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I know the following is true but couldn't convince me

$\mathbb{R}[x]/(x^2+1)\otimes_{\mathbb{R}} {\mathbb {C}}\cong \mathbb {C}[x]/(x^2+1)$ as a $\mathbb{R}$ module.

I need some help. Thank you.

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    You can define an explicit homomorphism by sending $(x^n + (x^2 + 1))\otimes c$ to $cx^n + (x^2 + 1)$ and extending by linearity. Then show that it is iso. – 2017-01-30

2 Answers 2

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More generally, for any two commutative, unital rings $ R \subset S $, and ideal $ I $ of $ R[X] $, we have that

$$ R[X]/I \otimes_R S \cong S[X]/I' $$

as $ S $-algebras, where $ I' $ denotes the ideal generated in $ S[X] $ by $ I $. Indeed, the map

$$ R[X]/I \times S \to S[X]/I' : (r + I, s) \to rs + I' $$

is well-defined and $ R $-bilinear, thus it factors through the tensor product to give a surjective $ R $-module homomorphism

$$ f : R[X]/I \otimes_R S \to S[X]/I' $$

This homomorphism is an $ S $-module homomorphism when the tensor product is equipped with the obvious $ S $-module structure. The homomorphism $ g : S[X] \to R[X]/I \otimes_R S $ given by sending $ x \to x \otimes 1_S $ and extending by the universal property is a $ S $-algebra homomorphism with $ I' $ in its kernel: indeed, if $ p \in I $, then one directly checks that $ g(p) = 0 $. Since $ I' $ is generated by the elements of $ I $, it follows that $ I' \subset \ker g $; and thus $ g $ descends to the quotient to give a map $ g' : S[X]/I' \to R[X]/I \otimes_R S $.

Now, one checks directly that $ g' $ is an inverse to $ f $, thus $ f $ is an $ S $-module isomorphism. It is an algebra isomorphism, as we have

$$ f((r_1 \otimes s_1) (r_2 \otimes s_2)) = f(r_1 r_2 \otimes s_1 s_2) = r_1 r_2 s_1 s_2 = r_1 s_1 r_2 s_2 = f(r_1 \otimes s_1) f(r_2 \otimes s_2) $$

Now, take $ R = \mathbb R $, $ S = \mathbb C $, $ I = (X^2 + 1) $ to obtain the claim in the question.

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This is just a dimension statement since both sides are finite dimensional vector spaces (i.e. $\Bbb R$-modules). But the LHS is just $\Bbb C\otimes_{\Bbb R}\Bbb C$ which is the tensor product of two $2$-dimensional vector spaces, hence has dimension $4$. For the RHS, this is--by the Chinese Remainder Theorem:

$$\Bbb C[x]/(x^2+1)\cong\Bbb C[x]/(x+i)\oplus\Bbb C[x]/(x-i)\cong \Bbb C\oplus\Bbb C$$

which also manifestly has dimension $4$ over $\Bbb R$. Since the vector spaces have the same dimension over the base field they are isomorphic.

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    I understand your arguments..but i wanted arguments using flatness of $\mathbb{C} $ over $\mathbb{R}$ – 2017-01-30
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    Thank you for your short answer..one more question ..are they isomorphic as a ring ? – 2017-01-30
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    They are isomorphic as $ \mathbb C $-algebras per my answer, so yes, they are. – 2017-01-30
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    @user371231 yes indeed, you can note that the LHS is $\Bbb C\otimes_{\Bbb R}\Bbb C$ which has a ring structure where you mutliply on pure tensors as $(a\otimes b)\cdot (c\otimes d) = (ac\otimes bd)$ so just use the map $$\begin{cases} 1\otimes 1\mapsto (1, 1) \\ i\otimes 1 \mapsto (i,1) \\ 1\otimes i \mapsto (1,i) \\ i\otimes i \mapsto (i,i)\end{cases}$$ which generate each side as rings, and extend by linearity. – 2017-01-30