Krull dimension and graded prime ideals

How can we show that $\dim R/p=0\Leftrightarrow p=(x_{1},\ldots,x_{n})\Leftrightarrow R/p\simeq\mathbb{K}$, where $R=\mathbb{K}[x_{1},\ldots,x_{n}]$ is considered graded with standard grading (i.e. $\deg(x_i)=1$) and $\mathbb{K}$ is an arbitrary field of characteristic zero. Also, $p$ is a graded prime ideal of $R$ and $\dim$ is the Krull dimension.

The Krull dimension of a ring $S$, written $\dim S$, is the supremum of the lengths of chains of prime ideals in $R$ (for example, the chain $p_0\subsetneq p_1\subsetneq \cdots\subsetneq p_n$ has length $n$)

I don't know if the following is of any help on this:

If $\dim R/p=0$, then $p$ is the unique graded maximal ideal of $R$. But how can we proceed to show that $p=(x_1,\ldots,x_n)$?

If anyone can help unstump me on this, I'd be grateful.

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$R/p$ is an integral domain. If $\dim R/p=0$, then it is a field, so $p$ is maximal. Since $p$ is graded it is contained in $(x_1,\dots,x_n)$, hence equality. The rest is easy.
You say that since $p$ is graded it is contained in ($x_1,\ldots,x_n$). Why is that? Can you please elaborate? – Vagerman Dec 5 '13 at 16:47
@Vagerman A graded ideal is generated by homogeneous elements. (If contains a homogeneous element of degree $0$, then the ideal is the whole ring which is not the case for prime ideals.) Furthermore, in $K[x_1,\dots,x_n]$ the ideal $(x_1,\dots,x_n)$ contains all homogeneous elements of degree greater than $0$. – user89712 Dec 5 '13 at 17:56