Finding all $\mathbb{C}$-algebra homomorphisms $\mathbb{C}[x^2, y^2]\to\mathbb{C}$ I am interested in calculating all $\mathbb{C}$-algebra homomorphisms from $\mathbb{C}[x^2, y^2]\to\mathbb{C}$. I understand that every $\mathbb{C}$-algebra homomorphism will fix $\mathbb{C}$ and send $x^2$ and $y^2$ somewhere in $\mathbb{C}$. But I don't exactly understand how do we "send" $x^2$ and $y^2$ to $\mathbb{C}$. Do we send $x$ and $y$ somewhere first? 
I know how to answer the exact same question for $\mathbb{C}[x,y]$. Namely, every $\mathbb{C}$-algebra homomorphism $\mathbb{C}[x,y]\to\mathbb{C}$ is determined by where the generators are sent. Since $x, y$ can freely go to any point in $\mathbb{C}$, we see that there is a one-to-one correspondence between the set of all $\mathbb{C}$-algebra homomorphisms $\mathbb{C}[x,y]\to\mathbb{C}$ and $\mathbb{C}^2$.
Could someone point me in the right direction?
 A: Edited to add: I forgot to directly answer the following question: "Do we send $x$ and $y$ somewhere first?" The answer is no: $x$ and $y$ aren't even in the ring $\mathbb{C}[x^2, y^2]$. By definition, this ring only has even powers of $x$ and $y$.
In your example of $\mathbb{C}[x,y]$, you mention "Since $x,y$ can freely go to any point in $\mathbb{C}$..." Why is this? It's because there are no nonzero relations (polynomials) that $x$ and $y$ satisfy. For $\mathbb{C}[x^2,y^2]$ perhaps this isn't as obvious. A common way to show this is by using dimension theory for rings, but perhaps one can prove it just using the definition.
Define the map $\varphi : \mathbb{C}[a,b] \rightarrow \mathbb{C}[x^2,y^2]$ by sending $a \rightarrow x^2$ and $b \rightarrow y^2$ which is a surjective homomorphism (show this if it isn't obvious!). Now we have to ask what the kernel is. It turns out to be $(0)$, and one can see this by noting that $\mathbb{C}[x^2, y^2]$ has dimension at least 2: $(0) \subset (x^2) \subset (x^2, y^2)$ being a chain of prime ideals. If the kernel was nonzero, then the dimension of $\mathbb{C}[a,b]/ \text{ker} \varphi \cong \mathbb{C}[x^2,y^2]$ would be less than 2, a contradiction. We have just shown
$$\mathbb{C}[a,b] \cong \mathbb{C}[x^2,y^2]$$
and from here the answer to your question should be easy.
The same technique as above works on other examples. Try $\mathbb{C}[x^2,xy,y]$ for fun. The chain of prime ideals in this case will be $(0) \subset (y) \subset (x^2,y)$.
