I am dealing with the following problem in commutative algebra.
Let $A\subset \mathbf{Q}[X]$ be the subring formed by all polynomials for which $f(0)=f(1)$.
$(1)\,\,$ Prove that the $\mathbf{Q}$-algebra $A$ is generated by the polynomials $X(X-1)$ and $X^2(X-1)$.
$(2)\,\,$ Prove that $A$ is Noetherian.
$A$ is clearly a subring and also a $\mathbf{Q}$-algebra, since it contains $\mathbf{Q}$ ($\iota:\mathbf{Q}\hookrightarrow A,x\mapsto x$).
For ($1$): It is clear that $X(X-1)$ and $X^2(X-1)$ are elements of $A$. A polynomial $a_0+a_1X+\ldots+a_nX^n$ is in $A$ if and only if $a_0=a_0+a_1+\ldots+a_n$ or equivalently $a_1+\ldots+a_n=0$. Hence, intuitively it is clear that we need $X^2(X-1)$ on top of $X(X-1)$, since there is no element of degree 1 in $A$. I don't see how to proceed, I thought of Euclidean division but this seems to be to brute force.
For ($2$): I know of some result which states that every finitely generated $R$-algebra ($R$ is a commutative ring) is some quotient of a polynomial ring $R[Y_1,\ldots,Y_n]$ on finitely many variables. This would also prove that $A$ is Noetherian by Hilbert's basis theorem and the theorem that the image of a Noetherian ring by a surjective ring morphism is Noetherian.
In addition, I would like to know how to write $A$ as a quotient of $\mathbf{Q}[U,V]$ by some ideal. Clearly we have the surjective ring morphism \begin{align*} \Phi:\mathbf{Q}[U,V] &\longrightarrow A \\ U &\longmapsto X(X-1) \\ V &\longmapsto X^2(X-1) \end{align*}but in order to use the first isomorphism theorem, we have to find the kernel.
Any help is appreciated.
(1) If $f\in A$, then $f(X)-f(0)$ vanishes at $X=0$ and at $X=1$. Then there is $g(X)=a_0+a_1X+\cdots+a_nX^n\in Q[X]$ such that $$f(X)=f(0)+X(X-1)g(X)$$
Divide $g(X)$ by $X(X-1)$, $g(X)=X(X-1)g_1(X)+aX+b$. So, $$X(X-1)g(X)=X^2(X-1)^2g_1(X)+X(X-1)(aX+b)$$
Now, $X(X-1)(aX+b)$ is in the algebra generated by $X(X-1)$ and $X^2(X-1)$ and $X(X-1)g_1(X)$ is $X(X-1)$ multiplied by a polynomial of smaller degree than $g$. Apply induction to show that $g_1$ is generated by $X(X-1)$ and $X^2(X-1)$. Note that (the base case) if $g_1$ has degree not greater than $1$ then it is by the same observation above.
Note that $A$ is the image of $Q[Y_1,Y_2]$ in $Q[X]$ by the homomorphism $\phi$ that sends $Y_1\mapsto X(X-1)$ and $Y_2\mapsto X^2(X-1)$. So, $A$ is isomorphic to $Q[Y_1,Y_2]/\ker(\phi)$. The kernel of $\phi$ are the polynomials $p(Y_1,Y_2)$ such that $p(X(X-1),X^2(X-1))=0$.
Since you added the question about the explicit form of the kernel and you were answered its form and how to confirm it, let me tell you how to find it.
Consider the ideal $(Y_1-X^2(X-1), Y_2-X^2(X-1))$ of $Q[X,Y_1,Y_2]$. Apply Buchberger's algorithm with a monomial order in which monomials containing $X$ are larger than monomials not containing $X$ to get the Groebner basis $$Y_1^3 + Y_1Y_2 - Y_2^2, X Y_2 - Y_1^2 - Y_2, XY_1 - Y_2, X^2 - X - Y_1$$ The elements of the basis not containing terms with $X$ generate the kernel. In this case $$Y_1^3 + Y_1Y_2 - Y_2^2$$ See also Implicitation of a rational curve.