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Zbl 1010.13004
Glasby, S.P.
On the tensor product of polynomials over a ring. (English)
[J] J. Aust. Math. Soc. 71, No.3, 307-324 (2001). ISSN 1446-7887; ISSN 1446-8107

Let $a_0,\dots, a_m,b_0, \dots,b_n$ be commuting indeterminants and let $\bbfZ_{m,n}= \bbfZ[a_0, \dots,b_n]$. Then $\bbfZ_{m,n}[X]$ contains the `generic' polynomials $a=a_0+ \cdots+ a_mX^m$ and $b=b_0+ \cdots+ b_nX^n$. Their tensor product is given by $a\otimes b=a^n_m b^m_n \prod_{i,j}(X-\alpha_i \beta_j)$ where $\alpha_1, \dots, \alpha_m$, $\beta_1, \dots, \beta_n$ are the roots of $a,b$ in their splitting field over the field of fractions of $\bbfZ_{m,n}$. An obvious specialization argument gives the definition of the tensor product of polynomials over any commutative ring. Writing $c(A)$ for the characteristic polynomial of a matrix $A$, we have $c(A\otimes B)= c(A)\otimes c(B)$.\par Let $R$ be an integral domain and let $R[X]^*$ denote the set of non-zero polynomials over $R$. It is shown that $R[X]^*$ is a Witt-type semi-ring, with `addition' given by multiplication, and `multiplication' by the tensor product. There is a canonical semi-ring homomorphism $\rho$ of $R[X]^*$, with $\rho^3= \rho$, and $\rho^2=\text {id}$ on the sub-semi-ring of polynomials with non-zero constant term.\par An ingeneous argument with cyclotomic polynomials shows that $a\otimes b$ is irreducible over $\bbfZ_{m,n}$ provided $m$, $n\ge 1$. In the case $a=X^m-a_0$, $b=x^n-b_0$, there is a discussion of the factorization of $a\otimes b$ over an arbitrary field.
[M.E.Keating (London)]

MSC 2000:: *13B25 Polynomials over commutative rings
13K05 Witt vectors and related rings
16Y60 Semirings
13P05 Polynomials, factorization

Keywords: tensor product of geneic polynomials; integral domain; Witt-type semi-ring

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