I'm doing problems in ring theory, and I am confused on a problem.

Let $k$ be a field of characteristic not equal to $2$ and $k[t]$ the polynomial ring over $k$. Show that $k[x,t]/(x^2 - t)$ is a flat $k[t]$-algebra. As a hint, it is said that this module is free of rank 2 (and free implies projective implies flat). So I am attempting to show freeness.

For the module to be free of rank 2, I need to find a free basis of 2 elements. Because we are working over a field characteristic not equal to $2$, the polynomials $x - \sqrt{t}$ and $x+\sqrt{t}$ are well-defined. But these are not polynomials in $k[x,t]$, so I am worried about this. I claim that their reductions mod $x^2 - t$ form the free basis (so $\overline{x+\sqrt{t}}$ and $\overline{x - \sqrt{t}}$).

Linear independence: If I have arbitrary $a(t),b(t) \in k[t]$ such that $\overline{a(x+\sqrt{t}) + b(x - \sqrt{t})} = \overline{0}$, then $x^2 - t$ divides $a(x + \sqrt{t}) + b(x - \sqrt{t})$. We need $a = b$, or else the sum will contain $\sqrt{t}$ terms that cannot be divided by $x^2 - t$. We also need $a = -b$, or else the sum will contain $x$ terms that cannot be divided by $x^2 - t$. Having $a = b$ and $a = -b$ simultaneously implies that $a = b = 0$.

Span: If $\overline{p(x,t)}$ is an arbitrary element of $A$, then I want to express $\overline{p(x,t)}$ as $\overline{a(x + \sqrt{t}) + b(x - \sqrt{t})}$ where $a$ and $b$ are uniquely chosen for $\overline{p(x,t)}$. I do not know where to begin here...but then, I may have just chosen a bad basis to begin with.

I appreciate all feedback. Thanks :-)

  • $\begingroup$ I am quite confused. In the quotient, $x = \sqrt{t}$, so I don't understand how $x - \sqrt{t}$ could be a basis element. Really $k[x,t]/(x^2 - t) \cong k[t][\sqrt{t}]$ under the map that sends $x$ to $\sqrt{t}$. I think a better choice for a basis would be $1, \sqrt{t}$. $\endgroup$ – Viktor Vaughn Nov 2 '16 at 0:53

You can work with $\sqrt t$, as $t$ is not a number, but an indeterminate.

It all relies on Euclidean division: for any polynomial $f(x,t)\in k[x,t]=k[t][x]\subset k(x)[t]$, there is a polynomial $q(x,t)$ and a polynomial $r(x,t)$ in $k[x,t]$ such that $$f(x,t)=q(x,t)(x^2-t)+r(x,t),\qquad\deg_x r(x,t)\le 1\enspace\text{or}\enspace r(x,t)=0.$$ Thus $f(x,t)\equiv r(x,t)\mod x^2-t$. As $\deg_x r(x,t)\le 1$, this means $f(x,t)$ is congruent to a polynomial $$a_0(t)+a_1(t)x,\quad a_o(t), a_1(t)\in k[t],$$ in other words, $\{1,x\}$ is a system of generators of the $k[t]$-module $k[x,t]/(x^2-t)$. This system of generators is free, because if $a_0(t)+a_1(t)x\equiv 0\mod x^2-1$, this means $x^2-t$ divides $a_0(t)+a_1(t)x$, which is impossible for degree reasons.

  • $\begingroup$ Thanks for this response. As I read this, it appears that you do not have to introduce any square root to show freeness. In the last line, you show linear independence. In the preceding argument you show spanning the $k[x,t]/(x^2 - t)$.Is the choice of $(a_0,a_1)$ unique, though? $\endgroup$ – user55912 Nov 2 '16 at 1:11
  • $\begingroup$ Since it' free, yes. Note the result is valid for any $R[x]/(p(x))$, where $R$ is a commutative ring (not necessarily an integral domain) and p(x) is a monic polynomial. $\endgroup$ – Bernard Nov 2 '16 at 1:17
  • $\begingroup$ Of course, but in the proof one would have to show uniqueness. That appears really easy to me now, though: $a_0(t) + a_1(t)x = b_0(t) + b_1(t)x$ would obviously imply that $a_0 = b_0$ and $a_1 = b_1$. $\endgroup$ – user55912 Nov 2 '16 at 1:20
  • $\begingroup$ It even suffices to show that $a_0(t)+a_1(t)x=0$ implies $a_0(t)=0$, $a_1(t)=0$. This results from $R[t]$ being a free $R$-module, with basis $\{1, x,x^2,\dots,x^n,\dots\}$. $\endgroup$ – Bernard Nov 2 '16 at 1:24

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