# Generating set for Polynomial rings

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.

• Hint for (a): it's probably easiest to use the fact that $\mathfrak{m}^n / \mathfrak{m}^{n+1}$ is a vector field over $R / \mathfrak{m} \simeq k$. – Daniel Schepler Jun 27 at 17:26
• @DanielSchepler I didn't get you. How does $\mathscr{m}^n/\mathscr{m}^{n+1}$ is a vector space? are we saying that $\mathscr{m}^{i+1} \subset \mathscr{m}^{i}$ $\forall 1\leq i < n$ ? – Kumar Jun 27 at 17:37
• 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. – Jyrki Lahtonen Jun 27 at 17:43
• 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. – ThorWittich Jun 27 at 17:46
• 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 $k$? – Jyrki Lahtonen Jun 27 at 19:56

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.

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