Prove that $R[x]$ is an integral domain if and only if $R$ is an integral domain. [duplicate]

Solution 1:

The first direction of the proof looks good - it gets the right idea that the leading coefficient of the product is the product of the leading coefficients of the factors - which of course is non-zero so long as neither factor was zero.

For the second direction, you just need to note that $R$ embeds into $R[x]$ as the constant polynomials. That is, for any $a,b\in R$, you can consider the polynomials $f(x)=a$ and $g(x)=b$. Since $R[x]$ is an integral domain, $f(x)g(x)\neq 0$, but of course, the left hand side is just $ab$ as a polynomial - implying that $ab$ is not zero in $R$ either. More generally, any subring of an integral domain is an integral domain by this reasoning.

Proof $\int_0^\infty \frac{\exp(-\sqrt{x^2+y^2})}{x^2+y^2}dx = \frac{\pi}{2}\left(\frac{1}{y} - K_0(y)L_{-1}(y) - K_1(y)L_{0}(y)\right)$

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