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Is it true that the units in $\mathbb{Q}[x,y]$ are precisely the constant polynomials? I believe this is true for $\mathbb{Q}[x]$.

A.B
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    Yes, obviously. The degree of the product of polynomials in an integral domain is the sum of the degrees, so since $1$ has degree $0$, it can only be the product of constants. – Rushabh Mehta Apr 05 '20 at 19:06

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Hint: Try to prove: If a polynomial $$f=a_0+a_1x+\cdots+a_nx^n\in R[x]$$ is invertible, then $a_0$ is invertible and $a_i$ for $1\leq i\leq n$ is nilpotent.

Now suppose $\mathbb{Q}[x,y]$ as the ring $\mathbb{Q}[x][y]$.

Qurultay
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  • "idempotent" or "nilpotent"? And why one needs to show such a strong result when the question asks for much less? If one assumes that $R$ is an integral domain (as in the question!), then using the degree one concludes in one line that the invertible polynomials are only the invertible elements of $R$. – user26857 Apr 05 '20 at 19:29
  • @user26857 Thank you. Was a typo. Also, yes you are right. But I think it worth to think on invertibles in $R[x]$. – Qurultay Apr 05 '20 at 19:33
  • IMO a hint should point out to a proof for what the OP has asked, and only then could mention a stronger result whose proof isn't trivial at all. Btw, the general question is answered here: https://math.stackexchange.com/questions/19132 – user26857 Apr 05 '20 at 19:35
  • @user26857 Yes, you are right. – Qurultay Apr 05 '20 at 19:36