No isomorphism between $k[x,y]/(x^{2}-y^{5})$ and ring of polynomials in one indeterminate Let $k$ be an algebraically closed field. Why there is no isomorphism (as finitely generated $k$-algebras) between the ring $k[x,y]/(x^{2}-y^{5})$ and $k[t]$?
 A: Let $R = k[x,y]/(x^2-y^5)$. The $k$-homomorphisms $R \to k$ correspond to pairs $a,b \in k$ with $a^2=b^5$. The kernel is a maximal ideal $\mathfrak{m}_{a,b} \subseteq R$, to which we may associate the dimension of the vector space $T^*_{a,b} := \mathfrak{m}_{a,b} / \mathfrak{m}_{a,b}^2$ over $k$ (this is the so-called cotangent space at the rational point $(a,b)$). One computes that $T^*_{0,0}$ has $k$-basis $x,y$, thus is $2$-dimensional. However, every cotangent space of $k[t]$ is $1$-dimensional: For $a \in k$ we $T^*_a=(t-a)k[t]/(t-a)^2 k[t]$, which has basis $t-a$.

This is, in fact, a geometric proof, which can be applied in other situations, too. $\mathrm{Spec}(R)$ is a singular curve (it has a cusp in the point $(0,0)$), whereas $\mathrm{Spec}(k[t]) = \mathbb{A}^1$ is just the affine line, which is a regular curve.
A: As $k[t]$ is a principal ideal domain, it is integrally closed in its field of fractions. However, $k[x,y]/(x^2-y^5)$ is not: the fraction $x/y$ is not in $k[x,y]/(x^2-y^5)$ but is a zero of the monic polynomial $f(s)=s^2 -y^3$.
A: We can take advantage that the ring $k[t]$ is a UFD. It seems to me that we don't
require $k$ to be algebraically closed, so I shall drop that assumption.
Assume contrariwise that there exists an isomorphism $f:k[x,y]/(x^2-y^5)\to k[t]$. Let
$P(t)=f(x)$ and $Q(t)=f(y)$ be the respective images of $x$ and $y$. Using the factorization we can write
$$
P(t)=a\prod_ip_i(t)^{n_i}\qquad\text{and}\qquad Q(t)=b\prod_jq_j(t)^{m_j},
$$
for some constants $a,b\in k$ and some irreducible monic polynomials $p_i,q_j\in k[t]$.
We must have $P(t)^2=f(x^2)=f(y^5)=Q(t)^5$, so by the uniqueness of factorization we get (after the usual renumbering) $a^2=b^5$, $p_i(t)=q_i(t)$, and $2n_i=5m_i$ for all $i$.
So all the exponents $n_i$ are divisible by five, and all the exponents $m_i$ are even.
Write $u=\prod_ip_i(t)^{n_i/5}\in k[t].$ We have
$$
P=a\cdot u^5,\qquad Q=b\cdot u^2,\qquad\text{and}\qquad a^2=b^5.
$$
We see that the image of $f$ is the subring $R=k[u^2,u^5]$ of $k[t]$. Clearly 
$$R=\bigoplus\sum_{i\in\mathbf{N},i\notin\{1,3\}}k\cdot u^i$$ is a proper subring of $k[u]$.
As $k[u]$ is a subring of $k[t]$ (these two rings could possibly be equal), we see that $f$ cannot be surjective.
A: Below is a very simple proof accessible at high-school level.
Theorem $\ $ If $\rm\: a^2 = b^5\:$ in  UFD $\rm\:R[x]\:$ then $\rm\:R[a,b]\:$ is a proper subring of $\rm\:R[x]\:$
Proof $\ $ Clear if $\rm\:a,b\in R.\:$ Else $\rm\:a\:$ or $\rm\:b\not\in R,\:$ so one is a nonunit, so has a prime factor $\rm\:p\not\in R.\:$ By $\rm\:a^2 = b^5,\:$ prime $\rm\:p\ |\ a\!\iff\! p\ |\ b,\:$ so $\rm\:p\ |\ a,b.\:$ Comparing powers of $\rm\:p\:$ in $\rm\:a^2 = b^5\:$ shows $\rm\:p^2\:|\ a,b.\:$ Thus $\rm\:p\not\in R[a,b]\:$ else $\rm\:p = r + a f + b g,\:$  $\rm r\in R\:$ so $\rm\:p^2\:|\:a,b\:$ $\Rightarrow$ $\rm\:p\ |\ r\:$ $\Rightarrow$ $\rm\:r=0\:$ $\Rightarrow$ $\rm\:p^2\ |\ p.$
A: The ring $k[T]$ is a PID, hence integrally closed, whereas  $A=k[x,y]/(x^2-y^5)=k[\xi, \eta]$ is not, so these rings are not isomorphic.  
To see that $A$ is not integrally closed, just observe that the element $\xi/ \eta^2\in Frac(A)$ is not in $A$ and nevertheless is integral over $A$ since it is a zero of the monic polynomial $X^2-y\in A[X]$.  
Generalization
Not only is the hypothesis "$k$ algebraically closed" irrelevant, but the exact same proof works if you only assume that $k$ is a  UFD rather than a field, since Gauss tells you that $k[T]$ is then also a UFD and UFD's are integrally closed. 
A: This is a minor (and nitpicking) complement to Georges and Sebastian's answers. (I up-voted the question and all answers.) I would state things as follows. Let me change the notation to 
$$
K[X,Y],\quad K[x,y]:=K[X,Y]/(X^2-Y^5).
$$
Assume by contradiction that $K[x,y]$ is a UFD. 
The square of $x/y$ being $y^3$, we see that $y$ divides $x$. 
Any element of $K[x,y]$ can be written in a unique way as $p(y)+x\,q(y)$ with $p,q\in K[Y]$. Thus we have 
$$
x=y\ (p(y)+x\ q(y))=y\ p(y)+ x\ (y\ q(y)),
$$
which implies $p(Y)=0$ and $Yq(Y)=1$, contradiction.
