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I try to understand the resolution of an algebra as described in https://projecteuclid.org/download/pdf_1/euclid.ijm/1255378502.

I don't understand how the multiplication is defined after adjoining a variable. Given, for instance, the exterior algebra $\Lambda_R(y)$ on a generator of degree one over a commutative ring $R$.

If I now adjoin a variables $T^{(i)}$ of degree $2i$ as in the paper I shall look at the free $\Lambda_R(y)$-module in a basis $\{T^{(i)}\}$. The multiplication of the $T^{i}$ is given in the paper. I just wonder how the product $yT\cdot yT$ is defined. As far as I understand it would be $yyTT$, but this is zero because $yy$ is.

However, $yT$ is of degree $3$ and the product with itself should land in degree $6$. Where is my misunderstanding? Thanks for your help!

ilil
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    This issue already arises in the exterior algebra: $y$ has degree one, and $y^2 = 0$. This is not a contradiction; there is a zero vector in every degree, if you like. – Qiaochu Yuan Mar 21 '17 at 00:47
  • Thanks for your comment! Can you be more explicit in what you mean with zero vector? Is it correct that $yT$ has degree 3? If so, how does $yT\cdot yT$ land in degree $6$? Do you state that a zero element can also be in degree $6$, or even in any degree? Then the algebra is not a direct sum, I'm afraid. – ilil Mar 21 '17 at 00:55
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    Yes, that's what I mean, and no, it does not imply that the algebra is not a direct sum. – Qiaochu Yuan Mar 21 '17 at 01:00
  • Ok, I see. Can we say that the multiplication is then given by the tensor product of algebras, the exterior algebra and the divided power one? – ilil Mar 21 '17 at 01:24
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    The graded tensor product, yes. (This introduces some signs into the multiplication.) – Qiaochu Yuan Mar 21 '17 at 01:26
  • Yes, the graded product but I don't get signs, do I? As $T$ is of even degree it holds that $(1\otimes T^{(i)})\cdot (y\otimes T^{(j)})=(-1)^{|y|*2}y\otimes T^{(i+j)}=y\otimes T^{(i+j)}$? There are never elements of odd degrees that pass each other... – ilil Mar 22 '17 at 03:33
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    I mean in general it's important to distinguish between the tensor product of graded algebras and the graded tensor product of graded algebras. It is sometimes the case that the signs don't end up mattering but that's no reason to get sloppy about what construction is relevant in general. – Qiaochu Yuan Mar 22 '17 at 03:36
  • Yes that's true and I thank you for your hint because I might have forgotten it! I just was wondering in this explicit situation that I might be missing a sign because there are none ;) – ilil Mar 22 '17 at 03:38

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