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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 $k[t]$?

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up vote 10 down vote accepted

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 $X^2-y\in A[X]$.

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.

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Or indeed $k$ is just some integrally closed commutative ring, not even a UFD :-) For example $k$ could be the ring of integers of some number field, and it is then allowed to have class number not equal to 1. – Alex B. Mar 28 '12 at 8:30
Dear @Alex, you are absolutely right and funnily enough that is the degree of generality I wanted to evoke in my post. However I would then have had to state that if $k$ is integrally closed, then so is $k[T]$. This is true (and proved for example in Lemma 7.31.8 of De Jong and collaborators' online book Stacks Project ) but I don't know an elementary reference, so I downgraded to UFD's. Do you know an elementary reference? – Georges Elencwajg Mar 28 '12 at 9:13
Unfortunately not. Indeed, I don't think I knew any reference until now, I just had this fact stored somewhere in my mind as "folklore wisdom", not having any idea where I had it from. So thank you! – Alex B. Mar 28 '12 at 9:17
Dear @Alex, then we'll just let interested readers check the reference to the online book I mentioned above. By the way, I think we all have some such "black boxes" stored somewhere in our brains: life is short and mathematics is vast... – Georges Elencwajg Mar 28 '12 at 9:23

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 $\rm\:p^2\:|\:a,b\:$ $\Rightarrow$ $\rm\:p\ |\ r\:$ $\Rightarrow$ $\rm\:r=0\:$ $\Rightarrow$ $\rm\:p^2\ |\ p.$

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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.

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I don't understand "we see that $y$ divides $x$". – Martin Brandenburg Mar 28 '12 at 10:12
Dear @Martin: Assume that $K[x,y]$ is a UFD, write $x/y=ua_1^{n_1}\cdots a_k^{k_1}$, with $u$ a unit, the $a_i$ non-associated irreducibles, $n_i\in\mathbb Z$, plug this into $(x/y)^2=y^3$, and note that $n_i\ge0$ for all $i$. – Pierre-Yves Gaillard Mar 28 '12 at 10:38
Alright this is just UFD => normal. – Martin Brandenburg Mar 29 '12 at 9:11

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 $f(s)=s^2 -y^3$.

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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 $t-a$.

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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.

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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.

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+1: This is a very nice proof since it is completely elementary in a logical sense, albeit complicated. By contrast it illustrates how introducing more sophisticated concepts like smoothness and integrality simplifies proofs. There is an interesting moral for beginners here: abstraction is for your good and not only for the pleasure of sadistic professors . – Georges Elencwajg Mar 28 '12 at 8:40
Everything that you say is very true, @Georges :-) OTOH I also enjoy the kind of exercises, where elementary methods are taken to their limits. After all, when (or, "if") we are pushing the limits of our understanding further out, we also need to know how to milk every last ounce out of the tools we have. Granted, the first year calculus students don't necessarily appreciate the kind of exercise, where they have to prove that $f(x)=x^3$ is an increasing function WITHOUT using the derivative. – Jyrki Lahtonen Mar 28 '12 at 9:21

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