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Question:
Let $k$ be a field and $R = k[x,y]$ and $\mathscr{m} = (x, y)$ and $n\in \mathbb{N}$.
$(a)$ Show that $\mathscr{m}^n$ cannot be generated by less than $n + 1$ elements.
$(b)$ Find a generating set of $\mathscr{m}^n$ containing $n + 1$ elements.

My Attempt:
$(a)$: (Proof by Induction:)
For $n=1$, It holds because $\mathscr{m}$ cannot be generated by less than two elements.

Induction Assumption:
Assume that the statement holds true for $n-1$. $\hspace{50pt}(*)$

Induction Step:
To Prove that the statement holds for $n$.

On Contrary, assume that $\mathscr{m}^n$ is generated by less than $n + 1$ elements.$\hspace{20pt}(**)$

Combining $(*)$ and $(**)$, we get that $\mathscr{m}^n$ is generated by $n$ elements.

In particular, this implies that $\mathscr{m}^2$ is generated by two $1$ element.
It is contradiction to the fact that $\mathscr{m}^2$ cannot be generated by less than three elements.

Hence, the statement holds for $n$, and the proof follows.


For $(b)$, I have no clue, how to solve.

Please check my $(a)$, and provide me a solution if it wrong, else please provide me another solution. Also, Please provide me hints for $(b)$.

Also, I am wondering what is the use of $k$ to be field?

I am referring to Dummit and Foote$(3^{ed})$.


Edit:
The italicized statements have been edited and added after the answer of ThorWittch

But I want a more elaborate answer for $(a)$ sticking to the reference, I am using.

Thanks.

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    $\begingroup$ Hint for (a): it's probably easiest to use the fact that $\mathfrak{m}^n / \mathfrak{m}^{n+1}$ is a vector field over $R / \mathfrak{m} \simeq k$. $\endgroup$ – Daniel Schepler Jun 27 at 17:26
  • $\begingroup$ @DanielSchepler I didn't get you. How does $\mathscr{m}^n/\mathscr{m}^{n+1}$ is a vector space? are we saying that $\mathscr{m}^{i+1} \subset \mathscr{m}^{i}$ $\forall 1\leq i < n$ ? $\endgroup$ – Kumar Jun 27 at 17:37
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    $\begingroup$ Correct, @Kumar. That is, indeed, what Daniel Schepler says. For example $\mathfrak{m}^2$ is generated by $x^2,xy,y^2$ (it consists of all the polynomials of degree two and higher). You should figure out the dimension of $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ over $k$ to get started. $\endgroup$ – Jyrki Lahtonen Jun 27 at 17:43
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    $\begingroup$ Yes @Kumar. For $n = 1$ that is for example used to define the cotangent space in algebraic and differential geometry at some point of a scheme or smooth manifold. $\endgroup$ – ThorWittich Jun 27 at 17:46
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    $\begingroup$ Correct, again. Have you covered Nakayama's lemma? Or, may be you can show directly that a generating set of $\mathfrak{m}^n$ (as an $R$-module) spans $\mathfrak{m}^n/\mathfrak{m}^{n+1}$ over $k$? $\endgroup$ – Jyrki Lahtonen Jun 27 at 19:56
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Your (a) is not clear at all to me.

Even for $n = 1$. The definition of $\mathcal{m}$ does not imply that the ideal is not principal (for example the ideal $(2,3) \subset \mathbb{Z}$ is principal, as it is generated by $1$), but you can get that with an easy argument.

Assume that $(f) = (x,y)$ for some polynomial $f \in k[x,y]$. Then we have $af = x$ and $bf = y$ for some $a,b \in k[x,y]$. This yields $\text{deg}(a) + \text{deg}(f) = 1$ and $\text{deg}(b) + \text{deg}(f) = 1$. Therefore either $a$ or $f$ has degree $0$. If $a$ has degree $0$, then $a \in k$ is just some non-zero scalar and we get $f = a^{-1}x$ and $\text{deg}(f) = 1$. Thus we have $\text{deg}(b) = 0$, such that $ba^{-1}x =y$ for some non-zero $a,b \in k$, which is a contradiction. If $\text{deg}(f) = 0$, then $(f) = k[x,y]$ as $f$ is a unit, which contradicts $(f) = (x,y)$. Therefore you need at least two generators for $\mathcal{m}$.

I also don't understand your argument for the rest.

(b)

For $n = 1$ we have the generators $x,y$.

For $n = 2$ we have the generators $x^2,xy,y^2$.

For $n = 3$ we have the generators $x^3,x^2y,xy^2,y^3$.

...

I guess we are working over a field in order for $\mathcal{m}$ to be a maximal ideal.

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  • $\begingroup$ If possible, can you please provide me, full solution for $(a)$? $\endgroup$ – Kumar Jun 27 at 17:29

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