Prove that $\mathbb{C}[x,y] \ncong \mathbb{C}[x]\oplus\mathbb{C}[y]$

$\mathbb{C}[x,y]$ is the polynomial ring of two variables over $\mathbb{C}$. I guess that we can consider images of $xy$ and $x+y$, but can't complete my argument. Can you help please?

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    $\begingroup$ Isomorphic as what sort of objects? $\endgroup$ Apr 22, 2012 at 10:54
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    $\begingroup$ If you mean isomorphic as rings you can show that one of the rings has non-trivial zero divisors whereas the other one is an integral domain. $\endgroup$
    – marlu
    Apr 22, 2012 at 10:58
  • $\begingroup$ @ChrisEagle as rings. $\endgroup$ Apr 22, 2012 at 11:09

1 Answer 1


As rings they cannot be isomorphic. The left hand side is an integral domain because:

$\Bbb{C}$ is an integral domain, hence the polynomial ring over it $\Bbb{C}[x]$. Since this is an integral domain, the polynomial ring over it in $y$ is an integral domain, viz. $\big(\Bbb{C}[x]\big)[y] \cong \Bbb{C}[x,y]$ is an integral domain. But for the right hand side we have

$$(x,0) \cdot(0,y) =(0,0)$$

so it cannot be an integral domain.

However if you view $\Bbb{C}[x,y]$ as a $\Bbb{C}$ - module, then we have the following $\Bbb{C}$ - module isomorphism:

$$\Bbb{C}[x,y] \cong \Bbb{C}[x] \otimes_\Bbb{C} \Bbb{C}[y]$$

To see this, note that $\Bbb{C}[x,y]$ is a free $\Bbb{C}$ - module with basis $x^iy^j$. Now for the right hand side $\Bbb{C}[x]$ is a free $\Bbb{C}$ - module with basis $x^i$ and for the left $\Bbb{C}[y]$ is a free $\Bbb{C}$ - module as well with basis $y^j$. Therefore (exercise) their tensor product has basis $x^i \otimes y^j$.

You can now do it as an exercise to prove that $\Bbb{C}[x,y]$ is isomorphic to $\Bbb{C}[x] \otimes_\Bbb{C} \Bbb{C}[y]$ via the $\Bbb{C}$ - module isomorphism that sends $x^iy^j$ to the elementary tensor $x^i \otimes y^j$.

$\textbf{Edit:}$ In fact let me prove to you directly that we have such an isomorphism. Now consider the map

$$B : \Bbb{C}[x] \times \Bbb{C}[y] \longrightarrow \Bbb{C}[x,y]$$ that sends $(p(x),q(y))$ to $p(x)q(y)$. It is easily checked that $B$ is well defined and bilinear. Therefore by the universal property of the tensor product, there exists a unique $\Bbb{C}$ - module homomorphism

$$L : \Bbb{C}[x] \otimes_\Bbb{C} \Bbb{C}[y] \longrightarrow \Bbb{C}[x,y]$$ such that $B = L \circ \pi $ (in other words $B$ factors through the tensor product) and on elementary tensors $L(x^i \otimes y^j) = B(x^i,y^j) = x^iy^j$. As usual $\pi$ is the canonical projection from $\Bbb{C}[x] \times \Bbb{C}[y]$ to the tensor product that is not necessarily surjective. We only need to define $L$ on elementary tensors because we can just extend additively. Now it is easy to see that $L$ is surjective. To see that $L$ is injective, suppose wlog that we have an element

$$\sum_{i,j} p_i(x) \otimes q_j(y)$$

in the kernel of $L$. Then by using the additivity of $L$ and the fact that $L$ is completely determined by the action of $B$ on a pair $(x^i,y^j)$ it is easy to see that this means that $\sum_{i,j} p_i(x)q_j(y)$ in $\Bbb{C}[x,y]$ must be $0$. This means that fixing an $\bar{i}$ and $\bar{j}$ that the coefficients of $p_{\bar{i}}q_{\bar{j}}$ are all zero, since we have noted that $\Bbb{C}[x,y]$ is a free $\Bbb{C}$ - module with basis as stated.

Now to show that $\sum_{i,j} p_i(x) \otimes q_j(y) = 0$ in the tensor product, it suffices to show that fixing some $\bar{i}$ and $\bar{j}$ that $p_{\bar{i}} \otimes q_{\bar{j}} = 0$.

Write $p_{\bar{i}} = p_0 + p_1x + \ldots p_nx^n$ and $q_{\bar{j}} = q_0 + q_1y + \ldots q_mx^m$. Then

$$\begin{eqnarray*} p_{\bar{i}} \otimes q_{\bar{j}} &=& (p_0 + p_1x + \ldots p_nx^n) \otimes (q_0 + \ldots + q_my^m) \\ &=& p_0 \otimes q_0 + \ldots + p_nx^n \otimes q_my^m \\ &=& p_0q_0 (1 \otimes 1) + \ldots + p_nq_m (x^n \otimes y^m). \end{eqnarray*}$$

But then as noted before, $p_0q_0 = 0$, $(p_1q_0 + q_1p_0) = 0, \ldots, p_nq_m =0$ so that $p_{\bar{i}} \otimes q_{\bar{j}}$ is zero. Since $\bar{i}$ and $\bar{j}$ were arbitrary, it follows that $p_i \otimes q_j$ is zero for all $i,j$ that appear in the sum

$$\sum_{i,j} p_i(x) \otimes q_j(y)$$

so that the sum itself is zero. It follows that $\ker L =\{0\}$ proving injectivity. Hence $L$ is a $\Bbb{C}$ - module isomorphism.

$\hspace{6.5in} \square$

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    $\begingroup$ thank you very much for a complete answer! Very enlightening. $\endgroup$ Apr 22, 2012 at 11:38
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    $\begingroup$ Nice answer, +1! $\endgroup$ Apr 22, 2012 at 11:54
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    $\begingroup$ @BenjaminLim There is a perhaps different reason why the equality $\mathbb{C}[x]\otimes_\mathbb{C}\mathbb{C}[y]$ should be "obvious". In the category of all commutative $\mathbb{C}$-algebras one has that $\mathbb{C}[x]$ is the free object on one generator. Moreover, $\otimes_\mathbb{C}$ is the coproduct and thus (at least with the intuition from normal coproducts, that they are "additive on dimension") the object $\mathbb{C}[x]\otimes_\mathbb{C}\mathbb{C}[y]$ should be the free object on two generators--or $\mathbb{C}[x,y]$. $\endgroup$ Apr 22, 2012 at 22:02
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    $\begingroup$ @AlexYoucis Thanks for sharing your insight. It was not so obvious to me, which is why I just decided to prove the isomorphims anyway. $\endgroup$
    – user38268
    Apr 23, 2012 at 3:29
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    $\begingroup$ @AlexYoucis I often use results about extension of scalars even now in learning about localisation. Your blog post on extension of scalars has been very helpful to me! $\endgroup$
    – user38268
    Apr 23, 2012 at 10:19

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