Editing Tensor product of fields (section)

==Examples==

To give an explicit example consider the fields <math>K=\mathbb{Q}[x]/(x^2-2)</math> and <math>L=\mathbb{Q}[y]/(y^2-2)</math>. Clearly <math>\mathbb{Q}(\sqrt{2})\cong K\cong L\neq K</math> are isomorphic but technically unequal fields with their (set theoretic) intersection being the prime field <math>N=\mathbb{Q}</math>. Their tensor product

: <math>K\otimes L=K\otimes_{N} L=\mathbb{Q}[x]/(x^2-2)\otimes_{\mathbb{Q}}\mathbb{Q}[y]/(y^2-2) \cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2})\otimes_{\mathbb{Q}}\mathbb{Q}(\sqrt{2})</math> 

is not a field, but a 4-dimensional <math>\mathbb{Q}</math>-algebra. Furthermore this algebra is isomorphic to a direct sum of fields

: <math>K\otimes L\cong \mathbb{Q}(\sqrt{2})[z]/(z^2-2)\cong \mathbb{Q}(\sqrt{2})\oplus\mathbb{Q}(\sqrt{2})</math> 

via the map induced by <math>1\mapsto (1,1), z\mapsto (\sqrt{2},-\sqrt{2})</math>.
Morally <math>\tilde{N}=\mathbb{Q}(\sqrt{2})</math> should be considered the ''largest common subfield up to isomorphism'' of ''K'' and ''L'' via the isomorphisms <math>\tilde{N}=\mathbb{Q}(\sqrt{2})\cong K\cong L</math>. When one performs the tensor product over this better candidate for the largest common subfield we actually get a (rather trivial) field

: <math>K\otimes_{\tilde{N}} L=\mathbb{Q}[x]/(x^2-2)\otimes_{\mathbb{Q}(\sqrt{2})}\mathbb{Q}[y]/(y^2-2) \cong \mathbb{Q}(\sqrt{2})=\tilde{N}\cong K\cong L</math>.

For another example, if ''K'' is generated over <math>\mathbb{Q}</math> by the [[cube root]] of 2, then <math>K \otimes_{\mathbb Q} K</math> is the sum of (a copy of) ''K'', and a [[splitting field]] of

:''X''<sup>3</sup> − 2,

of degree 6 over <math>\mathbb{Q}</math>. One can prove this by calculating the dimension of the tensor product over <math>\mathbb{Q}</math> as 9, and observing that the splitting field does contain two (indeed three) copies of ''K'', and is the compositum of two of them. That incidentally shows that ''R'' = {0} in this case.

An example leading to a non-zero nilpotent: let

:''P''(''X'') = ''X''<sup>''p''</sup> − ''T''

with ''K'' the field of [[rational function]]s in the indeterminate ''T'' over the [[finite field]] with ''p'' elements (see [[Separable polynomial]]: the point here is that ''P'' is ''not'' separable). If ''L'' is the field extension ''K''(''T''<sup>&thinsp;1/''p''</sup>) (the [[splitting field]] of ''P'') then ''L''/''K'' is an example of a [[purely inseparable field extension]]. In <math>L \otimes_K L</math> the element

:<math>T^{1/p}\otimes1-1\otimes T^{1/p}</math>

is nilpotent: by taking its ''p''th power one gets 0 by using ''K''-linearity.