Editing Tensor product (section)

== Tensor product of algebras ==
{{main|Tensor product of algebras}}
Let {{math|''R''}} be a commutative ring. The tensor product of {{math|''R''}}-modules applies, in particular, if {{math|''A''}} and {{math|''B''}} are [[Algebra (ring theory)|{{math|''R''}}-algebras]]. In this case, the tensor product <math>A \otimes_R B</math> is an {{math|''R''}}-algebra itself by putting:
<math display="block">(a_1 \otimes b_1) \cdot (a_2 \otimes b_2) = (a_1 \cdot a_2) \otimes (b_1 \cdot b_2).</math>
For example:
<math display="block">R[x] \otimes_R R[y] \cong R[x, y].</math>

A particular example is when {{math|''A''}} and {{math|''B''}} are fields containing a common subfield {{math|''R''}}. The [[tensor product of fields]] is closely related to [[Galois theory]]: if, say, {{math|1=''A'' = ''R''[''x''] / ''f''(''x'')}}, where {{math|''f''}} is some [[irreducible polynomial]] with coefficients in {{math|''R''}}, the tensor product can be calculated as:
<math display="block">A \otimes_R B \cong B[x] / f(x)</math>
where now {{math|''f''}} is interpreted as the same polynomial, but with its coefficients regarded as elements of {{math|''B''}}. In the larger field {{math|''B''}}, the polynomial may become reducible, which brings in Galois theory. For example, if {{math|1=''A'' = ''B''}} is a [[Galois extension]] of {{math|''R''}}, then:
<math display="block">A \otimes_R A \cong A[x] / f(x)</math>
is isomorphic (as an {{math|''A''}}-algebra) to the {{tmath|1= A^{\operatorname{deg}(f)} }}.