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This is a simple question about notation: Given two generators $x,y$ how does one denote the vector space spanned by all finite K-polynomials in $x$ and all finite polynomials in $y$. If I use K**$[x] \oplus$ *K*$[y]$, then I get two copies of **K. I could just quotient *K*$[x] \oplus$ *K*$[y]$ by *K*$[(1,0)-(0,1)]$, but this seems overly involved. Similarily, I could quotient *K*$[x,y]$ by <$xy,yx$>, but this too seems overly involved. Does any have any ideas?

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up vote 5 down vote accepted

If it is clear from context that you are working in the ambient setting of $k[x,y]$, then you can write $k[x] + k[y]$. Otherwise, I would spell it out in words.

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Yes, of course, an internal direct sum is what I'm looking for. Thanks. – Dyke Acland Jun 18 '11 at 12:32
@Dyke, as you point out in your question, it is only a sum, not a direct sum, because of the overlapping copy of $k$. – L Spice Jun 18 '11 at 15:19
Sorry, I meant internal sum. – Dyke Acland Jun 19 '11 at 11:40

I write $K[x]\oplus_K K[y]$; it's easy to understand, symmetric, pleasant, and agrees with category-theoretic convention.

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K[x]+yK[y] or K[y]+xK[x]

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Or $xk[x] \oplus yk[y] \oplus k$. – Gerald Edgar Jun 18 '11 at 16:11

It's the augemented product of $k[x]$ and $k[y]$.

Generally, in the category of augemented algebras, the product of $(A,\epsilon_A), (B,\epsilon_B)$ is defined as pullback of $\epsilon_A, \epsilon_B$, i.e. $A*B = \lbrace (a,b)\in A \times B \; | \; \epsilon_A(a) = \epsilon_B(b) \rbrace$ with the obvious induced augmentation.

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These answers remind me of this observation involving $0,1$ power series (it can be translated into results about finite sets which tile $\mathbb{N}$ or $\mathbb{N}^2$ by translation but I'll keep it short):

$\frac{1}{1-x}\frac{1}{1-y}=\frac{1}{1-xy}\left( \frac{1}{1-x}+\frac{1}{1-y}-1\right)=\frac{1}{1-xy}\left( \frac{1}{1-x}+\frac{y}{1-y}\right)=\frac{1}{1-xy}\left(1+ \frac{x}{1-x}+\frac{x}{1-y}\right)$

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