 01 '17 at 16:07