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Lets say you have an ideal in some algebra of characteristic p.

Yeah, so if you have a lie algebra with a field that is characteristic p. Can you cancel.

So for example if you have a vector space with $\{v_0,v_1,v_3\}$ and have say $2v_0 \in I$ where I is an ideal, then can you deduce that $v_0 \in I$?

As I'm stuck on a problem and I'm sort of assuming that if you have an element that if you have say $k v_0 \in I$, then $v_0 \in I$ if k isn't zero. But, I don't know if that is true if you are in characteristic p.

simplicity
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  • So if $2v_0\in I$ then any multiple of $2v_0$ is in $I$, in particular, $2^{-1}(2v_0)\in I$, and $2^{-1}$ exists since we're in a field with characteristic $p\neq 2$. Is this right? – Yongyi Chen Apr 17 '12 at 16:24
  • @YongyiChen Does characteristic p mean it's a prime. Like the characteristic can't be say 4 if it's char(P). – simplicity Apr 17 '12 at 16:26
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    I don't think you can have a non-prime characteristic, can you? For example, if $1+1+1+1=0$ then $2*(1+1)=0$ so since a field has no zero divisors, it must be that $1+1=0$. – Yongyi Chen Apr 17 '12 at 16:29

2 Answers2

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If $K$ is a field of whatever characteristic, any element $k\neq0$ in $K$ is invertible by definition (of field).

So, if you have a $K$-vector space $V$ and $kv$ belongs to some subspace $W$, by multiplying by the inverse of $k$ you certainly get $v\in W$.

If $K$ is just an algebra of characteristic $p$ then you cannot in general invert the non-zero elements. So, for instance, if you have $K=\Bbb F_p[X]$ ($\Bbb F_p$ being the field with $p$ elements) and $I=(X^2)$, then $X\cdot X\in I$ but $X\notin I$. (actually this examples remains true also in characteristic $0$).

Mind that any algebra $K$ of characteristic $p$ will contain $\Bbb F_p$, so that you will certainly be able to invert the elements $n\cdot1_K$ with $n\in\Bbb Z$ not a multiple of $p$.

Andrea Mori
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    I have assumed that $p$ is prime. This is always the case if $K$ is a field or more generally a domain. But sure there are rings $R$ of non-prime characteristic. In those you can invert $n\cdot1_R$ only when $n\in\Bbb Z$ is coprime with the characteristic. – Andrea Mori Apr 17 '12 at 16:35
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Hint $\ $ Units persist: $\rm\ 2\:(p+1)/2 = 1\:$ in $\rm\:\mathbb F_p$ persists under any ring hom that is unital $(1\to 1)$. Thus $2$ remains a unit in every $\rm\:\mathbb F_p$-algebra for odd $\rm p.$ For more on the persistence of units, coprimality, gcds, etc see this post and this and this.

darij grinberg
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Bill Dubuque
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