Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
Isomorphic as what sort of objects? – Chris Eagle Apr 22 '12 at 10:54
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. – marlu Apr 22 '12 at 10:58
@ChrisEagle as rings. – Sergey Filkin Apr 22 '12 at 11:09
up vote 24 down vote accepted

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.

$\hspace{6.5in} \square$

share|cite|improve this answer
thank you very much for a complete answer! Very enlightening. – Sergey Filkin Apr 22 '12 at 11:38
Nice answer, +1! – Rudy the Reindeer Apr 22 '12 at 11:54
@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 $\mathbb{C}[x,y]$. – Alex Youcis Apr 22 '12 at 22:02
@AlexYoucis Thanks for sharing your insight. It was not so obvious to me, which is why I just decided to prove the isomorphims anyway. – user38268 Apr 23 '12 at 3:29
@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! – user38268 Apr 23 '12 at 10:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.