# Understanding an Algebra over a Field

I am trying to understand the components of an algebra over a field.

an algebra over a field (often simply called an algebra) is a vector space equipped with a bilinear product.

I am familiar with the definition of a field, and sort of familiar with vector spaces. I am not familiar with bilinear products and the meaning of a "vector space over a field". Wikipedia says a bilinear map is a set of 3 vector spaces over the same field.

I am wondering the following:

1. What it means to be a vector space over a field. Wondering if it's different than just a vector space.
2. Since the 3 vector spaces are different, but they all "work on" the same underlying set, wondering how their operations are compatible. It would be helpful to see an example of a bilinear map/product as vector spaces over a field.
3. If "an algebra" in this context is the same as the generic definition of an algebra as "a set plus a set of operations on that set".
• The website you link to does NOT define an "algebra". It defines an "algebraic structure". The two are not at all the same any more than "an algebra" refers to the introductory Algebra course you took in secondary school! Commented Apr 28, 2018 at 20:39
• The word $\text{algebra}"$ has at least four different meanings. The first is the subject elementary algebra that everyone learned in secondary school, the second is the subject abstract algebra that is taught in university, ... Commented Apr 28, 2018 at 20:59
• ... the third is the algebraic structure $\sigma$-algebra which is used in measure theory and the fourth is the algebraic structure algebra which is a vector space equipped with a bilinear product. Commented Apr 28, 2018 at 21:00
• No one has mentioned it so I will, there is also Zhegalkin Algebra which is an instrument from the Boolean Logic area. The link is to the Wiki in Russian since the English version of it doesn't exist.
– Aelx
Commented May 10, 2023 at 7:36

An algebra over a field is a vector space along with a multiplication of your vectors. For this multiplication to be useful, you want it to be bilinear, that is if we multiply $$(a_{1}u_{1} + b_{1}v_{1})*(a_{2}u_{2}+b_{2}v_{2}),$$ we want this to be equal to $$a_{1}a_{2}(u_{1}*u_{2})+a_{1}b_{2}(u_{1}*v_{2})+b_{1}a_{2}(v_{1}*u_{2})+b_{1}b_{2}(v_{1}*v_{2}).$$ (here $a_{1}, b_{1}, a_{2}, b_{2}$ are scalars in my field, and $u_{1},v_{1}, u_{2}, v_{2}$ are vectors).
1. A vector space over a field $\mathbb{F}$ is just one level more of abstraction. $V = \mathbb{R}^n$ is a vector space over the field $\mathbb{R}$. A vector space $V = \mathbb{C}^n$ is a vector space over the field $\mathbb{C}$. We could have other fields like finite fields.
2. A bilinear product is a mapping involving three vector spaces. Think of matrix multiplication. A matrix $A \in \mathbb{R}^{n \times m}$ is a vector in the vector space $V=\mathbb{R}^{n \times m}$. A similar statement holds for another matrix $B \in \mathbb{R}^{m \times p}$. Multiplying them together, $AB$ together yields another matrix for which again a similar statement holds. In all this the field is $\mathbb{R}$.
3. No. An algebra over a field is defined to be exactly the structure you mention; its one vector space $V$ equipped with a bilinear product operation.
1. You wrote that you are “somewhat familiar with vector spaces”. In those vector spaces you can multiply vectors by scalars. Where do they come from? Are they real numbers? Then you have a vector space over the field $\mathbb R$ of real numbers. Are they complex numbers? Then you have a vector space over the field $\mathbb C$ of complex numbers. And so on.
2. A bilinear product is not “a set of 3 vector spaces over the same field”. It's a map from $V_1\times V_2$ into $W$ with certain properties, in which $V_1$, $V_2$ and $W$ are vector spaces over some field. In the case of an algebra, $V_1=V_2=W$.