Intereting Posts

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For every infinite $S$, $|S|=|S\times S|$ implies the Axiom of choice

Let $A$ be the polynomial $K$-algebra $K[t_1, t_2]$. Then the algebra $A$ is not local, why? More generally, $K[t_1, t_2, \cdots, t_n]$ is not local for any $n\geq 1$, why?

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- Can you construct a field with 6 elements?
- If $\phi(g)=g^3$ is a homomorphism and $3 \nmid |G|$, $G$ is abelian.
- Computing $\operatorname{Tor}_1^R(R/I,R/J)$
- Can this quick way of showing that $K/(Y-X^2)\cong K$ be turned into a valid argument?

How about the fact that you can find two distinct maximal ideals of $K[t_1, t_2]$? Like $(t_1,t_2)$ and $(t_1 – 1, t_2 – 1)$, for instance. In general, $K[t_1, \dots, t_n]$ is not local because of the ideals

$$

(t_1, \dots, t_n) \quad \text{ and } (t_1 – 1, \dots, t_n – 1).

$$

Hope that helps,

- If $A$ is local, then there does not exist a non-unit $f$ such that $f-1$ is also a non-unit.
- If $A$ is local, then it does not have two maximal ideals.
- If $A$ is local, then there can be at most one homomorphism $A \to K$ that is the identity on $K \subseteq A$.
- …

If you can manage to show any one of these conclusions false, you’ve proven $A$ is not local.

Try constructing a homomorphism from $K[x_1, …, x_n]$ into $K$ whose kernel is $(x_1 – k_1, …, x_n -k_n).$ Since $K$ is a field, what can we say about this ideal?

Seems like the easiest way to look at all of them is this way: $t_1$ and $1-t_1$ aren’t units, so they generate proper ideals $(t_1)$ and $(1-t_1)$. If there were a unique maximal proper ideal containing both, it would have to contain $t_1+(1-t_1)=1$, but that would contradict the fact the ideal is supposed to be *proper*.

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