# A problem on which of the following rings are integral domains?

Which of the following rings are integral domains?
(a) $\mathbb{R}[x]$, the ring of all polynomials in one variable with real coefficients.
(b) $M_n(\mathbb{R})$.
(c) the ring of complex analytic functions defined on the unit disc of the complex plane (with pointwise addition and multiplication as the ring operations).

Only (a) & (c) are correct. Am I right?

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Yes, you are right. –  user26857 Jan 19 '13 at 16:33

## 1 Answer

So for (a) is is true that whenever $D$ is an integral domain, $D[x]$ is also an integral domain. To see this, when you multiply two polynomials, look at the leading term of the product and observe that it has to be nonzero if you started with two nonzero polynomials.

On the other hand, (b) only holds for $n>1$, because $M_1(\mathbb R)$ is isomorphic to $\mathbb R$. For $n>1$, you can just do something silly like multiply a matrix with a $1$ in the upper left hand corner by a matrix with a $1$ in the bottom right hand corner and $0$'s elsewhere. Notice this breaks down for $n=1$.

(c) Follows because nonzero complex analytic functions have isolated zeros. Thus if you take two nonzero complex analytic functions on the unit disk, their zero sets are both finite in the disk, so there's a point in the disk where they both don't vanish, and their product won't vanish there either.

edit: For (c), to avoid the annoyance of zeroes accumulating at the boundary, we can just say that away from the boundary, for example, in the circle of radius $\frac{1}{2}$, this function definitely only has finitely many zeros, and then we can conclude.

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I think a complex analytic function can have infinitely many zeros in the disk, it is just that they only accumulate at the boundary (near an essential singularity). Maybe it depends on whether we mean the open or closed disk, or what one means by analytic on the boundary. –  Jack Schmidt Jun 15 '13 at 19:55
I was assuming that OP meant the closed unit disk and that it was analytic on the boundary. We can still salvage this, however. –  Dylan Yott Jun 16 '13 at 0:22
Yup, looks good. –  Jack Schmidt Jun 16 '13 at 4:24